REVERSE MATHEMATICS AND ORDERED GROUPS

A Dissertation

Presented to the Faculty of the Graduate School

of Cornell University

in Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy

by David Reed Solomon

REVERSE MATHEMATICS AND ORDERED GROUPS

May 1998 \begin{matrix}c \\ ◯ \end{matrix} David Reed Solomon 1998 ALL RIGHTS RESERVED David Reed Solomon, Ph.D. Cornell University 1998

We study theorems of ordered group theory from the viewpoints of reverse mathematics and computable mathematics. Reverse mathematics uses subsystems of second order arithmetic to determine which set existence axioms are required to prove particular theorems. We give equivalences of W KL_{0} (the orderability of direct products and nilpotent groups, the classical semigroup conditions for orderability), ACA_{0} (the existence of induced orders on quotient groups, the existence of the center of a group) and Π_{1}^{1} − CA_{0} (the classification of order types for countable fully ordered abelian groups). We show that RCA_{0} suffices to prove Hölder’s Theorem and the theorem that every ordered group is the quotient of an ordered free group by a convex normal subgroup. In computable mathematics, we examine the complexity of orders on computable abelian and nilpotent groups and we study the relationship between spaces of orders of computable groups and computably bounded Π_{1}^{0} classes. Finally, we show that every orderable computable abelian group has a computable presentation with a computable order. This work answers several open questions from Downey and Kurtz (1986).

Biographical Sketch

David Reed Solomon was born on July 9, 1970 in Mishawaka, Indiana, but soon moved to the neighboring city of South Bend. After attending Clay High School, he did his undergraduate work at the University of Notre Dame and received a BS in May, 1992.

Reed attended Cornell University from 1992 until 1998 when he earned a Doctor of Philosophy. In addition to studying mathematics, he learned to rock climb, to lindy hop and to play a few tunes on the mandolin. He spent many sunny spring and summer afternoons playing soccer and swimming in the gorges.

This fall, Reed will begin a Van Vleck Visiting Assistant Professorship at the University of Wisconsin.

To my parents, David and Mary Lou Solomon.

Acknowledgements

First and foremost, I would like to thank my advisor, Richard Shore, for all of his support and help. In addition to working with me on my thesis topic, he fostered a very lively atmosphere among the logic students and visitors. By encouraging our interaction and maintaining a dense seminar schedule, he helped all of us to mature as mathematicians.

I am grateful to the other members of my committee, Dexter Kozen and Anil Nerode, and to Tom Rishel for advice and encouragement. I am also indebted to Shirley Allen, Arletta Havlic and Joy Jones for a thousand little favors during my time here.

I have benefited from the friendship of far more people than I can thank here. Most especially, however, I owe thanks to Wicharn Lewkeeratiyutkul for becoming like a brother to me, to Leah Gold for teaching me to dance, and to Denis Hirschfeldt for being a running companion, a logic partner and a great friend.

Table of Contents

1 Computable and Reverse Mathematics 1 1.1 Computable Mathematics . . . . . . . . . . . . . . . . . . . . . . . 2 1.2 Reverse Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3 Summary of Results . . . . . . . . . . . . . . . . . . . . . . . . . . 15

2 Orders on Quotient Groups 21 2.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 2.2 Quotient Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.3 Induced Orders . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3 Group Conditions 35 3.1 Torsion Free Abelian and Nilpotent Groups . . . . . . . . . . . . . 36 3.2 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44 3.3 The Center . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.4 Spaces of Full Orders . . . . . . . . . . . . . . . . . . . . . . . . . . 58

4 Semigroup Conditions 63 4.1 Definition of the Conditions . . . . . . . . . . . . . . . . . . . . . . 63 4.2 Equivalence with W KL_{0} . . . . . . . . . . . . . . . . . . . . . . . . 69

5 Free Groups 79 5.1 Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 5.2 Fully Ordered Rings and Triangular Matrices . . . . . . . . . . . . . 84 5.3 Free Products of O-Groups . . . . . . . . . . . . . . . . . . . . . . . 92

6 Divisible Closures and Hölder’s Theorem 108 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 6.2 Hölder’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 6.3 Strong Divisible Closures . . . . . . . . . . . . . . . . . . . . . . . . 121 7 Order Types 128

7.1 Order Type of a Group . . . . . . . . . . . . . . . . . . . . . . . . . 128 7.2 The Reversal . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 138 7.3 Coding Turing Jumps . . . . . . . . . . . . . . . . . . . . . . . . . . 149

8 Spaces of Orders and Computable Presentations 157 8.1 Π_{1}^{0} Classes and Spaces of Orders . . . . . . . . . . . . . . . . . . . . 157 8.2 Extension to Nilpotent Groups . . . . . . . . . . . . . . . . . . . . . 166 8.3 Computable Presentations . . . . . . . . . . . . . . . . . . . . . . . 172

A Miscellaneous Definitions and Proofs 185 A.1 Free Groups and Free Products . . . . . . . . . . . . . . . . . . . . 185 A.2 Proofs for Tri_{ℚ[C]} . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 A.3 Proofs for Free Product Embedding . . . . . . . . . . . . . . . . . . 206

Chapter 1

Computable and Reverse

Mathematics

This chapter is an introduction to computable (or effective) mathematics and to reverse mathematics, both of which will later be used to study theorems of ordered groups. Computable mathematics involves applying ideas of computability theory to other areas of mathematics. By computable, we will always mean Turing computable. A typical question we might consider is whether the effective version of a particular theorem holds. That is, is the theorem true if all the objects in it are assumed to be computable? Precise definitions and concrete examples are given in Section 1.1.

Reverse mathematics is based more in proof theory, although as we shall see, it uses techniques from all areas of logic and has consequences in computable mathematics. The basic definitions and results are given in Section 1.2.

The last section of the chapter is a summary of the results contained later.

1.1 Computable Mathematics

We first give the general definition for a computable structure and then repeat it in the special case of groups. The languages involved are always assumed to be computable.

Definition 1.1. A model 𝔄 is computable if its domain A ⊆ ω is a computable set and its relations and functions are uniformly computable.

Definition 1.2. A computable group is a computable set G ⊆ ω together with a computable function ·_{G} which satisfies the usual group axioms. We will let 1_{G} (or sometimes 0_{G} in the abelian case) stand for the identity element.

We are interested in studying the conditions under which effective versions of theorems hold. For example, consider the following theorem:

Theorem 1.3. If H is a normal subgroup of a group G, then G/H is a group.

The effective version of this theorem says:

Theorem 1.4. If H is a computable normal subgroup of a computable group G, then G/H is a computable group.

If we are careful defining what we mean by G/H, the effective version will be true. Define G/H ⊆ ω by choosing the ω–least representative for each coset:

{ g ∈ G | ∀n < g(n ∉ G ∨ n^{−1}g ∉ H) }.

This definition picks out the least representative because nH = gH if and only if n^{−1}g ∈ H. To define multiplication ·_{G/H}, use the formula:

a ·_{G/H} b = c ↔ a, b, c ∈ G/H ∧ c^{−1} ·_{G} a ·_{G} b ∈ H.

Thus the effective version of the theorem holds.

Contrast this situation with the following example.

Definition 1.5. A binary branching tree is a set T ⊆ {0, 1}^{<ω} such that

∀σ, τ ∈ {0, 1}^{<ω} ( σ ⊆ τ ∧ τ ∈ T → σ ∈ T ).

A path through T is a function f : ω → {0, 1} such that f[n] ∈ T for all n, where f[n] is the sequence ⟨f(0), … f(n)⟩.

Weak König’s Lemma. Every infinite binary branching tree has a path.

The effective version of Weak König’s Lemma says that every infinite computable binary branching tree has a computable path. Unfortunately, this statement is false. Jockusch Jr. and Soare (1972), however, modified the effective version to discover some computational content in Weak König’s Lemma. Recall that a set A ⊆ ω is low if it has the lowest possible Turing jump, A′ = 0′.

Low Basis Theorem (Jockusch Jr. and Soare (1972)). Every infinite computable binary branching tree has a low path.

One of our goals will be to give a similar analysis for theorems about ordered groups. Computably bounded Π_{1}^{0} classes will play an important role in this analysis.

Definition 1.6. C ∈ {0, 1}^{ω} is a computably bounded (c.b.) Π_{1}^{0} class if there is a computable binary branching tree T such that C is the set of paths through T. Definition 1.7. Let A, B ∈ ω and A ∩ B = ∅. S is a separating set for A and B if either A ⊆ S ∧ S ∩ B = ∅ or B ⊆ S ∧ S ∩ A = ∅. Definition 1.8. C ⊆ {0, 1}^{ω} is a Π_{1}^{0} class of separating sets if there are computably enumerable (c.e.) sets A and B such that C is the class of characteristic functions for the separating sets of A and B.

Notice that not all c.b. Π_{1}^{0} classes are Π_{1}^{0} classes of separating sets. In particular, a c.b. Π_{1}^{0} class may have cardinality ω, but any Π_{1}^{0} class of separating sets have either finite cardinality or cardinality 2^{ω}. The Low Basis Theorem was originally stated as every c.b. Π_{1}^{0} class has a member of low degree. Jockusch Jr. and Soare (1972) proved a number of other results about c.b. Π_{1}^{0} classes which we will have reason to refer to later. Theorem 1.9 (Jockusch Jr. and Soare (1972)). There is an infinite c.b. Π_{1}^{0} class C such that for all f, g ∈ C, if f ≠ g, then f and g are Turing incomparable. Theorem 1.10 (Jockusch Jr. and Soare (1972)). There are c.e. sets A and B such that A∩B = ∅, A∪B is coinfinite and if D, E are separating sets for A and B, then either D and E have the same Turing degree or they are Turing incomparable.

To fix some notation before moving on, we will use ≤_{T} for Turing reducibility, ≡_{T} for Turing equivalence, deg(A) for the Turing degree of A, 0′ for the degree of the halting problem and 0^{(n)} for the n^{th} jump of ∅. This notation and others we will use follows Soare (1980).

1.2 Reverse Mathematics

Reverse mathematics is a branch of logic started by Harvey Friedman in the 1970’s. It seeks to answer the question: Which set existence axioms are required to prove the theorems of ordinary mathematics? In addition to being interesting in their own right, answers to this question often have consequences in both effective mathematics and the foundations of mathematics. Before discussing these consequences, we need to be more specific about the motivating question.

Ordinary mathematics refers to the areas of mathematics that remained unchanged by the introduction of abstract set theory. These areas concern either countable or essentially countable mathematics. Essentially countable mathematics is another vague term that is best explained by an example. The study of complete separable metric spaces involves essentially countable mathematics because, although the spaces may be uncountable, they can be understood in terms of a countable basis. Simpson (1985) gives the following list of areas included in ordinary mathematics: number theory, geometry, calculus, differential equations, real and complex analysis, combinatorics, countable algebra, separable Banach spaces, computability theory, and the topology of complete separable metric spaces. Ordinary mathematics usually does not include abstract functional analysis, abstract set theory, universal algebra, or general topology.

The rest of this chapter is an introduction to reverse mathematics and to some of the definitions and notation used later. For more information and results, in particular for more examples of equivalences between theorems and subsystems of second order arithmetic, see Friedman et al. (1983), Hirst (1994), Simpson (1984), Simpson (1985) or Simpson (Unpublished). We will mostly follow the notation found in Friedman et al. (1983) and Simpson (Unpublished).

The setting for reverse mathematics is second order arithmetic. Second order arithmetic uses a two sorted first order language, L_{2}. L_{2} has both number variables and set variables. The number variables are denoted by lower case letters and are intended to range over elements of ω. The set variables are denoted by capital letters and are intended to range over P(ω). Because L_{2} has two types of variables it also has two types of quantifiers: ∃X, ∀X and ∃x, ∀x. The terms of L_{2} are built from the number variables and the constants 0, 1 using the function symbols +, ·. Atomic formulas have the form t_{1} = t_{2}, t_{1} < t_{2} or t_{1} ∈ X where t_{1}, t_{2} are terms. General formulas are built from the atomic formulas using the standard logical connectives and the two types of quantifiers.

A model for L_{2} is a first order structure 𝔄 where

𝔄 = ⟨A, S_{A}, +_{A}, ·_{A}, 0_{A}, 1_{A}, <_{A}⟩.

The number variables range over A, the set variables range over S_{A} ⊆ P(A), and the functions, constants, and relations are interpreted as indicated.

The axioms for second order arithmetic fall into three categories. The basic axioms specify the properties of +, ·, 0, 1 and <.

\begin{matrix}n + 1 ≠ 0 & m + 1 = n + 1 → m = n \\ m + 0 = m & m + (n + 1) = (m + n) + 1 \\ m · 0 = 0 & m · (n + 1) = m · n + m \\ ¬m < 0 & m < n + 1 ↔ (m < n ∨ m = n) \end{matrix}

There is an induction axiom for sets,

( 0 ∈ X ∧ ∀n (n ∈ X → n + 1 ∈ X) ) → ∀n (n ∈ X)

and a comprehension scheme for forming sets,

∃X ∀n (n ∈ X ↔ φ(n))

where φ is any formula of L_{2} in which X does not occur freely. φ may contain other free variables as parameters. The following formula induction scheme is derivable from the comprehension scheme and the induction axiom:

( φ(0) ∧ ∀n (φ(n) → φ(n + 1) ) ) → ∀n φ(n).

The formal system of second order arithmetic is denoted by Z_{2}.

The intended model of Z_{2} is

⟨ω, P(ω), +, ·, 0, 1, <⟩.

Models of Z_{2} can be nonstandard in two different ways. The set over which the number variables range could be nonstandard or the nonstandard model could be of the form

⟨ω, S, +, ·, 0, 1, <⟩

where S ⊊ P(ω). Models in which the number variables range over ω are called ω–models.

The question of which set existence axioms are needed to prove a particular theorem is answered by examining which instances of the comprehension scheme are required. Hence, we examine subsystems of Z_{2} which arise from restricting the formulas over which the comprehension scheme applies. One consequence of limiting comprehension is that the formula induction scheme is also limited. We may add some extra formula induction, but we will not add full formula induction because this scheme is a disguised set existence principle. For a complete discussion of this issue, see Friedman et al. (1983) and Simpson (Unpublished). A surprising observation is that the motivating question can be answered for a remarkable number of theorems by examining only five subsystems of Z_{2}. These subsystems are presented below in increasing order of strength.

The weakest subsystem considered here is RCA_{0}. RCA_{0} stands for recursive comprehension axiom and is obtained by restricting the comprehension scheme to apply only to Δ_{1}^{0} formulas. This restricted scheme is called Δ_{1}^{0} comprehension. We also allow Σ_{1}^{0} induction:

( φ(0) ∧ ∀n (φ(n) → φ(n + 1) ) → ∀n φ(n)

where φ is Σ_{1}^{0}. Friedman et al. (1983) show that RCA_{0} proves the Π_{1}^{0} formula induction scheme as well. Because RCA_{0} contains Δ_{1}^{0} comprehension and the comprehension scheme allows parameters, it follows that every model of RCA_{0} is closed under Turing reduction. The minimum ω–model is

⟨ω, REC, +, ·, 0, 1, <⟩

where REC ⊆ P(ω) is the set of computable sets.

Proving theorems in RCA_{0} has consequences in computable mathematics. If a theorem is provable in RCA_{0} then its effective or computable version is true. We will see the following example of this phenomenon in Chapter 6. Given suitable definitions, Friedman et al. (1983) showed that RCA_{0} suffices to prove that every field has an algebraic closure. This result implies that every computable field has a computable algebraic closure, a result first proved earlier in Rabin (1960). Because of this connection to computable mathematics, we will be content once we have a proof in RCA_{0} and will not seek a proof in a weaker subsystem. For results regarding weaker subsystems, see Hatzikiriakou (1989)

RCA_{0} is strong enough to establish the basic facts about the number systems ℕ, ℤ, ℚ and ℝ. This fact makes RCA_{0} strong enough to be a reasonable base theory over which to do coding. The set of natural numbers, ℕ, is defined by ∀n (n ∈ ℕ) and the integers and rationals are defined with the help of the standard pairing function:

⟨x, y⟩ = \frac{1}{2}(i + j)(i + j + 1) + i.

Associated with this pairing function are two projection functions:

π_{1}(⟨x, y⟩) = x and π_{2}(⟨x, y⟩) = y.

Notice that we will use ω to denote the standard natural numbers and ℕ to denote the (possibly nonstandard) universe of a model of a subsystem of Z_{2}.

The real numbers do not form a set in Z_{2}; they must be represented by Cauchy sequences of rationals. Simpson (Unpublished) gives the following definitions.

Definition 1.11. (RCA_{0}) A sequence of rational numbers is a function f : ℕ → ℚ. Frequently we denote this sequence by ⟨q_{k} | k ∈ ℕ⟩.

Definition 1.12. (RCA_{0}) A real number is a sequence of rational numbers ⟨q_{k} | k ∈ ℕ⟩ such that

∀k ∀i ( | q_{k} − q_{k+i} | < 2^{−k} ).

We write x ∈ ℝ to mean that x is a sequence of rationals satisfying this convergence rate. 0_{ℝ}, or when it is not ambiguous just 0, denotes the sequence ⟨0 | k ∈ ℕ⟩ and 1_{ℝ}, or just 1, denotes ⟨1 | k ∈ ℕ⟩.

Definition 1.13. (RCA_{0}) Let x = ⟨q_{k} | k ∈ ℕ⟩ and y = ⟨q′_{k} | k ∈ ℕ⟩ be two real numbers. We say x = y if

∀k ( | q_{k} − q′_{k} | ≤ 2^{−k+1} ).

The sum x + y is the sequence

⟨q_{k+1} + q′_{k+1} | k ∈ ℕ⟩.

The product x · y is the sequence

⟨q_{n+k} · q′_{n+k} | k ∈ ℕ⟩

where n is the least natural number such that 2^{n} ≥ | q_{0} | + | q_{1} | + 2.

With these definitions it can be shown in RCA_{0} that the real number system obeys all the axioms of an Archimedean ordered field.

RCA_{0} is strong enough to define sets of unique codes for finite sets and finite sequences. For the details of this coding, see Simpson (Unpublished). We will denote the set of finite sequences of elements of X by Fin_{X}. RCA_{0} also suffices to define the length function lh : Fin_{X} → ℕ. For σ ∈ Fin_{X}, the i^{th} element in σ is denoted σ(i), and σ is written

⟨σ(0), …, σ(lh(σ) − 1)⟩.

If σ, τ ∈ Fin_{X}, then σ ⊆ τ if lh(σ) ≤ lh(τ) and for all i < lh(σ), σ(i) = τ(i). The concatenation of σ and τ is written σ^{⌢}τ. For any function f : ℕ → ℕ, f[n] is the sequence ⟨f(0), …, f(n − 1)⟩.

Definition 1.14. (RCA_{0}) A tree is a set T ⊆ Fin_{ℕ} such that T is closed under initial segments:

∀σ, τ ( (σ, τ ∈ Fin_{ℕ} ∧ σ ⊆ τ ∧ τ ∈ T) → σ ∈ T ).

T is finitely branching if each element of T has only finitely many successors:

∀σ ∃n ∀m (σ^{⌢}⟨m⟩ ∈ T → m < n).

T is binary branching if every element of T has at most two successors. T is bounded if there exists a function f : ℕ → ℕ such that for all τ ∈ T and all m < lh(τ), τ(m) < f(m). A path through T is a function g : ℕ → ℕ such that g[n] ∈ T for all n.

König’s Lemma. Every infinite finitely branching tree has a path. Weak König’s Lemma. Every infinite binary branching tree has a path. Bounded König’s Lemma. Every infinite bounded tree has a path.

Each of these lemmas is really a set existence principle; they each assert the existence of certain functions. König’s Lemma is strictly stronger than Weak König’s Lemma, and Weak König’s Lemma and Bounded König’s Lemma are equivalent. As we saw in Section 1.1, the effective version of Weak König’s Lemma fails.

The second subsystem of Z_{2} is called W KL_{0} or Weak König’s Lemma. It contains the axioms of RCA_{0} plus Weak König’s Lemma. Because the effective version of Weak König’s Lemma fails, W KL_{0} is strictly stronger than RCA_{0}. The best intuition for W KL_{0} is that Weak König’s Lemma adds a compactness principle to RCA_{0}.

We can now give a more formal explanation of the goals and methods of reverse mathematics. Consider a theorem T hm of ordinary mathematics which is stated in L_{2}. If RCA_{0} ⊢ T hm then from the viewpoint of reverse mathematics, we are satisfied. However, if W KL_{0} ⊢ T hm and we cannot find a proof of T hm in RCA_{0}, then we want to show that W KL_{0} is in some sense the weakest collection of set existence principles that proves T hm. To achieve this goal, we attempt to show for each axiom φ of W KL_{0}

RCA_{0} + T hm ⊢ φ.

We abbreviate this situation by RCA_{0}+T hm ⊢ W KL_{0} or RCA_{0} ⊢ T hm ↔ W KL_{0}. If we prove RCA_{0} + T hm ⊢ W KL_{0}, then we have shown that T hm is equivalent to the subsystem W KL_{0} and hence that no subsystem strictly weaker than W KL_{0} can prove T hm. In particular, we can stop looking for a proof of T hm in RCA_{0}. The proof that RCA_{0} + T hm ⊢ W KL_{0} is called the reversal of T hm and this process of proving axioms from theorems is the origin of the name reverse mathematics.

When trying to prove reversals, it is often helpful to use the following theorem.

Theorem 1.15. (RCA_{0}) The following are equivalent:

1. W KL_{0}

2. For every pair of functions f, g such that for all m, n, f(m) ≠ g(n), there exists a set X such that for all m, f(m) ∈ X and g(m) ∉ X.

One consequence of proving a theorem equivalent to W KL_{0} is that the effective version of the theorem must fail. Although I have presented results in computable mathematics as consequences of results in reverse mathematics, usually the results in computable mathematics come first. If the effective version of a theorem is known to hold, then the proof that it holds can often be translated into RCA_{0}. Similarly, if the computable version fails, then why it fails often gives a hint as to how to use coding methods to prove a reversal. Provability in W KL_{0} also has consequences for the foundations of mathematics. Because W KL_{0} is Π_{2}^{0} conservative over primitive recursive arithmetic, see Parsons (1970), it provides a modern rendering of Hilbert’s Program. For a discussion of these consequences see Simpson (1988), Drake (1989) and Feferman (1988).

The third subsystem is ACA_{0} or Arithmetic Comprehension Axiom. ACA_{0} has the axioms of RCA_{0} plus Σ_{1}^{0} comprehension. It is strong enough to prove König’s Lemma and hence is strictly stronger than W KL_{0}. Because ACA_{0} contains Σ_{1}^{0} comprehension, it can define the Turing jump of any set. Models of ACA_{0} are closed under the Turing jump and the minimum ω–model is

⟨ω, ARIT H, +, ·, 0, 1, <⟩

where ARIT H is the set of arithmetic sets.

The following theorem is helpful when proving reversals involving ACA_{0}.

Theorem 1.16. (RCA_{0}) The following are equivalent:

1. ACA_{0}

2. The range of every one–to–one function exists.

The fourth system is AT R_{0} or Arithmetic Transfinite Recursion. The description of AT R_{0} involves the notion of a well order.

Definition 1.17. (RCA_{0}) A linear order is a pair (X, ≤_{X}) such that X is a set and ≤_{X} is a binary relation on X such that:

∀y ∈ X (y ≤_{X} y)
∀y, z ∈ X (y ≤_{X} z ∧ z ≤_{X} y → y = z)
∀w, y, z ∈ X (w ≤_{X} y ∧ y ≤_{X} z → w ≤_{X} z).

Definition 1.18. (RCA_{0}) A well order is a pair (X, ≤_{X}) such that (X, ≤_{X}) is a linear order and

¬∃x ∈ X ∃f : ℕ → X (f(0) = x ∧ ∀i(f(i + 1) <_{X} f(i))).

AT R_{0} includes the axioms of ACA_{0} plus axioms that allow arithmetic comprehension to be iterated along any well order. While ACA_{0} is strong enough to prove that the n^{th} Turing jump 0^{(n)} exists, AT R_{0} is required to construct the uniform upper bound 0^{(ω)}. For a formal description of the axioms of AT R_{0} see Friedman et al. (1983).

The last and most powerful subsystem is Π_{1}^{1} − CA_{0}. Π_{1}^{1} − CA_{0} is ACA_{0} plus the scheme of Π_{1}^{1} comprehension. This system is strong enough to define Kleene’s O and hence models of Π_{1}^{1} −CA_{0} are closed under the hyperjump. Π_{1}^{1} −CA_{0} is strictly stronger than AT R_{0} and is useful for proving that certain recursions terminate. An example of this phenomenon is presented in Chapter 7. The following theorem is used to prove a reversal in that chapter.

Theorem 1.19. (RCA_{0}) The following are equivalent:

1. Π_{1}^{1} − CA_{0}

2. For any sequence of trees ⟨T_{k} | k ∈ ℕ⟩, there exists a set X such that k ∈ X if and only if T_{k} has a path.

1.3 Summary of Results

Now that we have the general definitions for reverse mathematics and computable mathematics, we can give a summary of the rusults to come.

In Chapter 2, we give the definitions and basic facts about partially and fully ordered groups. A partially ordered (p.o. ) group is a group together with a partial order such that the order is preserved under multiplication on the left and the right. If the order is linear, the group is called fully ordered (f.o. ).

A subgroup H of a p.o. group G is convex if for any a, b ∈ H and g ∈ G,

a ≤ g ≤ b ⇒ g ∈ H.

For a convex normal subgroup H of a p.o. group G, there is not only a natural group structure on G/H, but also an induced partial order. If G is fully ordered, then this induced quotient order is a linear and its existence is provable in RCA_{0}. However, if G is only known to be partially ordered, then the existence of the induced order is equivalent to ACA_{0}.

Throughout this text, we will provide results in computable mathematics as corollaries to theorems in reverse mathematics. Since RCA_{0} suffices to prove the existence of the induced order on the quotient of an f.o. group by a convex normal subgroup, it follows that if G is a computable fully ordered computable group and H is a computable convex normal subgroup, then the induced order on G/H is computable. By contrast, if G is a computably partially ordered computable group, then the induced order on G/H could be as complicated as 0′.

One of the fundamental questions in ordered group theory is how to tell if a particular group admits a full order. In Chapter 3, we discuss three of the classical group conditions that insure full orderability. The fact that every torsion free abelian group is fully orderable was shown to be equivalent to W KL_{0} by Hatzikiriakou and Simpson (1990). We extend this result to show the equivalence of W KL_{0} and the theorem that every tosion free nilpotent group is fully orderable. RCA_{0} suffices to show that the finite direct product of fully orderable groups is fully orderable, but W KL_{0} is required for the case of infinite direct products.

Notice that we can again derive corollaries in computable mathematics. Downey and Kurtz (1986) were the first to give an example of a computable torsion free abelian group with no computable order. There is a uniform sequence of computably fully orderable computable groups such that the direct product, which is computable, has no computable order.

The center of a group G, denoted C(G) or ζ_{1}(G), is defined by

C(G) = { g ∈ G|∀x(gx = xg) }.

The existence of the center is equivalent to ACA_{0}, from which it follows that the center of a computable group (in fact a computable nilpotent group) can be as complicated as 0′. In Chapter 8, this fact is used to explain why a particular method of construction for finitely generated nilpotent groups cannot be extended to infinitely generated nilpotent groups.

The last topic in Chapter 3 is the connection between computable bounded Π_{1}^{0} classes and spaces of full orders of orderable computable groups. Once a group G is known to be fully orderable, it is natural to study the space of all full orders on G, denoted 𝕏(G). How many orders are there? What do they look like? How complicated can they be? We show that for any orderable computable group, there is a c.b. Π_{1}^{0} class C and a Turing degree preserving bijection between C and 𝕏(G).

One alternative to looking for group conditions that imply full orderability is to look for semigroup conditions. In Chapter 4, we examine three classical semigroup conditions that imply full orderability and show that each is equivalent to W KL_{0}. The main trick here is to translate the conditions into the language of second order arithmetic is such a way that they can be studied inside RCA_{0}.

We next turn our attention to the analogue of the group theoretic result that every group can be represented as the quotient of a free group by a normal subgroup. In ordered group theory, every f.o. group is order isomorphic to the quotient of a f.o. free group by a convex normal subgroup. It turns out that RCA_{0} is strong enough to prove this theorem, which answers an open question from Downey and Kurtz (1986) by showing that the effective version of the theorem holds. The proof is split between Chapter 5 and Appendix A. In Chapter 5, we develope the main argument assuming various technical facts, mostly explicit formulas for the embedding of free products into infinite matrix groups. These formulas, together with the definitions and basic facts about free groups and free products, are relegated to Appendix A.

In Chapter 6, we examine divisible closures and Hölder’s Theorem. Hölder’s Theorem is the main tool for classifying full orders of a particular fully orderable group. It states that every Archimedean fully ordered group is order isomorphic to a subgroup of the additive group of real numbers. Once we define what we mean by a subgroup of the reals in second order arithmetic, it turns out that RCA_{0} suffices to prove Hölder’s Theorem.

The motivation for studying divisible closures of abelian groups comes from similiar studies of algebraic closures of fields and real closures of ordered fields in Friedman et al. (1983), Smith (1981) and Metakides and Nerode (1979). The main questions about these notions of closure are those of existence, uniqueness and strong existence. In this context, strong existence means that the original algebraic object is isomorphic to a subspace of the closure. That is, the range of the embedding into the closure exists.

Friedman et al. (1983) answered the existence and uniqueness questions for the divisible closure of abelian groups and Downey and Kurtz (1986) answered the same questions for ordered abelian groups. In Chapter 6, we show that even for Archimedean f.o. groups, the existence of a strong divisible closure is equivalent to ACA_{0}.

Classifying the full orders on an orderable group can be an extremely difficult task. There is, however, a nice result classifying all possible order types for countable f.o. groups. In Chapter 7, we show this classification is equivalent to Π_{1}^{1} − CA_{0}.

Chapter 8 is devoted to issues in computable mathematics. The first section returns to the connection between c.b. Π_{1}^{0} classes and spaces of full orders on fully orderable computable groups. If G is a fully orderable computable group and 𝕏(G) is the space of full orders on G, then we have already mentioned that there is a Turing degree preserving bijection between some c.b. Π_{1}^{0} class C and 𝕏(G). A question, asked in Downey and Kurtz (1986) and motivated by results in Metakides and Nerode (1979), is whether all c.b. Π_{1}^{0} classes can be represented by computable torsion free abelian groups. That is, for an arbitrary c.b. Π_{1}^{0} class C, is there a orderable computable abelian group G and a Turing degree preserving bijection from C to 𝕏(G). We show that the answer is no even if we weaken the notion of representation and restrict the class of c.b. Π_{1}^{0} classes even further. We also extend the result to include computable torsion free nilpotent groups.

In the second section, we examine computable presentations of orderable computable abelian groups. Downey and Kurtz (1986) showed that there is an orderable computable abelian group with no computable order. However, their example is isomorphic to a computable group with a computable order. We show that this phenomena is true in general. Every orderable computable abelian group is classically isomorphic to a computably orderable computable group.

Orders on Quotient Groups

2.1 Basic Definitions

This section lays out the basic definitions for partially and fully ordered groups. Fundamental notions such as convex subgroups, induced orders on quotient groups, and positive cones of partially ordered groups are defined. The main result is that RCA_{0} suffices to prove the existence of the induced order on the quotient of an fully ordered group, but ACA_{0} is required if the group is not known to be fully ordered. The notation and definitions below follow Kokorin and Kopytov (1974) and Fuchs (1963).

Definition 2.1. (RCA_{0}) A group is a set G ⊆ ℕ together with a constant, 1_{G} (or sometimes 0_{G}), and an operation, ·_{G}, which obey the usual group axioms.

∀a, b, c ∈ G (a ·_{G} (b ·_{G} c) = (a ·_{G} b) ·_{G} c)
∀a ∈ G (1_{G} ·_{G} a = a ·_{G} 1_{G} = a)
∀a ∈ G ∃a^{−1} ∈ G (a ·_{G} a^{−1} = a^{−1} ·_{G} a = 1_{G}).

Definition 2.2. (RCA_{0}) A partial order is a set X together with a binary relation ≤_{X} which satisfies the following axioms.

∀x ∈ X (x ≤_{X} x)
∀x, y ∈ X (x ≤_{X} y ∧ y ≤_{X} x → x = y)
∀x, y, z ∈ X (x ≤_{X} y ∧ y ≤_{X} z → x ≤_{X} z).

Definition 2.3. (RCA_{0}) A partially ordered (p.o.) group is a pair (G, ≤_{G}) where G is a group, ≤_{G} is a partial order on the set G, and for any a, b, c ∈ G, if a ≤_{G} b then a ·_{G} c ≤_{G} b ·_{G} c and c ·_{G} a ≤_{G} c ·_{G} b. If the order is a linear order, the pair (G, ≤_{G}) is called a fully ordered (f.o.) group. A group for which there exists some full order is called an O-group

It should be clear from Definition 1.1 what the definitions of the corresponding computable objects are. For example, a computably partially ordered computable group G is a computable group G together with a computable binary relation ≤_{G} on G such that ≤_{G} satisfies the axioms for a partial order and the order is preserved under multiplication on both the left and the right. Except for cases when they are needed to avoid confusion, the subscripts on ·_{G} and ≤_{G} are dropped.

Example 2.4. The additive groups (ℝ, +), (ℚ, +), and (ℤ, +) with the standard orders are all f.o. groups. Let ℚ^{+} and ℝ^{+} be the strictly positive rational and real numbers. The multiplicative groups (ℝ^{+}, ·) and (ℚ^{+}, ·) are f.o. groups under the standard orders.

Example 2.5. The most important example for our purposes is the free abelian group on ω generators. Let G be the free abelian group with generators a_{0}, a_{1}, . . .. Elements of G have the form ∑_{i∈I} r_{i}a_{i} where I ⊆ ω is a finite set, r_{i} ∈ Z and r_{i} ≠ 0. To compare the element above with ∑_{j∈J} q_{j}a_{j}, let K = I ∪ J. For each k ∈ K, define r_{k} = 0 if k ∈ J \ I and q_{k} = 0 is k ∈ I \ J. Let k be the maximum element of K such that r_{k} ≠ q_{k}. The order is given by: ∑_{i∈I} r_{i}a_{i} ≤ ∑_{j∈J} q_{j}a_{j} if and only if r_{k} ≤ q_{k}. This order makes G into an f.o. group.

As expected, RCA_{0} suffices to prove many basic facts about p.o. groups.

Lemma 2.6. (RCA_{0}) If (G, ≤) is a p.o. group and a < b, then for any c ∈ G, ac < bc and ca < cb.

Lemma 2.7. (RCA_{0}) Let (G, ≤) be a p.o. group.

1. If a < b then c^{−1}ac < c^{−1}bc.

2. If a < b then b^{−1} < a^{−1}.

3. If a < b and c < d then ac < bd.

Defining a partial order can sometimes be notationally complicated. It is frequently easier to specify only the elements which are greater than the identity element. It turns out that specifying these positive elements uniquely determines the order.

Definition 2.8. (RCA_{0}) The positive cone, P (G, ≤_{G}) of a p.o. group is the set of elements which are greater than or equal to the identity.

P (G, ≤_{G}) = { g ∈ G | 1_{G} ≤_{G} g }

Each element x ∈ P (G, ≤_{G}) is called positive.

When the intended order ≤_{G} is clear, P (G) is used instead of P (G, ≤_{G}). Because P (G) has a Σ_{0}^{0} definition, RCA_{0} is strong enough to prove the existence of P (G). Conversely, given the positive cone, P (G), of some partial order on G, the relationship between any two elements can be calculated. Because a ≤ b if and only if 1_{G} ≤ a^{−1}b, determining if a ≤ b requires only knowing whether a^{−1}b ∈ P (G). Thus, RCA_{0} can recapture the order ≤_{G} from P (G). Notice that if G is a computable group, these conditions mean that deg(P (G)) = deg(≤_{G}) for any partial order ≤_{G} and its associated positive cone.

Example 2.9. The complex numbers (ℂ, +) with the order determined by

P (G) = { x + yi | x > 0 ∨ ( x = 0 ∧ y ≥ 0 ) }

is an f.o. group. The group (ℚ^{+}, ·) with the order determined by P (G) = ℕ^{+} is a p.o. group. Unraveling the definition of the positive cone show that if a, b ∈ ℚ^{+} then a ≤ b if and only if a divides b. This order is not a full order but does form a lattice.

There are algebraic conditions which determine if an arbitrary subset of a group is the positive cone for some full or partial order on that group. Definition 2.10. (RCA_{0}) If G is a group and X ⊆ G, then

X^{−1} = { g^{−1} | g ∈ X }.

X is a full subset of G if X ∪ X^{−1} = G and X is a pure subset of G if X ∩ X^{−1} ⊆ {1_{G}}. Theorem 2.11. (RCA_{0}) A subset P of a group G is the positive cone of some partial order on G if and only if P is a normal pure semigroup with identity. Furthermore, P is the positive cone of a full order if and only if in addition P is full. Proof. Any of the standard proofs of this theorem carry through in RCA_{0}. For the details, see Kokorin and Kopytov (1974) or Fuchs (1963).

2.2 Quotient Groups

In the study of ordered groups, it is natural to ask which theorems of group theory hold for ordered groups and which theorems either fail completely or require extra conditions. For example, if H is a normal subgroup of G, then G/H inherits a group structure from G. However, if G is partially ordered, the normality of H is not strong enough for G to induce both a group structure and a partial order on G/H. H must also be convex for the partial order on G to induce a natural partial order on G/H. To formulate this statement in second order arithmetic, we first need a definition for the quotient group.

Let G be a group and H be a normal subgroup of G. As with computable groups in Chapter 1, unique representatives of each coset gH can be chosen by picking the ≤_{ℕ}–least element of gH. These choices can be made in RCA_{0} because mH = nH if and only if m^{−1}n ∈ H.

Definition 2.12. (RCA_{0}) If G is a group and H is a normal subgroup of G, then the quotient group G/H is defined by the set

{ n | n ∈ G ∧ ∀m < n (m ∉ G ∨ m^{−1} · n ∉ H) }

and the operation

a ·_{G/H} b = c ↔ a, b, c ∈ G/H ∧ c^{−1} ·_{G} a ·_{G} b ∈ H.

Definition 2.13. (RCA_{0}) A subset X of a partial order Y is convex if

∀a, b, x ∈ Y ( (a, b ∈ X ∧ a ≤ x ≤ b) → x ∈ X ).

A subgroup H of a p.o. group G is convex if it is convex as a subset of G.

Definition 2.14. Let (G, ≤) be a p.o. group and H a convex normal subgroup. The induced order, ≤_{G/H}, on G/H is given by:

{ ⟨a, b⟩ | a, b ∈ G/H ∧ ∃h ∈ H(a ≤_{G} bh) }.

That is, a ≤_{G/H} b if and only if ∃h ∈ H(a ≤_{G} bh)

There are two useful variants of this definition, both of which are seen to be equivalent by unraveling the definitions. The first is to define a ≤_{G/H} b if and only if ∃h ∈ H (a^{−1}bh ∈ P (G)). The second is to define P (G/H) as the image of P (G) under the canonical map G → G/H. This definition amounts to setting:

P (G/H) = { g ∈ G/H | ∃h ∈ H (gh ∈ P (G)) }.

As above, the subscript G/H will be dropped as long as it is clear whether a and b are being compared as elements of G or G/H. When confusion may arise, we will denote elements of G/H by aH and bH.

From the viewpoint of reverse mathematics, there are two questions to ask concerning this definition. Which set existence axioms are required to form the induced order on G/H, and if orders on G/H and H are given, how hard is it to put them together to give an order on G that makes H convex and gives G/H the appropriate induced order? The answer to the first question points out an example in which having a full order on G gives extra computational power. The condition in Definition 2.14 is Σ_{1}^{0}, so Σ_{1}^{0} comprehension suffices to define the induced order. In the next section, we will show that this is the best that can be done for p.o. groups. However, if the order is known to be total, then we can do better. Theorem 2.15. (RCA_{0}) Let (G, ≤) be an f.o. group and H a convex normal subgroup. The induced order on G/H exists. Proof. Let a, b ∈ G/H and a ≠ b. Because a and b are representatives of different cosets, ab^{−1} ∉ H. Claim. ∃h ∈ H (a ≤ bh) if and only if a ≤ b.

If a ≤ b then, because 1_{G} ∈ H, it follows that ∃h ∈ H (a ≤ bh). For the other direction, suppose ∃h ∈ H(a ≤ bh) and b < a. Then b < a ≤ bh and so 1_{G} < b^{−1}a ≤ h. Since H is convex, b^{−1}a ∈ H which gives a contradiction. The induced order can now be given by a Σ_{0}^{0} condition: aH ≤ bH if and only if aH = bH or a < b.

Corollary 2.16. If (G, ≤_{G}) is a computably fully ordered computable group and H is a computable convex normal subgroup, then the induced order on G/H is computable.

It is also worth noting that for a f.o. group, if aH < bH then ∀h ∈ H(ah < b). Indeed, if not, then ah ≥ b for some h ∈ H and hence bH ≤ aH. This fact is not true for general p.o. groups since it is possible that for some h, the elements a and bh are not comparable. The following example illustrates this point.

Example 2.17. Consider (ℚ^{+}, ·) with P (ℚ^{+}) = ℕ^{+}. Under this partial order, a ≤ b if and only if a divides b. Let H be the subgroup generated by 2. Elements of H are of the form 2^{n} for some n ∈ ℤ. The group is abelian, so H is normal. H is convex since if 2^{n} ≤ x ≤ 2^{m} then x must be of the form 2^{p} for some n ≤ p ≤ m. Let 3 represent the coset { …, 12, 6, 3, 3/2, 3/4, … } in ℚ^{+}/H and 9 represent the coset { …, 36, 18, 9, 9/2, 9/4, … }. Since 3 ≤ 9 in this group and 1 ∈ H, it is clear that ∃h ∈ H(3 ≤ 9h). However, 3 ≰ 9/2 and so it is not the case that ∀h ∈ H(3 ≤ 9h).

To answer the question about putting orders on G/H and H together, one additional condition is required. An order on H is not necessarily preserved under conjugation by arbitrary elements of G. However, any order on G must have this property. Thus for an order on H to extend to all of G, it must be that a ≤_{H} b implies gag^{−1} ≤_{H} gbg^{−1} for all g ∈ G. This condition turns out to be sufficient.

Definition 2.18. (RCA_{0}) Let H be a normal subgroup of G and ≤ a full order on H. (H, ≤) is fully G – ordered if for any a, b ∈ H and g ∈ G, a ≤ b implies gag^{−1} ≤ gbg^{−1}.

Theorem 2.19. (RCA_{0}) Let (H, ≤_{H}) be a fully G-ordered normal subgroup and (G/H, ≤_{G/H}) an f.o. group. G admits a full order under which the induced orders on H and G/H correspond to those given and H is convex.

Proof. Given a, b ∈ G, define a ≤_{G} b if and only if either aH ≤_{G/H} bH or aH = bH and a^{−1}b ∈ P (H). Verifying that this definition gives a full order on G is a straight forward matter of checking the axioms for a variety of cases. Notice that if a, b ∈ H then aH = bH and so they are compared in G using the order on H. Thus the restriction of ≤_{G} to H is ≤_{H}. To show H is convex under this order, suppose that a, b ∈ H, c ∉ H and a ≤_{G} c ≤_{G} b. Then aH ≠ cH, so a ≤_{G} c implies aH <_{G/H} cH. Similarly, cH <_{G/H} bH and so aH <_{G/H} bH which is a contradiction. To show that the induced order on the quotient is ≤_{G/H}, suppose that a, b ∈ G and a ≤_{G} bh for some h ∈ H. Either aH <_{G/H} bhH or aH = bhH and a^{−1}bh ∈ P (G). In either case aH ≤_{G/H} bH. Assuming aH ≤_{G/H} bH, either aH <_{G/H} bH, and hence a < b, or aH = bH and ∃h ∈ H(a = bh). In either case, ∃h ∈ H(a ≤_{G} b). Thus the induced order on G/H is correct.

2.3 Induced Orders

The goal of this section is to show that ACA_{0} is equivalent to the existence of the induced order on the quotient of a p.o. group. By Theorem 1.16, ACA_{0} is equivalent to the existent of the range of one-to-one functions. Given a one-to-one function, the strategy is to code the range into a group in such a way that it can be recovered from the order on the quotient group. The torsion free abelian group A on generators a_{i}, b_{i} for i ∈ ℕ is used to do the coding. The first step is to present this group formally.

The elements of A are quadruples of finite sets (I, q, J, p) where I and J are finite subsets of ℕ and p and q represent the functions

q : I → ℤ \ {0}
p : J → ℤ \ {0}.

The element represented by (I, q, J, p) is denoted

∑_{i∈I} q_{i}a_{i} + ∑_{j∈J} p_{j}b_{j}.

The elements represented by (I, q, J, p) and (I′, q′, J′, p′) are equal if and only if I = I′, J = J′, q = q′ and p = p′. The sum

( ∑_{i∈I} q_{i}a_{i} + ∑_{j∈J} p_{j}b_{j}) + ( ∑_{k∈K} r_{k}a_{k} + ∑_{l∈L} s_{l}b_{l})

is defined to be

∑_{m∈M} t_{m}a_{m} + ∑_{n∈N} u_{n}b_{n}

where

M = (I ∪ K) \ {x ∈ I ∩ K | q_{x} + r_{x} = 0}
N = (J ∪ L) \ {x ∈ J ∩ L | p_{x} + s_{x} = 0}

and t_{m} is defined to be q_{m} if m ∈ I \ K, r_{m} if m ∈ K \ I and q_{m} + r_{m} if m ∈ I ∩ K. u_{n} is defined similarly. The identity element, 0_{A}, is the element represented by (∅, ∅, ∅, ∅) and if g is represented by (I, q, J, p), then g^{−1} is the sum

∑_{i∈I} −q_{i}a_{i} + ∑_{j∈J} −p_{j}b_{j}.

Theorem 2.20. (RCA_{0}) The following are equivalent:

1. ACA_{0}

2. For every p.o. group (G, ≤_{G}) and every convex normal subgroup H, the induced order ≤_{G/H} on G/H exists.

Proof. Case. (1) ⇒ (2) :

For x, y ∈ G/H, use Σ_{1}^{0} comprehension in ACA_{0} to define the relation:

x ≤_{G/H} y ↔ ∃h ∈ H (x ≤_{G} yh).

Case. (2) ⇒ (1) :

Let f : ℕ → ℕ be a one-to-one function. By Theorem 1.16, it suffices to show that the range of f exists. Use Σ_{0}^{0} comprehension to define the partial order ≤_{A} by:

P (A) = { ∑_{i∈I} q_{i}a_{i} + ∑_{j∈J} p_{j}b_{j} | J = ∅ ∧ ∀i ∈ I(q_{i} > 0)} .

The idea here is that P (A) is the semigroup generated by the elements a_{i} for i ∈ ℕ. The definition is Σ_{0}^{0} because ∀i ∈ I is a bounded quantifier. Claim. P (A) is the positive cone for a partial order on A.

It suffices to show that P (A) is a pure normal semigroup with identity. By definition, 0_{A} ∈ P (A). P (A) is normal because it is a subset of an abelian group. P (A) is a semigroup since it is closed under componentwise addition. Finally, since P^{−1}(A) is defined by

P^{−1}(A) = { ∑_{i∈I} q_{i}a_{i} + ∑_{j∈J} p_{j}b_{j} | J = ∅ ∧ ∀i ∈ I(q_{i} < 0)} ,

it is clear that P (A) is pure.

Let H be the subgroup generated by elements of the form −a_{n} + b_{m} where f(n) = m. Formally, ∑_{i∈I} q_{i}a_{i} + ∑_{j∈J} p_{j}b_{j} is in H if and only if either I = J = ∅ or I ≠ ∅ and

∀i ∈ I(f(i) ∈ J ∧ q_{i} = −p_{f(i)}) ∧ ∀j ∈ J ∃i ∈ I(f(i) = j ∧ q_{i} = −p_{j}).

This condition is Σ_{0}^{0} since all the quantification is bounded. H is normal because the group is abelian. Claim. H is convex.

To prove this claim, it suffices to show that there are no nontrivial intervals in H. That is, for any c, d ∈ H, c ≤ d implies c = d. Notice that any c, d ∈ H can be expressed as

c = ∑_{i∈I} −q_{i}a_{i} + ∑_{i∈I} q_{i}b_{f(i)} d = ∑_{j∈J} −p_{j}a_{j} + ∑_{j∈J} p_{j}b_{f(j)}.

If c ≤ d, then −c + d ∈ P (A). Since P (A) is generated by the a_{i}’s, the b_{i} part of the sums must cancel out. Hence

∑_{i∈I} −q_{i}b_{f(i)} + ∑_{j∈J} p_{j}b_{f(j)} = 0.

Since 0 is represented by the quadruple (∅, ∅, ∅, ∅), we have I = J and q = p. Hence c = d as required.

Now that A, P (A), and H are defined, all that remains to show is how the range of f can be defined from the induced order ≤_{A/H} on A/H. This definition follows from the final two claims. Claim. The existence of ≤_{A/H} implies the existence of the set P (A) + H.

Given x ∈ A, we need to decide if x ∈ P (A) + H. Let n ∈ G/H be such that n + H = x + H. Since x and n differ by a element of H, x ∈ P (A) + H if and only if n ∈ P (A) + H. However,

0_{A/H} ≤_{A/H} n ↔ ∃h ∈ H (n + h ∈ P (A))
↔ n ∈ P (A) + H.

Thus, the set P (A) + H is definable from ≤_{A/H} in RCA_{0}. Claim. b_{m} ∈ P (A) + H ↔ m ∈ range(f)

First assume that b_{m} = p + h for some p ∈ P (A) and h ∈ H. Then b_{m} can be written as:

b_{m} = ∑_{i∈I} q_{i}a_{i} + ( ∑_{j∈J} −p_{j}a_{j} + ∑_{j∈J} p_{j}b_{f(j)}).

The parts of the equation with a_{i}’s must cancel out, leaving I = J. Furthermore, because only b_{m} appears on the left of the equation, J = {n} where f(n) = m and p_{n} = 1. Hence m is in the range of f.

For the other direction, assume that m is in the range of f. For some n, f(n) = m, and hence −a_{n} + b_{m} ∈ H and a_{n} ∈ P (A). Adding these equations shows that b_{m} ∈ P (A) + H.

Corollary 2.21. There is a computably partially ordered computable group (G, ≤_{G}) and a computable convex normal subgroup H such that the degree of the induced order on G/H is 0′.

Proof. Let f be a computable one-to-one function that enumerates 0′. Since f is computable, the p.o. group in the proof of Theorem 2.20 is a computably partially ordered computable group. The range of f is computable from the induced order on G/H, so deg(≤_{G/H}) = 0′.

Group Conditions

Now that we have the basic definitions for ordered groups, we would like to know which groups can be ordered. Notice that any group can be partially ordered: take the trivial partial order under which no two distinct elements are comparable. The question of when a group admits a full order is more complicated. Being torsion free is a necessary condition, but unfortunately not a sufficient one. Let G be the group presented by the letters a and b with the relation aba^{−1} = b^{−1}. G is torsion free but not orderable. Indeed, if b > 1_{G} then aba^{−1} = b^{−1} forces b^{−1} > 1_{G} and if b < 1_{G} then aba^{−1} = b^{−1} forces b^{−1} < 1_{G}. In Sections 3.1 and 3.2, three stronger conditions are presented that suffice to guarantee full orderability. In Section 3.3, we show that the existence of the center of a group is equivalent to ACA_{0}.

If G is an O-group, it is natural to study the space of all full orders on G. In Section 3.4, we present one connection between spaces of orders and c.b. Π_{1}^{0} classes. We will return to this theme in the last chapter once we have more tools from ordered group theory.

3.1 Torsion Free Abelian and Nilpotent Groups

Being torsion free and abelian is the simplest group condition that implies full orderability. A proof of this fact can be found in Fuchs (1963) or Kokorin and Kopytov (1974).

Theorem 3.1. Every torsion free abelian group is an O-group.

The effective content of Theorem 3.1 was first explored in Downey and Kurtz (1986). They constructed a computable group isomorphic to ⨁_{ω} ℤ which has no computable full order.

Theorem 3.2 (Downey and Kurtz (1986)). There is a computable torsion free abelian group with no computable full order.

Hatzikiriakou and Simpson (1990) used a similar proof in the context of reverse mathematics to show that Theorem 3.1 is equivalent to W KL_{0}. By the Low Basis Theorem, this fact implies that every computable torsion free abelian group must have a full order of low degree.

Theorem 3.3 (Hatzikiriakou and Simpson (1990)). (RCA_{0}) The following are equivalent:

1. W KL_{0}

2. Every torsion free abelian group is an O-group.

Theorem 3.1 is generalized in Kokorin and Kopytov (1974) to torsion free nilpotent groups. Theorem 3.4. Every torsion free nilpotent group is an O-group.

The goal of this section is to use arguments similar to those in Hatzikiriakou and Simpson (1990), to show that Theorem 3.4 is equivalent to W KL_{0}. Notice that as long as RCA_{0} suffices to prove that every abelian group is nilpotent, Theorem 3.3 already shows that Theorem 3.4 implies W KL_{0}. To state the result precisely, we need a formal definition of nilpotent groups in second order arithmetic.

In keeping with standard mathematical notation, if H is a normal subgroup of G, we will let π : G → G/H denote the projection function. That is, π picks out the <_{ℕ}-least representative of gH. Frequently, we will write gH instead of π(g). Definition 3.5. The center of G, C(G), is the set

{ g ∈ G | ∀x ∈ G (gx = xg) }.

In general, the existence of the center is equivalent to ACA_{0}, as we shall see in Section 3.3. However if C(G) is given, several properties of it can be proved in RCA_{0}. Lemma 3.6. (RCA_{0}) If C(G) exists then C(G) is a normal subgroup of G. Lemma 3.7. (RCA_{0}) If H is a normal subgroup of G, π : G → G/H and C(G/H) exists, then

K = {g ∈ G | π(g) ∈ C(G/H)} = π^{−1}(C(G/H))

is a normal subgroup of G.

Proof. It is straight forward to check that K is a subgroup. To show that K is normal, let k ∈ K and g ∈ G. Because π(k) ∈ C(G/H), it follows that k commutes modulo H with all elements of G. In particular, g^{−1}kH = kg^{−1}H or in the notation of π, π(g^{−1}k) = π(kg^{−1}). From here it follows that π(k) = π(gkg^{−1}). Thus π(gkg^{−1}) ∈ C(G/H) and gkg^{−1} ∈ K. Definition 3.8. Let G be a group. The upper central series of G is the series of subgroups

ζ_{0}G ≤ ζ_{1}G ≤ ζ_{2}G ≤ ⋯

defined by

ζ_{0}G = ⟨1_{G}⟩
ζ_{1}G = C(G)
ζ_{i+1}G = π^{−1}(C(G/ζ_{i}G))

where π : G → G/ζ_{i}G. G is nilpotent if ζ_{n}G = G for some n ∈ ω.

Notice that ζ_{i+1}G/ζ_{i}G = C(G/ζ_{i}G). In order to use nilpotent groups in reverse mathematics, we need to define a code for them that explicitly gives the information contained in the upper central series. Definition 3.9. The i^{th} column of a set X is denote by X_{i} and is defined by:

X_{i} = { n | ⟨n, i⟩ ∈ X }.

Definition 3.10. (RCA_{0}) The pair N ⊆ ℕ and n ∈ ℕ is a code for a nilpotent group G if the first n + 1 columns of N satisfy 1. N_{0} = ⟨1_{G}⟩

2. N_{1} = C(G)

3. N_{n} = G

4. For 0 ≤ i ≤ n, N_{i} is a normal subgroup of G.

5. For 0 ≤ i < n, if π : G → G/N_{i}, then N_{i+1} = π^{−1}(C(G/N_{i})). A group G is nilpotent is there is such a code (N, n) for G. Lemma 3.11. (RCA_{0}) Every abelian group is nilpotent. Proof. If G is abelian then we can define a code for G as a nilpotent group by setting n = 1 and N ⊆ ℕ with N_{0} = ⟨1_{G}⟩ and N_{1} = G. Lemma 3.12. (RCA_{0}) If (N, n) is the code for a nilpotent group G then for all 0 ≤ i < n, N_{i+1}/N_{i} is abelian. Proof. By definition, N_{i+1} = π^{−1}(C(G/N_{i})) with π : G → G/N_{i}. Therefore, N_{i+1}/N_{i} ≅ C(G/N_{i}). Theorem 3.13. (RCA_{0}) The following are equivalent.

1. W KL_{0}

2. Every torsion free nilpotent group is an O-group.

The goal of the rest of this section is to prove this theorem. The idea is that a nilpotent group is formed from a finite number of abelian quotients N_{i+1}/N_{i}. If these quotients are torsion free then each is fully orderable by Theorem 3.3 and so we only need to put these orders together in a nice way. The first step is to show that that if G is a torsion free nilpotent group then each N_{i+1}/N_{i} is also torsion free. Notice that if (N, n) is the code for a torsion free nilpotent group G and n ≥ 1, then N_{1} must also be torsion free since it is a subgroup of G.

Definition 3.14. The commutator of x and y, denoted [x, y], is the element x^{−1}y^{−1}xy.

Lemma 3.15. (RCA_{0}) Let (N, n) be a code for a nilpotent group G. If 0 ≤ i < n and x ∈ N_{i+1}, then [x, g] ∈ N_{i} for all g.

Proof. Notice that for i = 0, the lemma follows trivially because N_{1} is the center of G. Assume i ≥ 1. By definition, x ∈ N_{i+1} means xgN_{i} = gxN_{i} for all g. For any particular g, there is a c ∈ N_{i} such that xg = gxc and hence also cg^{−1}x^{−1} = x^{−1}g^{−1}. Let h be any element of G.

[x, g] · h = x^{−1}g^{−1}xg · h
= x^{−1}g^{−1}gxch
= ch

Since c ∈ N_{i}, we know that ch = hcc̃ for some c̃ ∈ N_{i−1}. We now have:

ch = hcc̃
= hcg^{−1}x^{−1}xgc̃
= hx^{−1}g^{−1}xgc̃.

Thus, we have [x, g] · h = h · [x, g] · c̃ for some c̃ ∈ N_{i−1}. Another way to write this equality is

[x, g] · hN_{i−1} = h · [x, g]N_{i−1}.

This equality means that [x, g]N_{i−1} is in the center of G/N_{i−1} and hence that [x, g] ∈ N_{i}.

Lemma 3.16. (RCA_{0}) Let (N, n) be a code for a nilpotent group G. If 1 ≤ i < n and x ∈ N_{i+1}, then for all m > 0

[x, g]^{m}N_{i−1} = [x^{m}, g]N_{i−1}.

Proof. Because [x, g]^{m}N_{i−1} = [x^{m}, g]N_{i−1} is a Σ_{0}^{0} statement, we can prove this lemma in RCA_{0} by induction on m. The case for m = 1 is trivial, so assume the equality holds for m and we prove it for m + 1. Since [x, g]^{m+1} = [x, g]^{m} · [x, g], we can apply the induction hypothesis in the form [x, g]^{m} = [x^{m}, g] · c for some c ∈ N_{i−1}. We now have:

[x, g]^{m+1} = [x^{m}, g] · c · [x, g]
= x^{−m}g^{−1}x^{m}gc · [x, g].

By Lemma 3.15, x ∈ N_{i+1} implies [x, g] ∈ N_{i} and so [x, g] commutes with elements of G modulo N_{i−1}. Therefore, for some c̃ ∈ N_{i−1} we have

x^{−m} · g^{−1}x^{m}gc · [x, g] = x^{−m} · [x, g] · g^{−1}x^{m}gcc̃
= x^{−m−1}g^{−1}xgg^{−1}x^{m}gcc̃
= [x^{m+1}, g] · cc̃.

Because cc̃ ∈ N_{i−1}, this calculation establishes the induction case.

Lemma 3.17 (Mal’cev). (RCA_{0}) Let (N, n) be a code for a torsion free nilpotent group G. For every 0 ≤ i < n, N_{i+1}/N_{i} is torsion free.

Proof. We prove this theorem by bounded induction on i. Because N_{0} = ⟨1_{G}⟩ we have N_{1}/N_{0} = N_{1}, which establishes the theorem for i = 0. Assume i ≥ 1 and the theorem holds for i − 1. The induction hypothesis tells us that N_{i}/N_{i−1} is torsion free. Let x ∈ N_{i+1} and suppose that x^{m} ∈ N_{i} for some m > 0. We need to show that x ∈ N_{i}. By Lemma 3.16, [x, g]^{m}N_{i−1} = [x^{m}, g]N_{i−1} for any g. By Lemma 3.15, x^{m} ∈ N_{i} implies that [x^{m}, g] ∈ N_{i−1}. Therefore, [x, g]^{m} ∈ N_{i−1}. Applying Lemma 3.15 to x ∈ N_{i+1} tells us that [x, g] ∈ N_{i}. Putting these facts together, we have:

[x, g]N_{i−1} ∈ N_{i}/N_{i−1}
[x, g]^{m}N_{i−1} = 1_{G}N_{i−1}.

Since N_{i}/N_{i−1} is torsion free, it must be that [x, g] ∈ N_{i−1}. However, this fact implies that xgN_{i−1} = gxN_{i−1} for all g and so x ∈ N_{i} as required.

Lemma 3.18. (W KL_{0}) Let (N, n) be a code for a torsion free nilpotent group. For every 0 ≤ i < n, N_{i+1}/N_{i} is a fully G/N_{i}-orderable group.

Proof. We need to show that there is a full order on N_{i+1}/N_{i} such that for all a, b ∈ N_{i+1}/N_{i} and g ∈ G/N_{i}, if aN_{i} < bN_{i} then gag^{−1}N_{i} < gbg^{−1}N_{i}. By Lemmas 3.12 and 3.17, N_{i+1}/N_{i} is a torsion free abelian group and hence by Theorem 3.3, W KL_{0} proves that it is fully orderable.

Let ≤ be any full order on N_{i+1}/N_{i}, let a < b be elements of N_{i+1}/N_{i} and let g ∈ G/N_{i}. Since N_{i+1}/N_{i} ≅ C(G/N_{i}), we have gag^{−1}N_{i} = aN_{i} and gbg^{−1}N_{i} = bN_{i}. Hence, aN_{i} < bN_{i} implies gag^{−1}N_{i} < gbg^{−1}N_{i}.

We are now ready to prove Theorem 3.13

Proof. Case. (2) ⇒ (1)

Assume every torsion free nilpotent group is an O-group. By Lemma 3.11, this assumption implies that every torsion free abelian group is an O-group. From here, Theorem 3.3 implies (1). Case. (1) ⇒ (2)

For each 1 ≤ i ≤ n, let P̄_{i} be the strict positive cone of a full G/N_{i−1}-order on N_{i}/N_{i−1}. Define P_{i} and P by:

P_{i} = {x ∈ N_{i} | xN_{i−1} ∈ P̄_{i}}
P = (∪_{i=1}^{n}P_{i}) ∪ {1_{G}}.

We prove that P is the positive cone for a full order on G in the following series of claims. Claim. P is a semigroup with identity.

It suffices to show P is closed under multiplication. Let x, y ∈ P with x, y ≠ 1_{G}. There are i, j such that x ∈ P_{i} and y ∈ P_{j}. If i = j then xN_{i−1}, yN_{i−1} ∈ P̄_{i} and so xyN_{i−1} ∈ P̄_{i} and xy ∈ P_{i}. If i ≠ j then, without loss of generality, assume that i < j. Since x ∈ P_{i}, it follows that x ∈ N_{i} and hence x ∈ N_{j−1}. But then, xyN_{j−1} = yN_{j−1} and so xy ∈ P_{j}.

Claim. P is normal.

Let x ∈ P, x ≠ 1_{G} and g ∈ G. There is an i such that x ∈ P_{i}. Since P̄_{i} is the strict positive cone of a full G/N_{i−1}–order on N_{i}/N_{i−1}, we have that xN_{i−1} ∈ P̄_{i} implies that gxg^{−1}N_{i−1} ∈ P̄_{i}. Hence gxg^{−1} ∈ P_{i}. Claim. P is pure.

Let x ∈ P and x ≠ 1_{G}. We need to show that x^{−1} ∉ P. There is an i such that x ∈ P_{i}. Because P̄_{i} is the strict positive cone on N_{i}/N_{i−1}, we know that x ∈ N_{i}, x ∉ N_{i−1}. Hence x^{−1} ∈ N_{i} and x^{−1} ∉ N_{i−1}. However, because xN_{i−1} ∈ P̄_{i}, it follows that x^{−1}N_{i−1} ∉ P̄_{i} and so x^{−1} ∉ P_{i}. To show x^{−1} ∉ P_{j} for j > i, notice that since x^{−1} ∈ N_{i}, we also have x^{−1} ∈ N_{j−1}. Therefore x^{−1}N_{j−1} = 1_{G}N_{j−1} and hence x^{−1} ∉ P_{j}. Finally, assume for a contradiction that j < i and x^{−1} ∈ P_{j}. It follows that x^{−1} ∈ N_{i−1}. However, above we showed that x^{−1} ∉ N_{i−1}. Thus, x^{−1} ∉ P_{j} for any j. Claim. P is full.

Let x ∈ P and x ≠ 1_{G}. We need to show that either x ∈ P or x^{−1} ∈ P. There is an i such that x ∈ N_{i} and x ∉ N_{i−1}. Since P̄_{i} is a full order on N_{i}/N_{i−1}, either xN_{i−1} ∈ P̄_{i} or x^{−1}N_{i−1} ∈ P̄_{i}. Thus, either x ∈ P_{i} or x^{−1} ∈ P_{i}.

3.2 Direct Products

Groups are frequently constructed by means of a direct product. These constructions preserve full orderability. A proof of the following theorem can be found in either Fuchs (1963) or Kokorin and Kopytov (1974).

Theorem 3.19. Any direct product of O-groups is an O-group.

To examine this theorem in reverse mathematics, we need to distinguish between finite and restricted countable direct products. The finite direct product A_{0} × A_{1} × …×A_{n−1} consists of sequences of length n such that the i^{th} element of each sequence is in A_{i}. Multiplication is componentwise. The elements of the restricted direct product of A_{i} for i ∈ ℕ are finite sequences σ such that for all i < lh(σ), σ(i) ∈ A_{i}. The idea is that the element represented by σ has 1_{A_{j}} as its j^{th} component for all j ≥ lh(σ). In order to make each sequence represent a distinct element, we add the requirement that the last element in the sequence not be an identity element. The formal definitions are given below. Definition 3.20. (RCA_{0}) If n ∈ ℕ and for all i < n, A_{i} is a group, then the finite direct product G = ∏_{i=0}^{n−1} A_{i} is defined by:

G = {σ ∈ Fin_{ℕ} | lh(σ) = n ∧ ∀i < n (σ(i) ∈ A_{i})}

1_{G} = ⟨1_{A_{0}}, 1_{A_{1}}, …, 1_{A_{n−1}}⟩
σ ·_{G} τ = ⟨σ(0) ·_{A_{0}} τ(0), …, σ(n − 1) ·_{A_{n−1}} τ(n − 1)⟩.

Theorem 3.21. (RCA_{0}) If n ∈ ℕ and for all i < n, A_{i} is an O–group, then G = ∏_{i=0}^{n−1} A_{i} is an O–group. Proof. Let P^{+}(A_{i}) be the strict positive cone of a full order on A_{i}. Order G lexicographically:

P^{+}(G) = {σ ∈ G | ∃i < n (σ(i) ∈ P^{+}(A_{i}) ∧ ∀j < i(σ(j) = 1_{A_{j}}))}

P (G) = P^{+}(G) ∪ { ⟨1_{A_{0}}, …, 1_{A_{n−1}}⟩ }.

From this definition, P (G) is clearly full, pure, and contains the identity. It remains to check that it is a normal semigroup. Since P (G) is closed under multiplication, it is a semigroup. To see that it is normal, let σ ∈ P (G) have its first non-identity element at σ(i). If τ = ⟨g_{0}, …, g_{n−1}⟩ ∈ G then τστ^{−1} is

⟨g_{0}, …, g_{n−1}⟩ ·_{G} ⟨1_{A_{0}}, …, 1_{A_{i−1}}, a_{i}, …, a_{n−1}⟩ ·_{G} ⟨g_{0}^{−1}, …, g_{n−1}^{−1}⟩.

The first non-identity element in this product is g_{i}a_{i}g_{i}^{−1}. Because a_{i} ∈ P^{+}(A_{i}), we have g_{i}a_{i}g_{i}^{−1} ∈ P^{+}(A_{i}) and hence τστ^{−1} ∈ P^{+}(G).

Definition 3.22. (RCA_{0}) Let A be a set such that for each i, the i^{th} column A_{i} is a group. The restricted direct product G = ∏_{n∈ℕ} A_{n} is defined by:

G = {σ ∈ Fin_{ℕ} | ∀i < lh(σ) (σ(i) ∈ A_{i}) ∧ σ(lh(σ) − 1) ≠ 1_{A_{lh(σ)−1}}}
1_{G} = ⟨⟩

where ⟨⟩ is the empty sequence. Multiplication is componentwise, removing any trailing identity elements.

Theorem 3.23. (RCA_{0}) The following are equivalent:

1. W KL_{0}

2. If ∀i (A_{i} is an O-group) then G = ∏_{i∈ℕ} A_{i} is an O-group.

Proof. Case. (1) ⇒ (2): We know ∀i ∃Y (Y is a positive cone on A_{i}). From the Theorem 3.21, if n ∈ ℕ then

RCA_{0} ⊢ ∃Y (Y is positive cone on ∏_{i=0}^{n−1} A_{i}).

A uniform (strict) order on the A′_{i}s is a set P such that P_{i} is the (strict) positive cone of a full order on A_{i}. To prove that G is an O-group, it suffices to prove the existence of a uniform order on the A_{i}. From a uniform order, we can define the lexicographic order on G as in Theorem 3.21. To show the existence of a uniform order, we build a tree T such that any path on the tree codes such an order. T is built in stages such that at the end of stage s, all nodes of length s are defined. Each node on T keeps a guess at an approximation to a uniform strict order. Suppose σ is a node on T at level s, s + 1 = ⟨e, i⟩, e ≠ 1_{A_{i}}, and P_{σ} is σ’s approximation. At stage s + 1 we check if 1_{A_{j}} ∈ P_{σ} for any j. Since P_{σ} is a finite set, this can be done computably. If 1_{A_{j}} ∈ P_{σ}, then P_{σ} cannot be a subset of a uniform strict order, so we terminate this branch. Otherwise, we define two extensions of P_{σ}: one by adding e ∈ A_{i} to P_{σ} and the other by adding e^{−1} ∈ A_{i} to P_{σ}. These sets are each closed under one step multiplication and conjugation by elements less than s. One extension becomes P_{σ^{⌢}0} and the other becomes P_{σ^{⌢}1}. This construction is presented formally below. T_{s} will be the set of nodes of T of length s. Construction Stage 0: Set T = {⟨⟩} and P_{⟨⟩} = ∅. Stage s + 1 : Assume s = ⟨e, i⟩. For each σ ∈ T_{s} do the following:

1. Check if 1_{A_{j}} appears in P_{σ} for any j. If so, σ has no extensions on T, so move on to the next node in T_{s}. If not, add σ^{⌢}0 and σ^{⌢}1 to T_{s+1} and move on to step 2.

2. If e = 1_{A_{i}} or e does not represent an element of A_{i}, then set P_{σ^{⌢}0} = P_{σ^{⌢}1} = P_{σ}

and move on to the next node in T_{s}.

3. If e ∈ A_{i} and e ≠ 1_{A_{i}} define

P̃_{σ^{⌢}0} = P_{σ} ∪ {⟨e^{−1}, i⟩}
P̃_{σ^{⌢}1} = P_{σ} ∪ {⟨e, i⟩}

Extend these by:

⟨k, j⟩ ∈ P_{σ^{⌢}0} ↔ ⟨k, j⟩ ∈ P̃_{σ^{⌢}0} ∨
∃⟨m, j⟩, ⟨n, j⟩ ∈ P̃_{σ^{⌢}0} (m ·_{A_{j}} n = k) ∨
∃n ≤ s ∃⟨m, j⟩ ∈ P̃_{σ^{⌢}0} (n ∈ A_{j} ∧ n ·_{A_{j}} m ·_{A_{j}} n^{−1} = k)
⟨k, j⟩ ∈ P_{σ^{⌢}1} ↔ ⟨k, j⟩ ∈ P̃_{σ^{⌢}1} ∨
∃⟨m, j⟩, ⟨n, j⟩ ∈ P̃_{σ^{⌢}1} (m ·_{A_{j}} n = k) ∨
∃n ≤ s ∃⟨m, j⟩ ∈ P̃_{σ^{⌢}1} (n ∈ A_{j} ∧ n ·_{A_{j}} m ·_{A_{j}} n^{−1} = k).

End of Construction Claim. T is infinite.

For a contradiction, suppose that T is not infinite and hence there is some level n at which T has no nodes. Notice that our coding for pairs satisfies the inequality ⟨x, y⟩ ≥ y. Therefore, if ⟨x, y⟩ occurs in the construction before stage n, we know that y ≤ n. That is, at stage n, T has only considered elements from A_{0} through A_{n}. By Theorem 3.21,

RCA_{0} ⊢ ∏_{i=0}^{n} A_{i} is an O-group.

Let X be the strict positive cone for a full order on this finite product and P^{+}(A_{i}) be defined by

x ∈ P^{+}(A_{i}) ↔ ⟨1_{A_{0}}, …, 1_{A_{i−1}}, x, 1_{A_{i+1}}, …, 1_{A_{n}}⟩ ∈ X.

For each k ≤ n, k = ⟨x, i⟩ for some i ≤ n. Define σ ∈ Fin_{ℕ} with lh(σ) = n by

σ(k) = \{ \begin{matrix}1 & if k = ⟨x, i⟩ ∧ x ∈ P^{+}(A_{i}) \\ 0 & otherwise \end{matrix}

From the definition it is clear that

σ(k) = 0 ↔ x = 1_{A_{i}} ∨ x^{−1} ∈ P^{+}(A_{i}) ∨ x ∉ A_{i}.  (3.1)

To prove the claim, it suffices to show that σ ∈ T. We show by induction that for all k ≤ n, σ|_{k} ∈ T and P_{σ|_{k}} ⊆ X. Clearly, σ|_{0} = ⟨⟩ ∈ T and P_{σ|_{0}} = ∅ ⊆ X. Assume that σ|_{k} ∈ T and P_{σ|_{k}} ⊆ X. Because 1_{A_{j}} ∉ P_{σ|_{k}} we know that σ|_{k+1} ∈ T. From the definition of σ and Equation 3.1, it is clear that P̃_{σ|_{k+1}} ⊆ X. Because P_{σ|_{k+1}} is obtained by multiplying and conjugating elements of P̃_{σ|_{k+1}}, it follows that P_{σ|_{k+1}} ⊆ X. Thus, σ|_{n} = σ ∈ T.

Since T is infinite, W KL_{0} provides a path S through T. Define:

Z̃ = ⋃_{σ∈S} P_{σ}

Z = Z̃ ∪ {⟨1_{A_{i}}, i⟩ | i ∈ ℕ}.

Z̃ has a Σ_{1}^{0} definition, but for x ≠ 1_{A_{i}}

⟨x, i⟩ ∈ Z̃ ↔ ⟨x^{−1}, i⟩ ∉ Z̃.

Thus, Z̃ has a Δ_{1}^{0} definition and so both Z̃ and Z exist. It remains to show that Z_{i} is the positive cone for a full order on A_{i}.

To show Z_{i} is full, consider any x ∈ A_{i}, x ≠ 1_{A_{i}}. Let σ ∈ S with lh(σ) = ⟨x, i⟩. Since S is a path, either σ^{⌢}0 ∈ S or σ^{⌢}1 ∈ S.

σ^{⌢}0 ∈ S ⇒ ⟨x, i⟩ ∈ P_{σ^{⌢}0} ⇒ x ∈ Z_{i}
σ^{⌢}1 ∈ S ⇒ ⟨x^{−1}, i⟩ ∈ P_{σ^{⌢}1} ⇒ x^{−1} ∈ Z_{i}

To show Z_{i} is pure, suppose x ≠ 1_{A_{i}} and x, x^{−1} ∈ Z_{i}. For some σ ∈ S, ⟨x, i⟩, ⟨x^{−1}, i⟩ ∈ P_{σ}. From the construction, 1_{A_{i}} appears in both P_{σ^{⌢}0} and P_{σ^{⌢}1} so neither σ^{⌢}0 nor σ^{⌢}1 has an extension. This condition means σ cannot be on a path which contradicts σ ∈ S.

Z_{i} is a semigroup since if x, y ∈ Z_{i} then there is a σ ∈ S such that ⟨x, i⟩, ⟨y, i⟩ ∈ P_{σ}. S is a path so P_{σ} has an extension τ in the tree. By the one step multiplicative closure, ⟨x ·_{A_{i}} y, i⟩ ∈ P_{τ} and hence x ·_{A_{i}} y ∈ Z_{i}. Showing Z_{i} is normal is similar but uses the one step closure under conjugates. Thus Z_{i} is a full order on A_{i} and we have constructed the desired uniform order. Case. (2) ⇒ (1):

Assume the restricted direct product of O-groups is an O-group. To prove W KL_{0}, it suffices by Theorem 1.15 to prove that a separating set exists for any two functions with disjoint ranges. Let f, g be functions such that for all n, m, f(n) ≠ g(m). We need to form a set S such that

range(f) ⊆ S ∧ range(g) ⊆ ℕ \ S.

Recall from the first half of this proof, that an order on the direct product is equivalent over RCA_{0} to a uniform order on the components A_{i}. The idea of this proof is to give abelian groups A_{n} each of which has two generators, a_{n} and b_{n}. If n is in the range of f, we force a_{n} and b_{n} to have the same sign in any order on A_{n}. That is, either both are positive or both are negative. If n is in the range of g, we force a_{n} and b_{n} to have different signs in any order. If neither of these holds, then we let A_{n} be a torsion free abelian group on two generators. Since the groups are abelian, we use additive notation. The groups look like:

A_{f(n)} = ⟨a_{f(n)}, b_{f(n)} | a_{f(n)} = p_{n}b_{f(n)}⟩
A_{g(n)} = ⟨a_{g(n)}, b_{g(n)} | a_{g(n)} = −p_{n}b_{g(n)}⟩

where p_{n} is the n^{th} prime starting with 3. If n is not in the range of f or g then

A_{n} = ⟨a_{n}, b_{n} | −⟩.

Formally, A_{n} is given by the elements ca_{n} + db_{n} where c, d ∈ ℤ and

¬∃i(p_{i} < 2|d| ∧ f(i) = n)
¬∃i(p_{i} < 2|d| ∧ g(i) = n).

To add ca_{n} + db_{n} and c′a_{n} + d′b_{n} we check whether (c + c′) a_{n} + (d + d′) b_{n} violates either of these conditions. If there is an i such that p_{i} < 2|d+d′| and f(i) = n, then we use the relation a_{n} = p_{i}b_{n} to rewrite (d + d′) b_{n} as c′′a_{n} + d′′b_{n} where |d′′| < p_{i}/2. If the second condition is violated, we do the same thing except we use the relation a_{n} = −p_{i}b_{n}. We start the numbering of the primes with 3 because if we started with 2 and f(2) = n then the elements a_{n} + b_{n} and 2a_{n} − b_{n} would be the same. Using only odd p_{i} makes the coding more convenient.

Because the definition of A_{n} is uniform in n, the sequence A_{n} exists. It remains to show that each A_{n} is orderable and that the separating set is given by a uniform order of the A_{n}. Claim. Each A_{n} is an O-group.

The proof of this claim splits into three cases. In RCA_{0}, we cannot tell which case holds, but we know that one of them must hold.

1. If f(i) = n then

P (A_{n}) = {ca_{n} + db_{n} | c > 0 ∨ (c = 0 ∧ d ≥ 0)}.

2. If g(i) = n then

P (A_{n}) = {ca_{n} + db_{n} | c > 0 ∨ (c = 0 ∧ d ≤ 0)}.

3. If n ∉ range(g) ∪ range(f) then P (A_{n}) = {ca_{n} + db_{n} | c > 0 ∨ (c = 0 ∧ d ≥ 0)}.

In each case it is easy to verify that the set given is the positive cone of a full order. This shows

RCA_{0} ⊢ ∀n(A_{n} is an O-group).

By assumption, there is a uniform order on the A_{n}. Let P be the uniform positive cone. That is, P_{n} is the positive cone of a full order on A_{n}. Define S by

S = {n | a_{n} ∈ P_{n} ↔ b_{n} ∈ P_{n}}.

S is the desired separating set since if n is in the range of f then a_{n} ∈ P_{n} ↔ b_{n} ∈ P_{n} while if n is in the range of g then a_{n} ∈ P_{n} ↔ −b_{n} ∈ P_{n}.

From the perspective of computable mathematics, uniformity is also the key issue in ordering direct products. If we are given computable f.o. groups uniformly, then the direct product can be computably fully ordered using the lexicographic order. However, if we have a sequence of computably fully orderable computable groups, it is not necessarily possible to compute a uniform sequence of computable full orders. We have the following corollaries to Theorems 3.21 and 3.23. Corollary 3.24. The direct product of a finite number of computably fully orderable computable groups is computably fully orderable. Corollary 3.25. There is a uniform sequence of computably fully orderable computable groups G_{i}, i ∈ ω, such that Π_{i∈ω}G_{i} is a computable group with no computable full order. Π_{i∈ω}G_{i} does have a full order of low degree.

3.3 The Center

The results in this section concern how complicated the center of a group can be. In terms of reverse mathematics, the existence of the center is equivalent to ACA_{0}. As a consequence, there is a computable group whose center is as complicated as 0′. However, this result can be refined to show that even for 2 step nilpotent groups, which are intuitively the simplest nonabelian groups, the center can still be as complicated as 0′. In order to prove these results, we need an introduction to 2 step nilpotent groups.

Definition 3.26. G is n step nilpotent, for n > 1, if ζ_{n}G = G. G is properly n step nilpotent if G is n step nilpotent and ζ_{n−1}G ≠ G.

According to the definition, G is properly 2 step nilpotent if C(G) ≠ G and G/C(G) is abelian. These groups can also be defined in terms of the lower central series. The following lemma states the essential property of this alternate definition. Lemma 3.27. G is 2 step nilpotent if and only if each commutator [x, y] commutes with all the elements of the group.

Lemma 3.27 can be used to establish the following identity for 2 step nilpotent groups.

[x^{−1}, y] = xy^{−1}x^{−1}y
= xy^{−1}x^{−1}yxx^{−1}
= x · [y, x] · x^{−1}
= [y, x]

Similarly, we have:

[x, y^{−1}] = [y, x]
[x^{−1}, y^{−1}] = [x, y]
[x, y]^{−1} = [y, x].

Let G be a free group on the generators a_{i}, i ∈ ω, subject to the relations [[g, h], k] = 1_{G} for all g, h, k ∈ G. This group is called the free 2 step nilpotent group on a_{i}. We have the following identity:

a_{i}a_{j} = a_{j}a_{i}a_{i}^{−1}a_{j}^{−1}a_{i}a_{j} = a_{j}a_{i} · [a_{i}, a_{j}].

Using the identities above and performing similar calculations, we get

a_{i}^{−1}a_{j} = a_{j}a_{i}^{−1} · [a_{j}, a_{i}]
a_{i}a_{j}^{−1} = a_{j}^{−1}a_{i} · [a_{j}, a_{i}]
a_{i}^{−1}a_{j}^{−1} = a_{j}^{−1}a_{i}^{−1} · [a_{i}, a_{j}].

Because these identities allow us to commute any pair of generators modulo a commutator of generators, we can write any element of G as

a_{j_{0}}^{k_{0}}a_{j_{1}}^{k_{1}} ⋯ a_{j_{l}}^{k_{l}} · c

where j_{0} < j_{1} < ⋯ < j_{l}, k_{i} ∈ ℤ \ {0} and c is a product of commutators. Furthermore, we can write c as a product of powers of commutators of the form [a_{i}, a_{j}] or [a_{i}, a_{j}]^{−1} with i < j. To get a unique normal form for each element, we arrange these commutators so that a power of [a_{i}, a_{j}] occurs to the left of a power of [a_{k}, a_{l}] if and only if i < k or i = k and j < l.

These normal forms give us a computable presentation of the free 2 step nilpotent group. Furthermore, since we can write down a description of the normal form using only bounded quantifiers, we can define the free 2 step nilpotent group on generators a_{i}, i ∈ ω, in RCA_{0}. Because an element is in the center if and only if it is a product of commutators, RCA_{0} suffices to prove that there is a nilpotent code for this group.

Theorem 3.28. (RCA_{0}) The following are equivalent:

1. ACA_{0}

2. For every group G the center of G, C(G), exists.

Proof. Case. (1) ⇒ (2)

The center of G is defined by a Π_{1}^{0} formula, so ACA_{0} suffices to prove its existence. Case. (2) ⇒ (1)

By Theorem 1.16, it suffices to prove the existence of the range of an arbitrary one-one function f. Let G be the free 2 step nilpotent group on generators a_{i} and b_{i} for i ∈ ℕ with the following extra relations:

a_{i}a_{j} = a_{j}a_{i} for all i, j ∈ ω
b_{i}b_{j} = b_{j}b_{i} for all i, j ∈ ω
a_{i}b_{j} = b_{j}a_{i} ⇔ ∀k ≤ i (f(k) ≠ j).

Elements of G have unique normal forms:

a_{i_{1}}^{n_{1}} ⋯ a_{i_{k}}^{n_{k}}b_{j_{1}}^{m_{1}} ⋯ b_{j_{l}}^{m_{l}} · c

where i_{1} < ⋯ < i_{k}, j_{1} < ⋯ < j_{l}, n_{p} ≠ 0 and m_{q} ≠ 0 for 1 ≤ p ≤ k and 1 ≤ q ≤ l, and c is a product of commutators with those which match the added relations removed. By the comments above, G exists as a group in RCA_{0}. However, as we are about to see, RCA_{0} is not strong enough to prove that there is a code for G as a nilpotent group.

Let C(G) be the center of G. To define the range of f we use the following equivalences:

b_{j} ∈ C(G) ⇔ ∀i (a_{i}b_{j} = b_{j}a_{i})
⇔ ∀i ∀k ≤ i (f(k) ≠ j)
⇔ ∀k (f(k) ≠ j).

Therefore, b_{j} ∈ C(G) if and only if j is not in the range of f. This equivalence allows us to give a Σ_{0}^{0} definition of the range of f.

range(f) = { j | b_{j} ∉ C(G) }

Corollary 3.29. There is a computable 2 step nilpotent group G such that C(G) ≡_{T} 0′.

Proof. Consider the group G constructed in the theorem when f is a computable one-one function enumerating the halting problem. G satisfies the conditions of the corollary. Since we can define the range of f from C(G), we have 0′ ≤_{T} C(G). However, because C(G) has a Π_{1}^{0} definition from G and G is computable, we know that C(G) ≤_{T} 0′.

In addition, the group G in Corollary 3.29 is computably fully orderable. Let H be the subgroup generated by the commutators. H is normal because G is 2 step nilpotent and H is computable because we can tell if an element is the product of commutators by looking at the normal form. H is generated by commutators of the form [a_{i}, b_{j}] for which ∃k ≤ i (f(k) = j). There are no relations between these generators, so H is a torsion free abelian group which can be computably fully ordered using the generators. Since G is 2 step nilpotent, the elements of H commute with all elements of G. Therefore, any full order on H is a full G-order. G/H is the abelianization of G, which is the torsion free abelian group generated by a_{i} and b_{j} for i, j ∈ ω. Again, there are no extra relations between these elements in G/H, so G/H can be computably fully ordered using these generators. Using Theorem 2.19, the orders on H and G/H can be combined into a computable full order on G.

Finitely generated nilpotent groups have been extensively studied from the viewpoint of computational algebra. These groups have very nice computational properties. For example, the word problem, the conjugacy problem and the isomorphism problem are all solvable. The key algebraic facts to establish these results are that finitely generated nilpotent groups are also finitely presented, that finitely generated nilpotent groups are residually finite and that every subgroup is finitely generated.

Theorem 3.30 (Baumslag et al. (1991)). The center of a finitely generated nilpotent group is computable.

Corollary 3.31 (Baumslag et al. (1991)). All the terms in the upper central series for a finitely generated nilpotent group are computable.

3.4 Spaces of Full Orders

Once a group G is known to be fully orderable, there are many natural questions to ask about the space of all full orders on G. For example, algebraists have tried to say explicitly what the full orders look like. This project turns out to be extremely difficult. An easier question is how computationally complicated can the orders be. Downey and Kurtz (1986) phrased this question in terms of the connection between c.b. Π_{1}^{0} classes and full orders on torsion free abelian groups. In Chapter 8, we will formulate their question explicitly and answer it.

A first step towards answering this question comes from looking at the proof of Theorem 3.23. In the proof of this theorem, we used nodes on a tree to guess at full orders and terminate branches when they violate certain algebraic conditions. We used the property that finite direct products are orderable to show, in W KL_{0}, that the tree is infinite and therefore has a path. This path codes a full order on the countable direct product.

In this section, we give a similar argument, only not in the context of reverse mathematics. Starting with a fully orderable computable group G, we build a computable binary branching tree, the paths of which code all the full orders on G. Since we are not restricted to a weak axiom system, the property that G is fully orderable is enough to guarantee that the tree is infinite. The paths of the tree correspond, up to Turing degree, exactly to the full orders of G. In this way, we show that up to Turing degree, the space of full orders on a computable group is a c.b. Π_{1}^{0} class.

Definition 3.32. Let G be a fully orderable computable group. The space of orders of G, denoted 𝕏(G), is

{ P ⊆ G | P is the positive cone of a full order on G }.

Metakides and Nerode performed a similar analysis for orderable computable fields. Their work on the connection between spaces of orders for such fields and c.b. Π_{1}^{0} classes was the motivation for Downey and Kurtz (1986) to ask about these connections for ordered groups. Theorem 3.33 (Metakides and Nerode (1979)). Let F be an orderable computable field. There is a c.b. Π_{1}^{0} class C and a Turing degree preserving bijection φ : 𝕏(F) → C. Theorem 3.34. Let G be a fully orderable computable group. There is a c.b. Π_{1}^{0} class C and a Turing degree preserving bijection φ : 𝕏(G) → C. Proof. Let G be a fully orderable computable group enumerated by g_{0}, g_{1}, … with g_{0} = 1_{G}. We build a computable binary branching tree T in stages. T_{n} denotes the part of T built at the end of stage n and includes all the nodes of T of length ≤ n. To each node σ ∈ T there is an associated finite set S_{σ}, which contains σ’s guess at a subset of a strict positive cone. Construction Stage 0: ⟨⟩ ∈ T_{0} and S_{⟨⟩} = ∅. Stage 1: ⟨1⟩ ∈ T_{1} and S_{⟨1⟩} = ∅. The purpose of this stage is to put a code for 1_{G} into every path but not to include 1_{G} in the strict positive cone. Stage n+1: Assume T_{n} is given and define T_{n+1} to include T_{n}. For each σ ∈ T_{n} with lh(σ) = n do the following:

1. If 1_{G} ∈ S_{σ}, then σ has no extensions in T_{n+1}.

2. Otherwise, both σ^{⌢}0 and σ^{⌢}1 are put into T_{n+1}.

We define the associated sets by:

S_{σ^{⌢}0} = S_{σ} ∪ {g_{n+1}^{−1}} ∪ {gh|g, h ∈ S_{σ}} ∪ {g_{i}^{−1}hg_{i}|i ≤ n + 1 ∧ h ∈ S_{σ}}

S_{σ^{⌢}1} = S_{σ} ∪ {g_{n+1}} ∪ {gh|g, h ∈ S_{σ}} ∪ {g_{i}^{−1}hg_{i}|i ≤ n + 1 ∧ h ∈ S_{σ}}.

End of Construction

Let C be the c.b. Π_{1}^{0} class of paths through T. For each P ∈ 𝕏(G), let f_{P} : ω → {0, 1} be the map that sends n to 0 if g_{n} ∉ P and sends n to 1 if g_{n} ∈ P. For each f ∈ C, let P_{f} be the set

P_{f} = { g ∈ G | f(g) = 1 } ∪ { g^{−1} | f(g) = 0 }.

We can now define the map φ : 𝕏 → C as the map that takes P to f_{P}. We need to verify that φ is a degree preserving bijection. Claim. For each f ∈ C, P_{f} ∈ 𝕏(G).

To prove this claim, we check the required algebraic properties of P_{f}. To see that P_{f} is full, notice that f(0) = 1, so 1_{G} ∈ P_{f}. For n > 0, either f(n) = 1, in which case n ∈ P_{f}, or f(n) = 0, in which case n^{−1} ∈ P_{f}. Therefore, P_{f} ∪ P_{f}^{−1} = G.

To show that P_{f} is a semigroup, we only need to show it is closed under multiplication. Assume g, h ∈ P_{f} and gh ∉ P_{f}. Since P_{f} is full, we know that h^{−1}g^{−1} ∈ P_{f}. Let n be the maximum of the indices for g, h and h^{−1}g^{−1}. By the construction we have the following:

g, h, h^{−1}g^{−1} ∈ S_{f|_{n}}
g, g^{−1} ∈ S_{f|_{n+1}}
1_{G} ∈ S_{f|_{n+2}}.

Therefore, f|_{n+2} has no extensions in T which contradicts the fact that f is a path through T. The proofs for normality and purity are similar. Claim. For each P ∈ 𝕏(G), f_{P} ∈ C.

We prove by induction that f_{P} |_{n} is on T for all n. The case for n = 0 follows from stage 1 of the construction because 1_{G} ∈ P, g_{0} = 1_{G} and ⟨1⟩ ∈ T. Also, notice that

S_{⟨1⟩} = ∅ ⊆ P^{+} = P \ {1_{G}}.

For the inductive step, assume that f_{P} |_{n} ∈ T and S_{f_{P} |_{n}} ⊆ P^{+}. Thus 1_{G} ∉ S_{f_{P} |_{n}} and so both f_{P} |_{n}^{⌢}0 and f_{P} |_{n}^{⌢}1 are in T. Without loss of generality, assume that f_{P} (n+1) = 1 and so f_{P} |_{n+1} = f_{P} |_{n}^{⌢}1. Since g_{n+1} ≠ 1_{G} and f_{P} (n+1) = 1, we know that g_{n+1} ∈ P^{+}. Also, since S_{f_{P} |_{n}} ⊆ P^{+} and P^{+} is closed under one multiplication and conjugation, we have S_{f_{P} |_{n+1}} ⊆ P^{+} as required.

To finish the proof, we make the following trivial observations:

1. If g ∈ C, then g = f_{P_{g}}.

2. If Q ∈ 𝕏(G), then Q = P_{f_{Q}}.

3. If P ≠ Q in 𝕏(G), then f_{P} ≠ f_{Q}.

4. If f ≠ g in C, then P_{f} ≠ P_{g}. These observations show that φ is a bijection and that φ^{−1} takes f ∈ C to P_{f} ∈ 𝕏(G). Finally, from the definitions it is clear that f_{P} ≤_{T} P for any P ∈ 𝕏(G) and P_{f} ≤_{T} f for any f ∈ C. Therefore, deg(P) = deg(f_{P}), so φ preserves Turing degrees.

Semigroup Conditions

4.1 Definition of the Conditions

In Chapter 3, we examined group conditions that imply full orderability. Semigroup conditions can also be analyzed to determine if a group is orderable. In this chapter, we study three theorems giving semigroup conditions. The versions stated in Kokorin and Kopytov (1974) are given below. In these theorems, S(a_{1}, …, a_{n}) denotes the normal semigroup generated by a_{1}, …, a_{n}. Theorem 4.1 (Fuchs (1958)). A partial order on G with positive cone P can be extended to a full order if and only if for any finite sequence of non-identity elements, a_{1}, …, a_{n} ∈ G, there is a sequence ϵ_{1}, …, ϵ_{n} with ϵ_{i} = ±1 such that

P ∩ S(a_{1}^{ϵ_{1}}, … a_{n}^{ϵ_{n}}) = ∅.

Theorem 4.2 (Łos (1954), Ohnishi (1952)). G is an O-group if and only if for any finite sequence of non-identity elements a_{1}, …, a_{n} there exists a sequence ϵ_{1}, …, ϵ_{n} such that

1_{G} ∉ S(a_{1}^{ϵ_{1}}, … a_{n}^{ϵ_{n}}).

Theorem 4.3 (Lorenzen (1949)). G is an O-group if and only if for any finite sequence of non-identity elements a_{1}, …, a_{n}

⋂ S(a_{1}^{ϵ_{1}}, … a_{n}^{ϵ_{n}}) = ∅

where the intersection extends over all sequences ϵ_{1}, …, ϵ_{n} with ϵ_{i} = ±1.

The first step in studying these theorems in reverse mathematics is to translate the semigroup conditions into the language of second order arithmetic. To do this, we need a definition for the normal semigroup generated by a finite number of elements. If A is the code of a finite sequence of elements of G, let S(A) be the normal semigroup generated by A. S(A) is built in stages with S(A) = ⋃_{n} S_{n}(A). The idea is to start with S_{0}(A) = A and at step n + 1, add the elements that can be formed by conjugating a member of S_{n}(A) or by multiplying two members of S_{n}(A).

S_{0}(A) = A
⋮
S_{n+1}(A) = S_{n}(A) ∪ {gag^{−1} | a ∈ S_{n}(A), g ∈ G} ∪ {ab | a, b ∈ S_{n}(A)}
⋮

It is clear that S(A) = ⋃_{n} S_{n}(A) is the desired semigroup. Formally, we define a function s such that:

x ∈ S_{n}(A) ↔ ∃m (s(A, n, m, x) = 1)

and for all n, m, x either s(A, n, m, x) = 0 or s(A, n, m, x) = 1. Define s by recursion on n with A and m as parameters.

s(A, 0, m, x) = \{ \begin{matrix}1 & if x ∈ A \\ 0 & otherwise \end{matrix}

s(A, n + 1, m, x) = \{ \begin{matrix}1 & if s(A, n, m, x) = 1 or \\ & ∃a, g ≤ m (s(A, n, m, a) = 1 ∧ x = gag^{−1}) or \\ & ∃a, b ≤ m (s(A, n, m, a) = s(A, n, m, b) = 1 ∧ \\ & ∧ x = ab) \\ 0 & otherwise \end{matrix}

Before proceeding with the main results of this chapter, we need to establish two preliminary facts. Definition 4.4. (RCA_{0}) If A is a code for a finite sequence of elements of G, let A^{−1} be the code for the finite sequence defined by A^{−1}(k) = A(k)^{−1} for 0 ≤ k ≤ lh(A). Lemma 4.5. (RCA_{0}) If A is a code for a finite sequence of elements of G and s(A, n, m, x) = 1, then ∃p (s(A^{−1}, n, p, x^{−1}) = 1). Proof. The proof is by Σ_{1}^{0} induction on n. Base case: Assume s(A, 0, m, x) = 1 and so x ∈ A. By definition, x^{−1} ∈ A^{−1} and s(A^{−1}, 0, m, x^{−1}) = 1. Induction case: Assume s(A, n + 1, m, x) = 1 and split into three subcases.

1. If s(A, n, m, x) = 1 then the induction hypothesis implies there is a p such that s(A^{−1}, n, p, x^{−1}) = 1 and hence s(A^{−1}, n + 1, p, x^{−1}) = 1.

2. There are g, a ≤ m with s(A, n, m, a) = 1 and x = gag^{−1}. By the induction hypothesis there is a p with s(A^{−1}, n, p, a^{−1}) = 1. Since x^{−1} = ga^{−1}g^{−1}, taking p̃ = max{p, g} gives s(A^{−1}, n + 1, p̃, x^{−1}) = 1.

3. There are a, b ≤ m with s(A, n, m, a) = s(A, n, m, b) = 1 and x = ab. By the induction hypothesis there are p_{1}, p_{2} such that s(A^{−1}, n, p_{1}, a^{−1}) = 1 and s(A^{−1}, n, p_{2}, b^{−1}) = 1. Let p = max{p_{1}, p_{2}}. Since x^{−1} = b^{−1}a^{−1}, if follows that s(A^{−1}, n + 1, p, x^{−1}) = 1.

Lemma 4.6. (RCA_{0}) Let P be the positive cone of a full order on G and A be a code for a finite sequence of nonidentity elements of P. If s(A, n, m, x) = 1 then x > 1_{G}.

Proof. The proof is by Σ_{1}^{0} induction on n. For the base case, assume s(A, 0, m, x) = 1. Since A ⊂ P and 1_{G} ∉ P, x > 1_{G}. For the induction case, use the same three subcases as in Lemma 4.5.

The next step is to write the semigroup conditions using s(A, n, m, x). Let Fin_{±1} be the set of codes for finite sequences of ±1. If A ∈ Fin_{G}, σ ∈ Fin_{±1}, and lh(A) = lh(σ), then let A_{σ} be the sequence of elements of G defined by:

lh(A_{σ}) = lh(A)
∀k < lh(A_{σ}) (A_{σ}(k) = A(k)^{σ(k)}).

For example, if A = ⟨1_{G}, a⟩ and σ = ⟨+1, −1⟩ then A_{σ} = ⟨1_{G}, a^{−1}⟩.

In the remaining equations in this section, it is assumed that A ranges over Fin_{G\1_{G}} and σ ranges over Fin_{±1}. The semigroup condition in Theorem 4.1 can be translated into the notation of s(A, n, m, x) by the following steps:

∀A ∃σ (lh(A) = lh(σ) ∧ P ∩ S(A_{σ}) = ∅)
∀A ∃σ ∀x (lh(A) = lh(σ) ∧ (x ∈ S(A_{σ}) → x ∉ P))
∀A ∃σ ∀x (lh(A) = lh(σ) ∧ (∃n, m s(A_{σ}, n, m, x) = 1 → x ∉ P))
∀A ∃σ ∀x (lh(A) = lh(σ) ∧ (∀n, m s(A_{σ}, n, m, x) = 0 ∨ x ∉ P)).

Since ∃σ ∈ Fin_{±1}(lh(σ) = lh(A)) is really a bounded quantifier, this condition is Π_{1}^{0}.

The semigroup condition in Theorem 4.2 can be translated as follows:

∀A ∃σ (lh(A) = lh(σ) ∧ 1_{G} ∉ S(A_{σ}))
∀A ∃σ (lh(A) = lh(σ) ∧ ∀n, m s(A_{σ}, n, m, 1_{G}) = 0).

Again, because ∃σ ∈ Fin_{±1}(lh(σ) = lh(A)) is a bounded quantifier, this condition is Π_{1}^{0}.

The semigroup condition in Theorem 4.3 can be written as follows:

∀A ∀x ∃σ (lh(A) = lh(σ) ∧ x ∉ S(A_{σ}))
∀A ∀x ∃σ (lh(A) = lh(σ) ∧ ∀m, n s(A_{σ}, n, m, x) = 0).

This sentence is also Π_{1}^{0}.

Theorems 4.1, 4.2 and 4.3 can now be stated in the language of second order arithmetic.

Theorem 4.7. (W KL_{0}) A partial order on G with positive cone P can be extended to a full order if and only if

∀A ∃σ ∀x, n, m (lh(A) = lh(σ) ∧ (s(A_{σ}, n, m, x) = 0 ∨ x ∉ P)).  (4.1)

Theorem 4.8. (W KL_{0}) G is an O-group if and only if

∀A ∃σ ∀n, m (lh(A) = lh(σ) ∧ s(A_{σ}, n, m, 1_{G}) = 0).  (4.2)

Theorem 4.9. (W KL_{0}) G is an O-group if and only if

∀A ∀x ∃σ ∀m, n (lh(A) = lh(σ) ∧ s(A_{σ}, n, m, x) = 0).  (4.3)

There are several connections between these theorems. G is an O-group if and only if the trivial order with positive cone P = {1_{G}} can be extended to a full order. By Theorem 4.7, this condition is equivalent to:

∀A∃σ∀x, n, m (lh(A) = lh(σ) ∧ (s(A_{σ}, n, m, x) = 0 ∨ x ≠ 1_{G}))

which in turn is equivalent to Equation (4.2). Hence, RCA_{0} proves that Theorem 4.8 is a special case of Theorem 4.7. Setting x = 1_{G} shows that (4.3) implies (4.2).

Showing that Equation (4.2) implies Equation (4.3) requires more work. For σ ∈ Fin_{±1}, let σ^{−1} have the same length as σ with σ^{−1}(k) = −σ(k). Notice that A_{σ^{−1}} = A_{σ}^{−1} and (A_{σ^{−1}})^{−1} = A_{σ}. For a contradiction, suppose that (4.2) holds and (4.3) does not. Because (4.3) fails, there are A and x such that

∀σ ∈ Fin_{±1}∃m, n (lh(σ) = lh(A) → s(A_{σ}, n, m, x) = 1).  (4.4)

Fix A and x. Because (4.2) holds, there is a σ such that

∀n, m (s(A_{σ}, n, m, 1_{G}) = 0).  (4.5)

Fix σ. Applying (4.4) with σ^{−1}, we have s(A_{σ^{−1}}, n, m, x) = 1 for some m, n and hence by Lemma 4.5, s(A_{σ}, n, p, x^{−1}) = 1. Applying (4.4) with σ we have s(A_{σ}, ñ, m̃, x) = 1 for some m̃, ñ. Without loss of generality, assume n ≥ ñ. By definition, s(A_{σ}, n, m̃, x) = 1 and so if k > n, m̃ then s(A_{σ}, n, k, 1_{G}) = 1. This fact contradicts (4.5).

4.2 Equivalence with W KL_{0}

The goal for the rest of this chapter is to prove the following theorem. Theorem 4.10. (RCA_{0}) The following are equivalent:

1. W KL_{0}

2. Theorem 4.7

3. Theorem 4.8

4. Theorem 4.9 By the comments at the end of the last section, we know that (2) implies (3)

and that (3) and (4) are equivalent. In this section, we will show that (1) implies (2) and that (3) implies (1). Proposition 4.11. (RCA_{0}) If a partial order on G with positive cone P can be extended to a full order, then Equation (4.1) holds for P.

Proof. Assume Q is the positive cone of a full order extending P. Given any A ∈ Fin_{G\1_{G}}, let σ ∈ Fin_{±1} be such that lh(σ) = lh(A) and for every k < lh(σ), A(k)^{−σ(k)} ∈ Q. For a contradiction, assume for some x, n, m we have

s(A_{σ}, n, m, x) = 1 ∧ x ∈ P.

Because P ⊆ Q, we have that x ∈ Q. Applying Lemma 4.5 to s(A_{σ}, n, m, x) = 1, we have s(A_{σ^{−1}}, n, p, x^{−1}) = 1 for some p. However by our choice of σ, A_{σ^{−1}} must be contained in Q \ 1_{G} and hence x^{−1} > 1_{G} by Lemma 4.6. Thus x, x^{−1} ∈ Q and so x = 1_{G}. This conclusion contradicts x^{−1} > 1_{G}.

Proposition 4.12. (W KL_{0}) If P ⊂ G and Equation (4.1) holds for P then P can be extended to the positive cone of a full order on G.

Proof. This proof is similar to the proof of Theorem 3.23. Suppose G is enumerated as g_{0}, g_{1}, . . .. We build a binary branching tree T which codes the positive cone of a full order along every path. Equation (4.1) will imply that T is infinite and so W KL_{0} guarantees that it has a path. To simplify the notation we construct T ⊆ Fin_{±1} instead of T ⊆ Fin_{{0,1}}. For each σ ∈ T with lh(σ) = k, let Q_{σ} ∈ Fin_{G\1_{G}} be

Q_{σ} = ⟨g_{1}^{σ(1)}, …, g_{k−1}^{σ(k−1)}⟩.

For example, if σ = ⟨+1, −1, −1⟩ then Q_{σ} = ⟨g_{1}^{−1}, g_{2}^{−1}⟩. The reason for not including g_{0} in Q_{σ} is so that 1_{G} ∉ Q_{σ}. Q_{σ} represents σ’s guess as a subset of a strict positive cone extending P.

As in the earlier constructions, T is built in stages and after stage k, no new nodes of length k enter T. T_{k} denotes the nodes of T after stage k.

Construction Stage 0: Set T_{0} = {⟨⟩} and Q_{⟨⟩} = ⟨⟩. Stage 1: Set T_{1} = {⟨⟩, ⟨−1⟩} and Q_{⟨−1⟩} = ⟨⟩. The purpose of this stage is to code 1_{G} into every path. Stage s = k+1: For each σ ∈ T_{k} check if Equation (4.1) has been violated by a number below k:

∃x, n, m ≤ k (s(Q_{σ}, n, m, x) = 1 ∧ x ∈ P )

There are two possible answers to this question.

1. If YES: Equation (4.1) has been violated, so neither σ^{⌢}⟨−1⟩ nor σ^{⌢}⟨+1⟩ enters T_{k+1}.

2. If NO: Extend σ by putting both σ^{⌢}⟨−1⟩ and σ^{⌢}⟨+1⟩ in T_{k+1}.

End of Construction

We need to verify various properties of the construction. Let [k] = ⟨g_{1}, …, g_{k−1}⟩ and [k]_{σ} = ⟨g_{1}^{σ(1)}, …, g_{k−1}^{σ(k−1)}⟩. Lemma 4.13. (RCA_{0}) T is infinite.

Proof. It suffices to show that for each k there is an element of T of length k. Fix k > 0. Since P satisfies Equation (4.1), there is a σ ∈ Fin_{±1} with lh(σ) = k and

∀x, n, m (s([k]_{σ}, n, m, x) = 0 ∨ x ∉ P ).

In particular, this condition holds if we bound the quantifiers by k. From the definition of T, it follows that for all i ≤ k, σ|_{i} ∈ T and hence σ ∈ T.

By Weak König’s Lemma there is a path h through T. Let

h[n] = ⟨h(0), …, h(n − 1)⟩ ∈ Fin_{±1}
h̃[n] = [n]_{h[n]} = ⟨g_{1}^{h(1)}, …, g_{n−1}^{h(n−1)}⟩.

Lemma 4.14. (RCA_{0}) For any x ∈ G \ 1_{G}, h(x) = 1 ↔ h(x^{−1}) = −1.

Proof. If h(x) = h(x^{−1}) = 1 and k is the maximum of the indices for x and x^{−1} and the values of x and x^{−1} as natural numbers, then x, x^{−1} ∈ h̃[k + 1]. Since s(h̃[k + 1], 0, 0, x) = 1 and s(h̃[k + 1], 0, 0, x^{−1}) = 1, it follows that s(h̃[k + 1], 1, k, 1_{G}) = 1. But, 1_{G} ∈ P and so by the construction of T, h[k + 1] has no extensions. This statement contradicts the choice of h as a path. The case for h(x) = h(x^{−1}) = −1 is similar.

We are now in a position to define Q and verify that it is a full order extending P.

x ∈ Q ↔ h(x) = −1

Q exists by Δ_{1}^{0} comprehension. It contains 1_{G} since the only node of length 1 in T is ⟨−1⟩ and it is both full and pure by Lemma 4.14. Claim. P ⊂ Q

Suppose g_{i} ∈ P \ 1_{G} and h(g_{i}) = 1. By definition, g_{i} ∈ h̃[i + 1] and so s(h̃[i + 1], 0, 0, g_{i}) = 1. As in Lemma 4.14, s(h̃[i + 1], 0, 0, g_{i}) = 1 and g_{i} ∈ P contradicts the fact that h is a path. Claim. Q is closed under multiplication.

Suppose that a, b ∈ Q and ab ∉ Q. From Lemma 4.14 and the definition of Q, it follows that h(a^{−1}) = 1, h(b^{−1}) = 1 and h(ab) = 1. For a large enough k, we have a^{−1}, b^{−1}, ab ∈ h̃[k] and hence if m is the maximum of the values of a^{−1}, b^{−1}, ab, b^{−1} and a^{−1} as natural numbers and their indices as elements of G, then s(h̃[k], 2, m, 1_{G}) = 1. Since 1_{G} ∈ P, this statement contradicts the fact that h is a path. Claim. Q is normal.

Suppose q ∈ Q, g ∈ G and gqg^{−1} ∉ Q. As above, h(q^{−1}) = 1, h(gqg^{−1}) = 1 and there is a k with q^{−1}, gqg^{−1} ∈ h̃[k]. There is an m such that s(h̃[k], 2, m, 1_{G}) since the definition of s yields the normal semigroup. As above, h[k] cannot be on a path. This claim completes the proof that Q is a full order extending P.

Together Propositions 4.11 and 4.12 show (1) implies (2) in Theorem 4.10. The next step is to show that (3) implies (1) in the theorem.

Proposition 4.15. (RCA_{0}) For an abelian group G the following are equivalent:

1. Equation (4.2) holds.

2. G is torsion free.

Proof. Case. (1) ⇒ (2) :

For a contradiction assume that Equation (4.2) holds and a ≠ 1_{G} is a torsion element of G. Claim. For all k ≥ 1, ∃p[s(⟨a⟩, k − 1, p, a^{k}) = 1].

The claim is proved by Σ_{1}^{0} induction on k. If k = 1, then a ∈ ⟨a⟩ implies s(⟨a⟩, 0, 0, a) = 1. For k + 1, the induction hypothesis states that there are p, p′ such that s(⟨a⟩, k − 1, p, a^{k}) = 1 and s(⟨a⟩, k − 1, p′, a) = 1. If p′′ = max{p, p′} then s(⟨a⟩, k, p′′, a^{k+1}) = 1, which proves the claim.

If a is a torsion element then for some k, a^{k} = (a^{−1})^{k} = 1_{G}. Equation (4.2) for the sequence ⟨a⟩ says that either

∀n, m (s(⟨a⟩, n, m, 1_{G}) = 0)
or ∀n, m (s(⟨a^{−1}⟩, n, m, 1_{G}) = 0).

But the claim implies that

∃p (s(⟨a⟩, k − 1, p, 1_{G}) = s(⟨a⟩, k − 1, p, a^{k}) = 1)
∃p (s(⟨a^{−1}⟩, k − 1, p, 1_{G}) = s(⟨a^{−1}⟩, k − 1, p, (a^{−1})^{k}) = 1).

Case. (2) ⇒ (1)

The first step of this direction is to show that for an abelian group G the normal semigroup generated by A ∈ Fin_{G} is the same as the semigroup generated by A. That is, if A = ⟨a_{1}, … a_{n}⟩ then any element of S(A) can be written as a_{1}^{k_{1}} ⋯ a_{n}^{k_{n}} for some choice of k_{1}, … k_{n} ∈ ℕ with at least one k_{i} > 0. Informally this statement is clear because any subset of an abelian group is normal. However, we need the formal fact that every element can be written in this form. We define a function prod(A, σ) that takes A = ⟨a_{1}, …, a_{n}⟩ and σ = ⟨σ_{1}, …, σ_{n}⟩ to a_{1}^{σ_{1}} ⋯ a_{n}^{σ_{n}}. For A ∈ Fin_{A} and σ ∈ Fin_{ℕ} with lh(A) = lh(σ), define prod(A, σ) by recursion on lh(A): prod(A, σ) = 0 if lh(A) = 0, prod(A, σ) = A(0)^{σ(0)} if lh(A) = 1 and if lh(A) > 1 then

prod(A, σ) = \{ \begin{matrix}prod(A′, σ′) · prod(A′′, σ′′) & if lh(σ) = lh(A) \\ 0 & if lh(σ) ≠ lh(A) \end{matrix}

where

lh(A′) = lh(σ′) = lh(A) − 1
lh(A′′) = lh(σ′′) = 1
∀k < lh(A′)[A′(k) = A(k) ∧ σ′(k) = σ(k)]
[A′′(0) = A(lh(A) − 1)] ∧ [σ′′(0) = σ(lh(σ) − 1)].

There are two lemmas to prove about this formal notation. Lemma 4.16. (RCA_{0}) If A ∈ Fin_{G}, σ, τ ∈ Fin_{ℕ} and lh(A) = lh(σ) = lh(τ) then

prod(A, σ) · prod(A, τ) = prod(A, σ + τ)

where σ + τ ∈ Fin_{ℕ} is defined by (σ + τ)(k) = σ(k) + τ(k). Proof. This lemma is proved by induction on lh(A). If lh(A) = 1:

prod(A, σ) · prod(A, τ) = A(0)^{σ(0)}A(0)^{τ(0)}
= A(0)^{σ(0)+τ(0)}
= prod(A, σ + τ).

If lh(A) > 1 then we rewrite prod(A, σ) · prod(A, τ) as:

prod(A′, σ′) · prod(A′′, σ′′) · prod(A′, τ′) · prod(A′′, τ′′)
= prod(A′, σ′) · prod(A′, τ′) · prod(A′′, σ′′) · prod(A′′, τ′′)
= prod(A′, σ′ + τ′) · prod(A′′, σ′′ + τ′′)
= prod(A, σ + τ).

The second line uses the fact that G is abelian and the third line uses the induction hypothesis.

Lemma 4.17. If A ∈ Fin_{G}, n ∈ ℕ, x ∈ G and ∃m [s(A, n, m, x) = 1] then there is a σ ∈ Fin_{ℕ} with lh(σ) = lh(A) and at least one k < lh(σ) with σ(k) > 0 such that x = prod(A, σ).

Proof. This claim is proved by induction on n. For the base case, assume s(A, 0, m, x) = 1. Then x = A(j) for some j. Define σ by σ(j) = 1 and σ(i) = 0 for i ≠ j. For the induction case, assume s(A, n + 1, m, x) = 1 and split into three subcases.

1. If s(A, n, m, x) = 1, then we are done by the induction hypothesis.

2. If x = ab with s(A, n, m, a) = s(A, n, m, b) = 1, then by the induction hypothesis, a = prod(A, σ) and b = prod(A, τ). By Lemma 4.16, x = prod(A, σ + τ).

3. If x = gag^{−1} with s(A, n, m, a) = 1 and g ≤ m, then since G is abelian, x = a. Hence s(A, n, m, x) = 1 and the induction hypothesis applies.

We can now prove that if G is torsion free abelian then Equation (4.2) holds by Π_{1}^{0} induction on lh(A). Base case: We need to show that for each a ∈ G \ 1_{G} either

∀n, m (s(⟨a⟩, n, m, 1_{G}) = 0)
or ∀n, m (s(⟨a^{−1}⟩, n, m, 1_{G}) = 0).

Suppose that neither equation holds and that s(⟨a⟩, n, m, 1_{G}) = 1. By Lemma 4.17, 1_{G} = prod(⟨a⟩, σ) for some σ and 1_{G} = a^{σ(0)} by the definition of prod. Therefore a is a torsion element which contradicts the fact that G is torsion free. Induction step: This case will be presented less formally to avoid an undue amount of notational baggage. Assume Equation (4.2) holds for ⟨a_{1}, …, a_{n}⟩ and fails for ⟨a_{1}, …, a_{n}, b⟩. Let ⟨ϵ_{1}, …, ϵ_{n}⟩ be the exponents in Equation (4.2) for ⟨a_{1}, …, a_{n}⟩. By assumption, there are n_{1}, m_{1}, n_{2}, m_{2} such that

s(⟨a_{1}^{ϵ_{1}}, …, a_{n}^{ϵ_{n}}, b⟩, n_{1}, m_{1}, 1_{G}) = 1
s(⟨a_{1}^{ϵ_{1}}, …, a_{n}^{ϵ_{n}}, b^{−1}⟩, n_{2}, m_{2}, 1_{G}) = 1.

By Lemma 4.17, there are k_{1}, …, k_{n+1} and l_{1}, …, l_{n+1} such that

a_{1}^{ϵ_{1}k_{1}} ⋯ a_{n}^{ϵ_{n}k_{n}}b^{k_{n+1}} = 1_{G}
a_{1}^{ϵ_{1}l_{1}} ⋯ a_{n}^{ϵ_{n}l_{n}}b^{−l_{n+1}} = 1_{G}

which gives

a_{1}^{ϵ_{1}(k_{1}l_{n+1}+k_{n+1}l_{1})} ⋯ a_{n}^{ϵ_{n}(k_{n}l_{n+1}+k_{n+1}l_{n})} = 1_{G}.

This equation contradicts Equation (4.2) for ⟨a_{1}, …, a_{n}⟩.

Proposition 4.15 shows that statement (3) in Theorem 4.10 implies that every torsion free abelian group is an O–group. By Theorem 3.3, this statement implies W KL_{0}. We have now completed the proof of Theorem 4.10.

Free Groups

There is a well known result in group theory that every group can be written as a quotient of a free group. A similar result holds for ordered groups. For every f.o. group G, there is an f.o. free group F and a convex normal subgroup N ⊆ F such that G ≅ F/N by an order preserving isomorphism. It is assumed that F/N has the induced quotient order. Surprisingly, this theorem is provable in RCA_{0} and that proof is the main result of this chapter. This result answers an open question from Downey and Kurtz (1986) by showing that the effective version of the theorem holds. There is one technical lemma required to prove the main theorem. We need to know that RCA_{0} suffices to show that the free group on two generators is an O-group. In the first section, we assume this fact and show that RCA_{0} proves the main theorem. The last two sections are devoted to proving the technical lemma. In Section 2, we introduce ordered rings and a special class of matrices. In Section 3, we use this class of matrices to order the free product of O-groups. As a corollary, we prove that the free group on two generators is an O-group.

5.1 Main Result

The formal definitions and properties of free groups and free products are given in Appendix A. There we show that RCA_{0} is strong enough to prove the existence of the set of reduced words and define a group structure on them. In this chapter, we will use the standard mathematical notation for free groups. In general, if X = {x_{0}, x_{1}, … } is the set of generators of a free group, then Word_{X} denotes the set of words in X and the elements of the group are written as x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} with n_{i} ∈ ℤ \ 0. Here, if n_{i} > 0, then x_{i}^{n_{i}} refers to x_{i}^{1} repeated n_{i} times. If n_{i} < 0, then x_{i}^{n_{i}} refers to x_{i}^{−1} repeated |n_{i}| times. Using this notation, we prove that every group is isomorphic to the quotient of a free group by a normal subgroup.

Theorem 5.1. (RCA_{0}) Every group is the epimorphic image of a free group.

Proof. Let F be the free group on the generators X = {x_{0}, x_{1}, … } and suppose G is enumerated as g_{0}, g_{1}, . . .. Define a map ψ : X → G by x_{n} ↦ g_{n} and extend ψ to a map from Word_{X} → G by sending x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} ↦ g_{i_{1}}^{n_{1}} ⋯ g_{i_{k}}^{n_{k}}. This map respects the equivalence of words and hence restricts to an epimorphism ψ̃ : F → G. This argument can be made formal by using notions similar to the Prod notation from Chapter 4.

If N is the kernel of ψ̃ then G ≅ F/N. The main technical facts needed to prove the version of this theorem for f.o. groups are stated as Theorem 5.2 and Corollary 5.3. The proofs of these results will be given in Section 3 of this chapter. Using these tools, we can prove that the free group on a countable number of generators is an O-group. ℕ^{+} denotes the set of strictly positive natural numbers. Theorem 5.2. (RCA_{0}) The free product of two O-groups is an O-group. Corollary 5.3. (RCA_{0}) The free group on two generators is an O-group. Lemma 5.4. (RCA_{0}) Let F be the free group on the two generators x, y. For each i ∈ ℕ^{+} let α_{i} = x^{i}y^{i}.

1. The word α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} with i_{j+1} ≠ i_{j}, n_{j} ∈ ℤ \ {0}, and k > 0 freely reduces to a word ending in x^{ϵ}y^{i_{k}} if n_{k} > 0 and y^{ϵ}x^{−i_{k}} if n_{k} < 0.

2. No product α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} with the above restrictions is the identity element. Proof. Assuming the first property holds, there is either an x or a y with a nonzero exponent in the reduced form of α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}}. Therefore, the second property follows immediately from the first.

The first property is proved by induction on k. If k = 1 then

α_{i_{1}}^{n_{1}} = (x^{i_{1}}y^{i_{1}})^{n_{1}}

which satisfies the first property. If k > 1, then split into four cases depending on the signs of n_{k} and n_{k−1}. Case. n_{k} > 0 and n_{k−1} > 0

By the induction hypothesis, α_{i_{1}}^{n_{1}} ⋯ α_{i_{k−1}}^{n_{k−1}} reduces to a word ending in x^{ϵ}y^{i_{k−1}}. Thus α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} reduces to a word ending in x^{ϵ}y^{i_{k−1}}(x^{i_{k}}y^{i_{k}})^{n_{k}}. Since n_{k} > 0, this word ends in x^{i_{k}}y^{i_{k}}.

Case. n_{k} > 0 and n_{k−1} < 0

By the induction hypothesis, α_{i_{1}}^{n_{1}} ⋯ α_{i_{k−1}}^{n_{k−1}} reduces to a word ending in y^{ϵ}x^{−i_{k−1}}. Hence α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} reduces to a word ending in y^{ϵ}x^{−i_{k−1}}(x^{i_{k}}y^{i_{k}})^{n_{k}}. If n_{k} = 1 then we have a word ending in y^{ϵ}x^{i_{k}−i_{k−1}}y^{i_{k}}. By assumption, i_{k} − i_{k−1} ≠ 0 so we have satisfied the first property. If n_{k} > 1, then this word ends in x^{i_{k}}y^{i_{k}}. Case. n_{k} < 0 and n_{k−1} < 0

This case is similar to the first case. Case. n_{k} < 0 and n_{k−1} > 0

This case is similar to the second case.

Proposition 5.5. (RCA_{0}) The free group on a countable number of generators is an O-group.

Proof. Let F and α_{i} be as in Lemma 5.4 and let P (F) be the positive cone for some full order on F. Let G be the free group on the generators x_{0}, x_{1}, . . .. Define the homomorphism:

ψ : G → F
x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} ↦ α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}}.

If x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} is fully reduced and x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} ≠ 1_{G}, then α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} satisfies the hypotheses of the previous lemma. Hence α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} ≠ 1_{F} and so ψ is a monomorphism. The order on G is defined from P (F).

P (G) = { x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} | α_{i_{1}}^{n_{1}} ⋯ α_{i_{k}}^{n_{k}} ∈ P (F) }

It is straightforward to verify that P (G) is the positive cone for a full order on G.

Definition 5.6. (RCA_{0}) If G_{1} and G_{2} are p.o. groups, then a map ψ : G_{1} → G_{2} is called an o-homomorphism if ψ is an order preserving homomorphism. If ψ is onto, then ψ is called an o-epimorphism.

We can now prove the main result of the chapter. Standard proofs of this result can be found in Fuchs (1963) or Kokorin and Kopytov (1974). These proofs cannot be done in RCA_{0}. The proof used here comes from Revesz (1986).

Theorem 5.7. (RCA_{0}) Any fully ordered group is the o-epimorphic image of a fully ordered free group.

Proof. Let G be an f.o. group, P (G) be the positive cone of a full order on G, and g_{0}, g_{1}, … be an enumeration of G. Let F be the free group on the generators x_{0}, x_{1}, … and P (F) be the positive cone of some full order on F. As in Theorem 5.1, define the epimorphism:

φ : F → G
x_{i_{1}}^{n_{1}} ⋯ x_{i_{k}}^{n_{k}} ↦ g_{i_{1}}^{n_{1}} ⋯ g_{i_{k}}^{n_{k}}.

We need to produce a new order P̃(F) on F under which φ is order preserving. Define the embedding:

ψ : F → G × F
a ↦ ⟨φ(a), a⟩.

Order G × F lexicographically:

⟨a, b⟩ ∈ P (G × F) ↔ (a ∈ P (G) ∧ a ≠ 1_{G}) ∨ (a = 1_{G} ∧ b ∈ P (F)). As in the proofs on direct products from Chapter 3, this set gives a full order on G × F. Since the map ψ : F → G × F is a monomorphism, we can use it to define a new order on F:

P̃(F) = {a | ψ(a) ∈ P (G × F)}.

All that is left to show is that φ is order preserving under P̃(F). Rewriting the definition of P̃(F) we have that:

a ∈ P̃(F) ↔ (φ(a) ∈ P (G) ∧ φ(a) ≠ 1_{G}) ∨ (φ(a) = 1_{G} ∧ a ∈ P (F)). Let ≤_{F} be the order corresponding to P̃(F). Suppose that a ≤_{F} b. It follows that a^{−1}b ∈ P̃(F) and hence φ(a^{−1}b) = φ(a)^{−1}φ(b) ∈ P (G). This calculation shows that φ(a) ≤_{G} φ(b). Now, suppose that c, d ∈ G and c <_{G} d. Since φ is onto, there are a, b ∈ F with φ(a) = c and φ(b) = d. Since c <_{G} d, we have that c^{−1}d ∈ P (G) and c^{−1}d ≠ 1_{G}. Because φ(a^{−1}b) = c^{−1}d and c^{−1}d ≠ 1_{G}, we know that a ≠ b. By the definition of P̃(F), we have that a^{−1}b ∈ P̃(F) and so a <_{F} b. Therefore, φ is an o-epimorphism from F with the order P̃(F) onto G.

5.2 Fully Ordered Rings and Triangular Matrices

Definition 5.8. (RCA_{0}) A ring is a set R together with two functions +_{R}, ·_{R} and two constants 0_{R}, 1_{R} which satisfy the usual axioms for a commutative ring with identity.

As with groups, the subscripts will be dropped when the context is clear. Notice that we are using the term ring to mean a commutative ring with identity.

Definition 5.9. (RCA_{0}) A partially ordered ring (p.o. ring) is a ring R together with a binary relation ≤_{R} such that:

1. (R, ≤_{R}) is a partial order.

2. a ≤_{R} b implies a + c ≤_{R} b + c for all a, b, c ∈ R.

3. a ≤_{R} b and c >_{R} 0 implies ca ≤_{R} cb and ac ≤_{R} bc.

If ≤_{R} is linear, then (R, ≤_{R}) is a fully ordered ring (f.o. ring).

As with a p.o. group, we define the positive and negative cones of a p.o. ring:

P = {r ∈ R | r ≥ 0}
−P = {−r | r ∈ P }.

We can verify properties of P similar to those of the positive cone of a p.o. group. For example, P ∩ −P = {0} and if P is the positive cone for a full order, then P ∪ −P = R. These are two of the four required properties for a set P to be the positive cone of some full order on R. The other two defining properties are that P + P ⊆ P and P P ⊆ P. RCA_{0} is not strong enough to prove that the sets P + P and P P exist. However, RCA_{0} is strong enough to show

∀x, y ∈ R (x ∈ P ∧ y ∈ P → x + y ∈ P)
∀x, y ∈ R (x ∈ P ∧ y ∈ P → xy ∈ P).

In the context of RCA_{0}, we take P + P ⊆ P to stand for the top formula and P P ⊆ P to stand for the bottom formula. With this convention, we can state the next theorem in its standard notation.

Theorem 5.10. (RCA_{0}) A subset P of a ring R is the positive cone of some partial order on R if and only if the following conditions are satisfied:

1. P ∩ −P = {0}

2. P + P ⊆ P

3. P P ⊆ P

Furthermore, P is the positive cone of some full order on R if and only if in addition P satisfies P ∪ −P = R.

Proof. We have already mentioned that any positive cone satisfies these requirements. Conversely, if P is a set with these properties, then the order can be defined by:

a ≤ b ↔ b − a ∈ P.

RCA_{0} suffices to verify that this gives an order on R.

To prove that the free product of two O-groups is an O-group, we will embed the free product into a group of infinite matrices over an f.o. ring. Hence, we need to develop the definitions and the tools to handle such matrices. Given a f.o. ring K, we are interested in upper triangular matrices whose rows and columns are indexed by the elements of ℕ^{+}. Furthermore, we want the elements along the main diagonal to be positive and invertible. Such a matrix resembles:

( \begin{matrix}k_{11} & k_{12} & k_{13} & k_{14} & … \\ 0 & k_{22} & k_{23} & k_{24} & … \\ 0 & 0 & k_{33} & k_{34} & … \\ ⋮ & ⋮ & ⋮ & ... & ... \end{matrix} )

where each k_{ii} is in the positive cone of K and has a multiplicative inverse. Obviously there are an uncountable number of such matrices and hence in second order arithmetic, the best we can do is to represent them as a class of functions. Definition 5.11. (RCA_{0}) Let (K, ≤) be a fully ordered ring with positive cone P. The function f : ℕ^{+} × ℕ^{+} → K is in the class Tri_{K} if and only if it satisfies the following conditions:

1. For all i > j, f(i, j) = 0_{K}.

2. For all i, f(i, i) ∈ P and ∃x ∈ K(f(i, i) · x = 1_{K}).

The first of these conditions says that the matrices are upper triangular and the second says that the entries along the main diagonal are positive and invertible. We will use f ∈ Tri_{K} as shorthand to mean that f is a function that satisfies these two conditions.

Since our goal is to use this matrix group to order free products, we need to define both an order and a group structure on Tri_{K}. First we define the order. Given f, g ∈ Tri_{K}, we say that f < g if and only if for some pair ⟨i, j⟩ ∈ ℕ^{+} × ℕ^{+} with i ≤ j the following two conditions hold:

1. f(i, j) <_{K} g(i, j).

2. f(k, k + s) = g(k, k + s) for all k, s such that i + s < j or i + s = j and k < i.

A pair ⟨i, j⟩ for which these conditions hold is called a witness for f < g. These conditions are much easier to understand if f and g are viewed as matrices as opposed to as functions. They mean that we compare f and g down the diagonals, starting with the main diagonal, then the diagonal to its right, and so on, until we find the first place that f and g differ. The entries of f and g are compared in the order indicated in this picture:

( \begin{matrix}1 & ω & ω + ω & ω + ω + ω & … \\ · & 2 & ω + 1 & ω + ω + 1 & ... \\ · & · & 3 & ω + 2 & ... \\ · & · & · & 4 & ... \\ ⋮ & ⋮ & ⋮ & ⋮ & ... \end{matrix} )

From this picture, it is clear that we are comparing f and g by comparing two ordered sequences of elements of K each with order type ω^{ω}. If f ≠ g, then the relationship between f and g is determined by the relationship between the elements of K at the first place where these ordered sequences differ. Unfortunately, RCA_{0} is not strong enough to prove that ω^{ω} is well ordered, and so it cannot prove that there is a least place where they differ. In the next section, we will define a countable subgroup of Tri_{K} for which the order is much easier to determine. For now, our goal is to define the group structure and to prove that the elements of Tri_{K} satisfy the axioms for a partially ordered group with this order.

Given f, g ∈ Tri_{K}, we define the product f · g to be the function:

f · g : ℕ^{+} × ℕ^{+} → K

f · g(i, j) = ∑_{n=0}^{∞} f(i, n)g(n, j).

This definition exactly matches the definition for multiplication of infinite matrices. The first thing to check is that this sum converges. By definition, f(i, n) = 0 for n < i and g(n, j) = 0 for n > j. Hence, if n is not between i and j, then f(i, n)g(n, j) = 0. This has two consequences. First, if i > j, then the sum is 0. Second, if i ≤ j, then the infinite sum reduces to the finite sum:

∑_{n=i}^{j} f(i, n)g(n, j).

Thus, RCA_{0} proves that f · g is a well defined function. Furthermore, f · g(i, i) = f(i, i)g(i, i) and so f · g(i, i) is both positive and invertible. Hence, f · g is in Tri_{K}. The matrix I ∈ Tri_{K} defined by I(i, i) = 1_{K} and I(i, j) = 0_{K} for i ≠ j plays the role of the identity element in Tri_{K}.

The next two lemmas show that RCA_{0} proves the associativity of the multiplication and the existence of inverses. We prove Lemma 5.12 to give an example of how to work in this formalism, but the proof of Lemma 5.13 is presented in Appendix A.

Lemma 5.12. (RCA_{0}) (f · g) · h = f · (g · h)

Proof. For j < i, both of these products have the value 0. For i ≤ j, we perform two computations.

((f · g) · h)(i, j) = ∑_{m=i}^{j}(f · g)(i, m) h(m, j)

= ∑_{m=i}^{j} ∑_{n=i}^{m} f(i, n)g(n, m)h(m, j)

= ∑_{i≤n≤m≤j} f(i, n)g(n, m)h(m, j)

(f · (g · h))(i, j) = ∑_{n=i}^{j} f(i, n) (g · h)(n, j)

= ∑_{n=i}^{j} ∑_{m=n}^{j} f(i, n)g(n, m)h(m, j)

= ∑_{i≤n≤m≤j} f(i, n)g(n, m)h(m, j)

Lemma 5.13. (RCA_{0}) If f ∈ Tri_{K}, then f has an inverse g ∈ Tri_{K}, in the sense that f · g = g · f = I, given by:

g(i, j) = \{ \begin{matrix}0 & j < i \\ f(i, j)^{−1} & i = j \\ −\frac{f(i,j)}{f(i,i)f(j,j)} + ∑_{i<k_{1}<j} \frac{f(i,k_{1})f(k_{1},j)}{f(i,i)f(k_{1},k_{1}),f(j,j)} − \\ − ∑_{i<k_{1}<k_{2}<j} \frac{f(i,k_{1})f(k_{1},k_{2})f(k_{2},j)}{f(i,i)f(k_{1},k_{1})f(k_{2},k_{2})f(j,j)} + ⋯ & i < j \\ ⋯ + (−1)^{j−i} \frac{f(i,i+1)···f(j−1,j)}{f(i,i)f(i+1,i+1)···f(j,j)} \end{matrix}

Since f(n, n) is invertible, we write it in the denominator of a fraction as shorthand for f(n, n)^{−1}.

Now that we have both a group structure and an order on Tri_{K}, we need to check that they interact as in an ordered group. Instead of verifying directly that the functions in Tri_{K} satisfy properties similar to the axioms for a partially ordered group, we give a condition for elements of Tri_{K} to be in the positive cone and verify that these functions satisfy the appropriate properties. If f ∈ Tri_{K}, we say f ∈ P (Tri_{K}) if and only if f = I or I < f in the order given above. If f ≠ I this is equivalent to either

∃i [f(i, i) > 1 ∧ ∀j < i(f(j, j) = 1)]

or

∀i (f(i, i) = 1) ∧ ∃i, j ( i < j ∧ f(i, j) > 0 ∧
∧ ∀k ∀s > 0 ((i + s < j ∨ (i + s = j ∧ k < i)) → f(k, k + s) = 0) ).

Lemma 5.14. (RCA_{0})

1. If f, g ∈ P (Tri_{K}) then f · g ∈ P (Tri_{K})

2. If f ∈ P (Tri_{K}) and f ≠ I then f^{−1} ∉ P (Tri_{K})

3. If f ∈ P (Tri_{K}) and g ∈ Tri_{K} then gfg^{−1} ∈ P (Tri_{K})

One requirement for a fully ordered group is missing: for each f ∈ Tri_{K}, either f ∈ P (Tri_{K}) or f^{−1} ∈ P (Tri_{K}). This requirement in fact holds, but it is not provable in RCA_{0}. All that we will need for the next section, however, is that the conditions in the lemma are satisfied. There is a proof of Lemma 5.14 in Appendix A.

5.3 Free Products of O-Groups

In this section we prove Theorem 5.2. The proof has several steps, so we outline them here. Given two fully ordered groups A, B, we form a larger group, C, of which A and B are direct summands. We take the group ring ℚ[C] and use the orders on A and B to fully order ℚ[C]. Using the definitions introduced in the previous section, we form the ordered matrix group Tri_{ℚ[C]}. The free product A ∗ B can be embedded in Tri_{ℚ[C]} and we examine this embedding in detail. The order on A ∗ B is defined using the properties of Tri_{ℚ[C]} proved in the previous section.

Let A and B be fully ordered groups. We first define a larger ordered group C. For each pair ⟨i, j⟩ ∈ ℕ^{+} × ℕ^{+}, let x_{ij} and y_{ij} generate copies of ℤ ordered such that x_{ij}^{n} and y_{ij}^{n} are positive if and only if n ≥ 0. For each i ∈ ℕ^{+}, let u_{i} and v_{i} generate copies of ℤ ordered in the same way. The notation ⟨x_{ij}⟩ is used for the group generated by x_{ij}, and similarly for ⟨y_{ij}⟩, ⟨u_{i}⟩, and ⟨v_{i}⟩.

The group C is defined as the restricted direct product:

C = A × B × ∏_{i,j=1}^{∞}⟨x_{ij}⟩ × ∏_{i,j=1}^{∞}⟨y_{ij}⟩ × ∏_{i=1}^{∞}⟨u_{i}⟩ × ∏_{i=1}^{∞}⟨v_{i}⟩.

Since there is a uniform order on the factors of C, C can be lexicographically ordered in RCA_{0}. It is important to realize that C is written multiplicatively instead of additively, even though many of the summands are normally written additively. As a notational convenience, we use x_{ij} to denote the element of C which is the identity in all components of C except the ⟨x_{ij}⟩ component and has value x_{ij} in the ⟨x_{ij}⟩ component. We abuse notation similarly for a ∈ A, b ∈ B and the generators u_{i}, v_{i}, y_{ij}.

Let ℚ[C] be the group ring of C over ℚ. Formally, the elements of ℚ[C] are the finite sums ∑ α_{i}c_{i} with α_{i} ∈ ℚ \ {0}, c_{i} ∈ C and all the c_{i} distinct. Addition is defined by:

∑_{i∈I} α_{i}c_{i} + ∑_{j∈J} β_{j}c_{j} =

∑_{i∈I\J} α_{i}c_{i} + ∑_{j∈J\I} β_{j}c_{j} + ∑_{i∈I∩J}(α_{i} + β_{i})c_{i}

with the stipulation that any terms in the third sum with α_{i} + β_{i} = 0 are removed. Multiplication is defined by:

(∑_{i∈I} α_{i}c_{i}) (∑_{j∈J} β_{j}c_{j}) = ∑_{i∈I} ∑_{j∈J}(α_{i}β_{j})c_{i}c_{j}

where the terms with the same value from C in this finite sum are collected and any term with coefficient 0 is dropped. The additive identity here is the empty sum I = ∅, and the multiplicative identity is the sum with one element 1_{ℚ}1_{C}. RCA_{0} proves that ℚ[C] exists.

The next goal is to order ℚ[C]. The positive cone P (ℚ[C]) is defined from the order ≤_{C} on C. The sum ∑_{i∈I} α_{i}c_{i} is in P (ℚ[C]) if and only if I = ∅ or α_{j} >_{ℚ} 0 where j is such that c_{j} is the ≤_{C}-least element among the c_{i} with i ∈ I. Since I is finite there is such a ≤_{C}-least element. RCA_{0} suffices to prove that this gives a full order on ℚ[C].

Now that we have a fully ordered ring, we can use the machinery of the previous section to work with Tri_{ℚ[C]}. The goal is to embed A ∗ B into Tri_{ℚ[C]} and then use our formal ordering of Tri_{ℚ[C]} to order A ∗ B. The embedding is given by uniformly associating to each element of A∗B a function in Tri_{ℚ[C]}. To do this we specify four matrices in Tri_{ℚ[C]} and denote them by X, Y, U and V. In the definitions, 0 and 1 refer to the additive and multiplicative identities respectively in ℚ[C].

X(i, j) = \{ \begin{matrix}1 & i = j \\ 0 & i > j \\ x_{ij} & i < j \end{matrix}

Y (i, j) = \{ \begin{matrix}1 & i = j \\ 0 & i > j \\ y_{ij} & i < j \end{matrix}

U(i, j) = \{ \begin{matrix}u_{i} & i = j \\ 0 & i ≠ j \end{matrix}

V (i, j) = \{ \begin{matrix}v_{i} & i = j \\ 0 & i ≠ j \end{matrix}

It is useful to see what these functions look like as matrices.

X = ( \begin{matrix}1 & x_{12} & x_{13} & … \\ 0 & 1 & x_{23} & … \\ 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

U = ( \begin{matrix}u_{1} & 0 & 0 & … \\ 0 & u_{2} & 0 & … \\ 0 & 0 & u_{3} & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

In these definitions, u_{i} denotes the element of ℚ[C] that is represented by the sum with the single element 1_{ℚ}u_{i}. U is upper triangular and has positive elements on the diagonal since u_{i} is positive in our order on ⟨u_{i}⟩. Also, since 1_{ℚ}u_{i}·1_{ℚ}u_{i}^{−1} = 1_{ℚ[C]}, U has invertible elements along the diagonal. This point is where it is important to realize that we are using multiplicative instead of additive notation for the groups. Thus, U ∈ Tri_{ℚ[C]}. Similarly, X, Y, V ∈ Tri_{ℚ[C]}.

These matrices are used to define the embedding in several steps. For each a ∈ A, define α(a) : ℕ^{+} × ℕ^{+} → ℚ[C] by:

α(a)(i, j) = \{ \begin{matrix}1 & i = j and i is odd \\ a & i = j and i is even \\ 0 & i ≠ j \end{matrix}

As a matrix, this looks like:

α(a) = ( \begin{matrix}1 & 0 & 0 & … \\ 0 & a & 0 & … \\ 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

As above, 1 stands for 1_{ℚ[C]} and a denotes the sum with one element 1_{ℚ}a. Regardless of whether a is positive or negative in A, 1_{ℚ}a is positive in ℚ[C] since 1_{ℚ} > 0_{ℚ}. Also, a is invertible because

(1_{ℚ}a)(1_{ℚ}a^{−1}) = 1_{ℚ}1_{C} = 1_{ℚ[C]}.

Hence, α(a) ∈ Tri_{ℚ[C]}.

For each b ∈ B define β(b) ∈ Tri_{ℚ[C]} similarly:

β(b)(i, j) = \{ \begin{matrix}1 & i = j and i is odd \\ b & i = j and i is even \\ 0 & i ≠ j \end{matrix}

As a matrix, β(b) looks just like α(a), except it has b’s instead of a’s. We define two more maps on each of A and B. For each a ∈ A define:

α′(a) = X^{−1} · α(a) · X

α′′(a) = U^{−1} · α′(a) · U.

For each b ∈ B define:

β′(b) = Y^{−1} · β(b) · Y
β′′(b) = V^{−1} · β′(b) · V.

Later, we will use results from the previous section to produce explicit formulas for the entries in these matrices.

Because RCA_{0} proves that Tri_{ℚ[C]} is closed under inverses and products, α′′(a) and β′′(b) are both in Tri_{ℚ[C]}. Also, since we have explicit formulas for inverses and products in Tri_{ℚ[C]}, α′′(a) and β′′(b) can be given uniformly from A and B. The embedding of A∗B into Tri_{ℚ[C]} is given by associating to each word a_{1}b_{1} ⋯ a_{n}b_{n} the product α′′(a_{1})β′′(b_{1}) ⋯ α′′(a_{n})β′′(b_{n}) in Tri_{ℚ[C]}. Notice that the term embedding is being used very loosely here. Tri_{ℚ[C]} is not a set, so the correspondence is really a uniform construction of a function in Tri_{ℚ[C]} for each word over A, B. That said, we will continue to use the term embedding and will use γ(w) to denote the element of Tri_{ℚ[C]} which corresponds to the word w.

We need to describe and check the properties of this embedding. If a ∈ A, then from the formula for α(a), it is clear that α(a)^{−1} = α(a^{−1}). Examining γ(a) reveals

γ(a)^{−1} = (U^{−1}X^{−1}α(a)XU)^{−1}
= U^{−1}X^{−1}α(a)^{−1}XU
= U^{−1}X^{−1}α(a^{−1})XU
= γ(a^{−1}).

The same property, γ(b)^{−1} = γ(b^{−1}) holds for b ∈ B. In fact, if a_{1}b_{1} ⋯ a_{n}b_{n} is any, not necessarily reduced, words over A, B, then

γ((a_{1}b_{1} ⋯ a_{n}b_{n})^{−1}) = γ(a_{1}b_{1} ⋯ a_{n}b_{n})^{−1}.

This equation shows that for every reduced word w ∈ A ∗ B, γ(w)^{−1} = γ(w^{−1}). It also shows that γ respects the reduction of words and hence is a group homomorphism. If w_{1}, w_{2} are reduced words in A ∗ B, then

γ(w_{1}w_{2}) = γ(w_{1})γ(w_{2}).

It is much more important and non trivial to check that γ is one-to-one.

Proposition 5.15. (RCA_{0}) If w_{1} ≠ w_{2} in A ∗ B, then γ(w_{1}) ≠ γ(w_{2}) in Tri_{ℚ[C]}.

In order to prove this proposition, we need several lemmas. The proofs of these lemmas are presented in Appendix A. Throughout these lemmas a is an arbitrary element of A, b is an arbitrary element of B and w_{1}, w_{2} are arbitrary words in A ∗ B. Our first goal is to explore α′(a), and by analogy β′(b). Let f = α(a) · X and g = X^{−1}. We are interested in deriving formulas for α′(a) = g · f ∈ Tri_{ℚ[C]}. More explicitly, g can be given by: g(i, i) = 1, g(i, j) = 0 for i > j and for i < j:

g(i, j) = −x_{ij} + ∑_{i<k_{1}<j} x_{ik_{1}}x_{k_{1}j} −

− ∑_{i<k_{1}<k_{2}<j} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}j} + ⋯ + (−1)^{j−i}(x_{i(i+1)} ⋯ x_{(j−1)j}).

As a matrix, this looks like:

g = ( \begin{matrix}1 & −x_{12} & −x_{13} + x_{12}x_{23} & … \\ 0 & 1 & −x_{23} & … \\ 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

f can be given explicitly by:

f(i, j) = \{ \begin{matrix}1 & i = j ∧ i is odd \\ a & i = j ∧ i is even \\ x_{ij} & i < j ∧ i is odd \\ ax_{ij} & i < j ∧ i is even \\ 0 & i > j \end{matrix}

f = ( \begin{matrix}1 & x_{12} & x_{13} & … \\ 0 & a & ax_{23} & … \\ 0 & 0 & 1 & … \\ ⋮ & ⋮ & ⋮ & ... \end{matrix} )

Lemma 5.16. (RCA_{0})

α′(a)(i, i) = \{ \begin{matrix}1 & i is odd \\ a & i is even \end{matrix}

Lemma 5.17. (RCA_{0}) If i < j and i, j are both even, then

α′(a)(i, j) = (1 − a) ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) −

− ∑_{i<k_{1}<k_{2}<n}(x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}n}x_{nj}) + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj})

Lemma 5.18. (RCA_{0}) If i < j, i is even, and j is odd then

α′(a)(i, j) = (1 − a)(−x_{ij}) + (1 − a) ∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (x_{in}x_{nj} − ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) +

+ ∑_{i<k_{1}<k_{2}<n} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{n}n}x_{nj} − ⋯ + (−1)^{n−i}x_{ii+1} ⋯ x_{n−1n}x_{nj})

Lemma 5.19. (RCA_{0}) If i < j and both i, j are odd then

α′(a)(i, j) = (1 − a) ∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (x_{in}x_{nj} − ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) +

+ ∑_{i<k_{1}<k_{2}<n}(x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}n}x_{nj}) + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj})

Lemma 5.20. (RCA_{0}) If i < j, i is odd, and j is even then

α′(a)(i, j) = (1 − a)(x_{ij}) + (1 − a) ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) −

− ∑_{i<k_{1}<k_{2}<n} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{n}n}x_{nj} + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj})

The same results hold for β′(b) with b substituted into the formulas for a. From these formulas, it is clear that if a = 1_{A} then α′(a) = I in Tri_{ℚ[C]}, and similarly for b. Also, if a ≠ 1_{A}, then in particular, the diagonal elements of α′(a) are not all 1, so α′(a) ≠ I. A closer look at these formulas reveals the following lemma. Lemma 5.21. (RCA_{0}) If a ≠ 1_{A} and b ≠ 1_{B} then for any i, j with i ≤ j, α′(a) ≠ 0 and β′(b) ≠ 0.

We are now ready to go back and prove Proposition 5.15. Proof. To show that w_{1} ≠ w_{2} in A ∗ B implies that γ(w_{1}) ≠ γ(w_{2}) in Tri_{ℚ[C]}, it suffices to show that γ(w) ≠ I for an arbitrary nonidentity element w. Let w = a_{1}b_{1} ⋯ a_{t}b_{t} be an arbitrary nonidentity word in A ∗ B that is reduced, except that possibly a_{1} = 1_{A} or b_{t} = 1_{B}. It suffices to show for i < j that γ(w)(i, j) ≠ 0.

The multiplication formula in Tri_{ℚ[C]} extends to the following formula for the product of m functions f_{1}, …, f_{m} in Tri_{ℚ[C]}: for i > j, f_{1} ⋯ f_{m} = 0 and for i ≤ j

f_{1} ⋯ f_{m}(i, j) = ∑_{i≤k_{1}≤···≤k_{m−1}≤j} f_{1}(i, k_{1})f_{2}(k_{1}, k_{2}) ⋯ f_{m}(k_{m−1}, j).

We consider the case in which a_{1} ≠ 1_{A} and b_{t} ≠ 1_{B}. By the extended multiplication formula, if

c = α′′(a_{1})β′′(b_{1}) ⋯ α′′(a_{t})β′′(b_{t})

then

c(i, j) = ∑_{i≤k_{i}≤···≤k_{2t−1}≤j} (α′′(a_{1})(i, k_{1})β′′(b_{1})(k_{1}, k_{2}) ⋯

⋯ α′′(a_{t})(k_{2t−2}, k_{2t−1})β′′(b_{t})(k_{2t−1}, j)).

Applying the formulas for multiplication and inverses, we can show that:

α′′(a)(i, j) = \frac{1}{u_{i}} α′(a)(i, j)u_{j}

β′′(b)(i, j) = \frac{1}{v_{i}} β′(b)(i, j)v_{j}.

As before, the notation \frac{1}{u_{i}} stands for u_{i}^{−1} Putting these formulas together gives us:

c(i, j) = ∑_{i≤k_{i}≤···≤k_{2t−1}≤j} (\frac{u_{k_{1}}}{u_{i}} \frac{v_{k_{2}}}{v_{k_{1}}} \frac{u_{k_{3}}}{u_{k_{2}}} ⋯ \frac{v_{j}}{v_{k_{2t−1}}} α′(a_{1})(i, k_{1}) ·

· β′(b_{1})(k_{1}, k_{2}) ⋯ α′(a_{t})(k_{2t−2}, k_{2t−1})β′(k_{2t−1}, j)).

Viewing c(i, j) as a polynomial in u_{i}, v_{i}, 1/u_{i} and 1/v_{i}, it is clear that none of the terms in the polynomial cancel. Also, since α′(a_{m})(i, j) ≠ 0, β′(b_{m})(i, j) ≠ 0, and any group ring has no zero divisors, none of the terms drop out because they are zero. The remaining cases, a_{1} = 1_{A}, b_{t} ≠ 1_{B} etc., are similar. Thus c ≠ I.

Recall that comparing elements of Tri_{ℚ[C]} involved comparing sequences with order type ω^{ω}. One of the keys to proving Theorem 5.2 is to show that if w_{1} ≠ w_{2} ∈ A ∗ B then comparing γ(w_{1}) and γ(w_{2}) requires only comparing sequences of elements of ℚ[C] with order type ω. Definition 5.22. (RCA_{0}) If r ∈ ℚ[C] then define r^{+n} to be the element of ℚ[C] that looks just like r except the subscripts on x_{ij}, y_{ij}, u_{i} and v_{i} are all adjusted by +n. That is, x_{ij} ↦ x_{(i+n)(j+n)}, u_{i} ↦ u_{i+n}, etc. Proposition 5.23. (RCA_{0}) If f ∈ Tri_{ℚ[C]} is in the image of γ then

f(1, j)^{+2n} = f(1 + 2n, j + 2n)

f(2, j)^{+2n} = f(2 + 2n, j + 2n).

Definition 5.24. (RCA_{0}) If the conditions in the conclusion of Proposition 5.23 hold for f, then we say f possesses the shift property.

The proof of Proposition 5.23 is broken into several lemmas. Lemma 5.25. (RCA_{0}) If f, g ∈ Tri_{ℚ[C]} possess the shift property, then so does f · g. Proof. Consider (f · g)(1, j). If j = 1, then we have:

f · g(1 + 2n, 1 + 2n) = f(1 + 2n, 1 + 2n)g(1 + 2n, 1 + 2n)
= f(1, 1)^{+2n}g(1, 1)^{+2n}
= (f · g(1, 1))^{+2n}.

If j > 1 then we have:

f · g(1 + 2n, j + 2n) = ∑_{m=1+2n}^{j+2n} f(1 + 2n, m)g(m, j + 2n)
= ∑_{m=1}^{j} f(1 + 2n, m + 2n)g(m + 2n, j + 2n)
= ∑_{m=1}^{j} f(1, m)^{+2n}g(m, j)^{+2n}
= ∑_{m=1}^{j} (f(1, m)g(m, j))^{+2n} = (f · g(1, j))^{+2n}.

The cases for f · g(2, j) are similar. Lemma 5.26. (RCA_{0}) If a ∈ A then α′(a) and α′(a^{−1}) have the shift property.

Proof. This proof utilizes the formulas which we derived for α′(a). Along the principle diagonal, we have:

α′(a) = \{ \begin{matrix}1 & i is odd \\ a & i is even \end{matrix}

This satisfies the shift property for the cases α′(a)(1, 1) and α′(a)(2, 2). If j > 1 and odd, then using our formulas:

α′(a)(1, j) = (1 − a) ∑_{\begin{matrix}m=2 \\ m even \end{matrix}}^{j−1} (x_{1m}x_{mj} − ∑_{1<k_{1}<m}(x_{1k_{1}}x_{k_{1}m}x_{mj}) +

+ ∑_{1<k_{1}<k_{2}<m}(x_{1k_{1}}x_{k_{1}k_{2}}x_{k_{2}m}x_{mj}) + ⋯ + (−1)^{m−1}x_{12} ⋯ x_{m−1m}x_{mj}) .

When we write the formula for α′(a)(1 + 2n, j + 2n) instead of letting m range from 2 + 2n to j − 1 + 2n, we let it range from 2 to j − 1 and adjust the subscripts inside the sum.

α′(a)(1 + 2n, j + 2n) = (1 − a) ∑_{\begin{matrix}m=2 \\ m even \end{matrix}}^{j−1} (x_{(1+2n)(m+2n)}x_{(m+2n)(j+2n)} −

− ∑_{1<k_{1}<m}(x_{(1+2n)(k_{1}+2n)}x_{(k_{1}+2n)(m+2n)}x_{(m+2n)(j+2n)}) + ⋯

+ ⋯ (−1)^{m+2n−(1+2n)}(x_{(1+2n)(1+2n+1)} ⋯ x_{(m+2n)(j+2n)}))

Once you observe that m + 2n − (1 + 2n) = m − 1, it is clear that these two sums can be obtained from one another by a shift in the indices of +2n. The other cases follow similarly using the formulas for α′(a) and α′(a^{−1}).

Lemma 5.27. (RCA_{0}) If a ∈ A then α′′(a) and α′′(a^{−1}) have the shift property.

Proof. This follows from the fact that

α′′(a)(i, j) = \frac{u_{j}}{u_{i}} α′(a)(i, j)

We have

α′′(a)(i + 2n, j + 2n) = \frac{u_{j+2n}}{u_{i+2n}} α′(a)(i + 2n, j + 2n)  (5.1)
= \frac{u_{j+2n}}{u_{i+2n}} α′(a)(i, j)^{+2n}  (5.2)
= α′′(a)(i, j)^{+2n}.  (5.3)

The case for α′′(a^{−1}) is similar.

Lemma 5.28. (RCA_{0}) If b ∈ B then β′(b), β′(b^{−1}), β′′(b) and β′′(b^{−1}) have the shift property.

Proof. The proof is the same as for α′(a) and α′′(a).

We can now prove Proposition 5.23.

Proof. By assumption γ(w) = f for some w ∈ A ∗ B. From the facts that w is a word over A and B, that γ(a), γ(a^{−1}), γ(b) and γ(b^{−1}) have the shift property for all a ∈ A and b ∈ B, and that the shift property is preserved under multiplication, it follows that f has the shift property.

It remains to show how to pull the order on Tri_{ℚ[C]} back to A ∗ B. Suppose that f ∈ Tri_{ℚ[C]}, f ≠ I, and f has the shift property. Since f ≠ I, there is some pair ⟨i, j⟩ such that f(i, j) ≠ I(i, j). In order to tell if f ∈ P (Tri_{ℚ[C]}) we need to look down the diagonals until we find the first such pair. However, because f has the shift property, if f and I agree on the first two entries in any diagonal, they will agree on all entries in that diagonal. Comparing f and I is now easy. Thinking of them as matrices, we compare the entries in the following order:

( \begin{matrix}1 & 3 & 5 & 7 & … \\ · & 2 & 4 & 6 & … \\ − − irrelevent − − \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \end{matrix} )

We only need to search through a sequence of elements with order type ω. If we know that f ≠ I then we can find the first place they differ in this sequence. We finally show how to define P (A ∗ B) from P (Tri_{ℚ[C]}).

P (A ∗ B) = {⟨⟩} ∪ {x ∈ A ∗ B | x ≠ 1_{A∗B} ∧ γ(x) ∈ P (Tri_{ℚ[C]})}

RCA_{0} proves the existence of this set because for any x ≠ 1_{A∗B}, we know that γ(x) ≠ I and γ(x) has the shift property. Therefore, RCA_{0} proves there is a computable procedure to determine if γ(x) ∈ P (Tri_{ℚ[C]}). It remains to show that this set is in fact the positive cone of a full order on A ∗ B. Claim. P (A ∗ B) is closed under multiplication.

Assume x, y ∈ P (A ∗ B). Since P (Tri_{ℚ[C]}) is closed under multiplication and γ(x), γ(y) ∈ P (Tri_{ℚ[C]}), we have γ(x)γ(y) = γ(xy) ∈ P (Tri_{ℚ[C]}). Also, assuming that at least one of x, y is not 1_{A∗B}, then x and y cannot be inverses because P (Tri_{ℚ[C]}) is pure. Thus, γ(xy) ∈ P (Tri_{ℚ[C]}) implies that xy ∈ P (A ∗ B) and so P (A ∗ B) is closed under multiplication. Claim. P (A ∗ B) is pure.

Assume x ∈ P (A ∗ B) and x ≠ 1_{A∗B}. Since P (Tri_{ℚ[C]}) is pure, we have γ(x)^{−1} = γ(x^{−1}) ∉ P (Tri_{ℚ[C]}). Therefore, x^{−1} ∉ P (A ∗ B), and so P (A ∗ B) is pure. Claim. P (A ∗ B) is normal.

Assume x ∈ P (A ∗ B) and y ∈ A ∗ B. Since γ(x) ∈ P (Tri_{ℚ[C]}) and P (Tri_{ℚ[C]}) is normal,

γ(y)γ(x)γ(y)^{−1} = γ(yxy^{−1}) ∈ P (Tri_{ℚ[C]}).

Thus, yxy^{−1} is in P (A ∗ B) and P (A ∗ B) is normal. Claim. P (A ∗ B) is full.

Assume γ(x) ∉ P (Tri_{ℚ[C]}). We need to show that γ(x)^{−1} = γ(x^{−1}) ∈ P (Tri_{ℚ[C]}). Notice that γ(x) ≠ I. We split this proof into two cases. Case. Either γ(x)(1, 1) ≠ 1 or γ(x)(1, 1) = 1 and γ(x)(2, 2) ≠ 1.

Assume that γ(x)(1, 1) ≠ 1. The other case is similar. Since γ(x) ∉ P (Tri_{ℚ[C]}) it must be that γ(x)(1, 1) < 1. From the definition of γ(x)^{−1}:

γ(x)^{−1}(1, 1) = γ(x)(1, 1)^{−1} > 1.

Thus, γ(x^{−1}) ∈ P (Tri_{ℚ[C]}). Case. γ(x)(1, 1) = γ(x)(2, 2) = 1

Because γ(x) has the shift property, there is a least j > 1 such that either γ(x)(1, j) ≠ 0 or γ(2, j) ≠ 0 and γ(x)(1, j) = 0. Assume that γ(x)(1, j) ≠ 0. The other case is similar. Since γ(x) ∉ P (Tri_{ℚ[C]}) it must be that γ(x)(1, j) < 0. Using the fact that γ(x)(n, n) = 1 for all n, the formula for γ(x)^{−1}(1, j) gives:

γ(x)^{−1}(i, j) = −γ(x)(1, j) + ∑_{1<k_{1}<j} γ(x)(1, k_{1})γ(x)(k_{1}, j) − ⋯

⋯ + (−1)^{j−1}(γ(x)(1, 2) ⋯ γ(x)(j − 1, j)).

All the terms drop out except for the first one because γ(x)(1, k) = 0 for any 1 < k < j. Thus, γ(x)^{−1}(1, j) = −γ(1, j) > 0. The check that γ(x)^{−1}(k, k + s) = 0 for the appropriate k, s is similar.

We have completed the proof of Theorem 5.2.

Divisible Closures and Hölder’s

Theorem

Three naturally occurring notions of closure in algebra are the algebraic closure of a field, the real closure of an ordered field and the divisible closure of an abelian group. In this chapter, a survey of results on these closure operations is presented along with a proof in RCA_{0} of Hölder’s Theorem.

6.1 Introduction

Friedman et al. (1983) give definitions for algebraic, real and divisible closures in RCA_{0}.

Definition 6.1. (RCA_{0}) A field is a set K ⊆ ℕ together with two binary operations, +_{K}, ·_{K}, a unary operation, −_{K}, and two constants, 0_{K}, 1_{K} which obey the standard field axioms (see Hungerford (1974)).

If K is a field, then the polynomial ring K[x] is given by

{ σ ∈ Fin_{K} | σ(lh(σ) − 1) ≠ 0_{G} }.

Intuitively, ⟨k_{0}, …, k_{n}⟩ represents the polynomial

k_{0} + k_{1}x + … + k_{n}x^{n}.

The restriction that the last element of σ not be 0_{G} insures that each polynomial has a unique sequence representative. Addition and multiplication of sequences are defined to mimic the corresponding operation on polynomials. If f ∈ K[x] and a ∈ K, then f(a) ∈ K is the element

k_{0} + k_{1}a + k_{2}a^{2} + … + k_{n}a^{n}.

f ∈ K[x] is nonconstant if lh(f) > 1 and a is a root of f if f(a) = 0_{K}. If f = k_{0} + k_{1}x … + k_{n}x^{n} and h : K → K̃ is a field homomorphism, then h(f) ∈ K̃[x] is

h(k_{0}) + h(k_{1})x + … + h(k_{n})x^{n}.

Definition 6.2. (RCA_{0}) A field K is algebraically closed if every nonconstant polynomial f(x) ∈ K[x] has a root in K. An algebraic closure of a field F consists of a monomorphism h : F → K where K is an algebraically closed field and for each a ∈ K there is a nonzero polynomial f(x) ∈ F [x] such that h(f)(a) = 0.

Classically, every field has an algebraic closure which is unique up to isomorphism. There are several question to ask about the computability of the algebraic closure. Is it effective? That is, does every computable field have a computable algebraic closure? Is the uniqueness effective? That is, if a computable field has two computable algebraic closures, is there a computable isomorphism between them? Because the range of the function h is the definition of algebraic closure need not be computable, does every computable field have a computable algebraic closure such that the original field is isomorphic to a computable subfield of the closure?

Rabin (1960) proved that every computable field has a computable algebraic closure. Friedman et al. (1983) used Rabin’s idea to show that the existence of an algebraic closure is provable in RCA_{0}. They also gave precise definitions to address the other two questions. Definition 6.3. (RCA_{0}) A field F has a unique algebraic closure if whenever h_{i} : F → K_{i}, i = 1, 2, are two algebraic closures of F, there exists an isomorphism k : K_{1} → K_{2} such that k(h_{1}(a)) = h_{2}(a) for all a ∈ F. Definition 6.4. (RCA_{0}) Let F be a field. A strong algebraic closure of F is an algebraic closure h : F → K such that h is an isomorphism between F and a subfield of K. Theorem 6.5 (Friedman et al. (1983)). (RCA_{0})

1. Every field has an algebraic closure.

2. W KL_{0} is equivalent to the statement that every field has a unique algebraic closure.

3. ACA_{0} is equivalent to the statement that every field has a strong algebraic closure.

Computationally, this theorem says that computable fields do not necessarily have either a computably unique algebraic closure or a computable strong algebraic closure.

The second type of closure is the real closure of an ordered field. Classically, every formally real field is orderable and every ordered field has a unique real closure. Carrying these notions over into reverse mathematics gives the following definitions and theorems.

Definition 6.6. A field K is formally real if K does not contain a finite sequence of nonzero elements c_{0}, …, c_{n} such that c_{0}^{2} + ⋯ + c_{n}^{2} = 0.

Definition 6.7. (RCA_{0}) An ordered field is a field K together with a linear order ≤_{K} which satisfies the axioms for an ordered field.

Theorem 6.8 (Friedman et al. (1983)). (RCA_{0}) The following are equivalent:

1. W KL_{0}

2. Every formally real field is orderable.

Definition 6.9. An ordered field K is real closed if for all g(x) ∈ K[x] and a < b ∈ K such that g(a) < 0 < g(b), there exists c ∈ K such that g(c) = 0 and a < c < b. A real closure of an ordered field F consists of a real closed ordered field K and a monomorphism h : F → K such that for each b ∈ K, there exists a nonconstant polynomial f(x) ∈ K[x] for which h(f)(b) = 0.

The definitions for a strong real closure and a unique real closure are analogous to the same definitions for algebraic closures.

Theorem 6.10 (Friedman et al. (1983)). (RCA_{0})

1. Every ordered field has a real closure.

2. Every ordered field has a unique real closure.

3. ACA_{0} is equivalent to the statement that every ordered field has a strong real closure.

On the computational side, these results show that every computably ordered computable field F has a computably unique computable real closure, but F need not have a computable strong real closure.

The third notion of closure is the divisible closure of an abelian group. Friedman et al. (1983) give the following definitions.

Definition 6.11. (RCA_{0}) Let D be an abelian group. D is divisible if for all d ∈ D and all n ≥ 1 there exists a c ∈ D such that nc = d. Here, we are using the additive notation of abelian groups, so nc refers to c added to itself n times.

Definition 6.12. (RCA_{0}) Let A be an abelian group. A divisible closure of A is a divisible abelian group D together with a monomorphism h : A → D such that for all d ∈ D, d ≠ 1_{D}, there exists n ∈ ℕ such that nd = h(a) for some a ∈ A, a ≠ 1_{A}.

Smith (1981) proved that every computable abelian group has a computable divisible closure and that this divisible closure is unique if and only if there is a uniform algorithm which for each prime p decides if an arbitrary element of the original group is divisible by p. Using the ideas in these proofs, Friedman et al. (1983) proved the following theorem.

Theorem 6.13 (Friedman et al. (1983)). (RCA_{0})

1. Every abelian group has a divisible closure.

2. ACA_{0} is equivalent to the statement that every abelian group has a unique divisible closure.

We will extend these results to strong divisible closures in Section 6.3. Downey and Kurtz (1986) considered another possible extension. They proved that every computably fully ordered computable abelian group has a computably unique divisible closure. An examination of their proof shows that RCA_{0} suffices to prove the uniqueness of the divisible closure for fully ordered abelian groups.

Theorem 6.14 (Downey and Kurtz (1986)). (RCA_{0}) Every f.o. abelian group G has a f.o. divisible closure h : G → D such that h is order preserving. This divisible closure is unique up to order preserving isomorphism.

In Section 6.3, we will consider the strong divisible closure not only for fully ordered groups, but also for the much smaller class of Archimedean fully ordered groups.

Definition 6.15. (RCA_{0}) If G is an f.o. group, then the absolute value of x ∈ G is given by:

| x | = max {x, x^{−1}}

Definition 6.16. (RCA_{0}) If G is an f.o. group, then a ∈ G is Archimedean less that b ∈ G, denoted a ≪ b, if | a^{n} | < | b | for all n ∈ ℕ. If there are n, m ∈ ℕ such that | a^{n} | ≥ | b | and | b^{m} | ≥ | a |, then a and b are Archimedean equivalent, denoted a ≈ b. The notation a ⪅ b means a ≈ b ∨ a ≪ b. G is an Archimedean fully ordered group if G is fully ordered and for all a, b ≠ 1_{G}, a ≈ b.

It is not hard to check that ≈ is an equivalence relation and that ≪ is transitive, antireflexive, and antisymmetric. The next lemma lists several other straightforward properties of ≈ and ≪. For proofs, see Fuchs (1963). Lemma 6.17. (RCA_{0}) If G is a f.o. group, then the following conditions hold for all a, b, c ∈ G.

1. Exactly one of the following holds: a ≪ b, b ≪ a, or a ≈ b.

2. a ≪ b implies that xax^{−1} ≪ xbx^{−1} for all x ∈ G.

3. a ≪ b and a ≈ c imply that c ≪ b.

4. a ≪ b and b ≈ c imply that a ≪ c.

An early conjecture about ordered groups was that the number of full orders of a given O-group was always a power of 2. This conjecture also stated that a group could not have a countable number of orders. Buttsworth (1971) showed part of this conjecture was false by constructing a group with a countably infinite number of orders and Kargapolov et al. (1965) showed it was false for groups with a finite number of orders. Classifying all possible full orders for a given class of O-groups is a harder problem than just counting them. One of the few classes for which this problem has been solved is the class of free abelian groups of finite rank. These results can be found in several places, including Teh (1960). The key ingredient in these results about counting or classifying full orders is Hölder’s Theorem. For a more in depth discussion, see either Kokorin and Kopytov (1974) or Mura and Rhemtulla (1977).

Hölder’s Theorem states that every f.o. Archimedean group can be embedded in the naturally ordered additive group of the reals. Before examining strong divisible closures, we show that Hölder’s Theorem is provable in RCA_{0}.

6.2 Hölder’s Theorem

Because real numbers are given by functions from ℕ to ℚ, the first step towards proving Hölder’s theorem is to decide what is meant by a subgroup of the real numbers in second order arithmetic.

Definition 6.18. (RCA_{0}) A nontrivial subgroup of the additive real numbers (ℝ, +_{ℝ}) is a sequence of reals A = ⟨ r_{n} | n ∈ ℕ ⟩ together with a function +_{A} : ℕ × ℕ → ℕ and a distinguished number i ∈ ℕ such that

1. r_{i} = 0_{ℝ}

2. n +_{A} m = p if and only if r_{n} +_{ℝ} r_{m} = r_{p}

3. (ℕ, +_{A}) satisfies the group axioms with i as the identity element.

Recall that a real number r is a sequence of rationals, r = ⟨q_{n}|n ∈ ℕ⟩, such that

∀k ∀i ( | q_{k} − q_{k+i} | ≤ 2^{−k} ).

Thus A is a double indexed sequence of rationals

A = ⟨ q_{n,m} | n, m ∈ ℕ ⟩

where

r_{n} = ⟨ q_{n,m} | m ∈ ℕ ⟩.

Let (G, ≤) be an Archimedean fully ordered group. Because G must be abelian, see Lemma 6.19, we use additive notation for G. The idea of the proof of Hölder’s theorem is to pick an element a ∈ P (G), a ≠ 1_{A}, and define the subgroup of (ℝ, +) by using a to approximate the other elements of G. For now, assume that 2^{n} divides a in G for all n. That is, assume that for all n there exists c ∈ G such that 2^{n}c = a. To construct the real to which an element g ≠ 1_{G} is sent, we first find p_{0} ∈ ℤ such that

p_{0} a ≤ g < (p_{0} + 1) a.

Such a p_{0} exists by the assumption that G is Archimedean. Next we find p_{1} such that

p_{1} \frac{a}{2} ≤ g < (p_{1} + 1) \frac{a}{2}

and continue to find p_{i} such that

p_{i} \frac{a}{2^{i}} ≤ g < (n_{i} + 1) \frac{a}{2^{i}} .

The real corresponding to g will be ⟨p_{i}/2^{i}|i ∈ ℕ⟩. Because the elements a/2^{i} may not exist, we achieve the same effect by choosing p_{i} such that

p_{i} a ≤ 2^{i} g < (p_{i} + 1) a.

Lemma 6.19. (RCA_{0}) Every Archimedean fully ordered group is abelian.

Proof. The standard proof goes through in RCA_{0}. For the details, see Kokorin and Kopytov (1974).

Hölder’s Theorem. (RCA_{0}) Every nontrivial Archimedean f.o. group is order isomorphic to a nontrivial subgroup of the naturally ordered additive group (ℝ, +).

Proof. Let (G, ≤) be an Archimedean f.o. group and g_{0}, g_{1}, … be an enumeration of G with no repetitions such that g_{0} = 1_{G} and g_{1} ∈ P (G). We construct a subgroup A of (ℝ, +) by constructing r_{n} = ⟨ q_{n,m} | m ∈ ℕ ⟩ uniformly in n from g_{n} and g_{1}. For simplicity of notation, let a = g_{1}. The first two elements of A are

r_{0} = ⟨ 0 | m ∈ ℕ ⟩
r_{1} = ⟨ 1 | m ∈ ℕ ⟩.

To construct r_{n} for n > 1, define p_{n,m} ∈ ℕ and q_{n,m} ∈ ℚ by

p_{n,m} a ≤ 2^{m}g_{n} < (p_{n,m} + 1) a

q_{n,m} = \frac{p_{n,m}}{2^{m}} .

Because G is Archimedean, such p_{n,m} exist and are uniquely determined by the inequality. The real r_{n} is

r_{n} = ⟨ q_{n,m} | m ∈ ℕ ⟩.

It remains to show that A = ⟨ r_{n} | n ∈ ℕ ⟩ is a subgroup of (ℝ, +) and that the map from G to A that sends g_{n} to r_{n} is an order preserving isomorphism.

Claim. Each r_{n} is a real number.

To prove this claim we must show

∀m ∀k ( | q_{n,m} − q_{n,m+k} | ≤ 2^{−m} ).

Notice that since

p_{n,m} a ≤ 2^{m}g_{n} < (p_{n,m} + 1) a

it follows that

2p_{n,m} a ≤ 2^{m+1}g_{n} < (2p_{m,n} + 2) a.

Hence, for all m, either p_{n,m+1} = 2p_{n,m}, and hence q_{n,m+1} = q_{n,m}, or p_{n,m+1} = 2p_{n,m} + 1, and hence q_{n,m+1} = q_{n,m} + 1/2^{m+1}. Thus,

| q_{n,m} − q_{n,m+k} | ≤ ∑_{i=1}^{i=k} \frac{1}{2^{m+i}} < \frac{1}{2^{m}} .

Claim. If g_{n} + g_{m} = g_{k} then r_{n} + r_{m} = r_{k}.

As above, this claim reduces to checking convergence rates. By definition,

r_{n} + r_{m} = ⟨ q_{n,i+1} + q_{m,i+1} | i ∈ ℕ ⟩.

To prove r_{n} + r_{m} = r_{k} we need to show that for every i ∈ ℕ

| q_{n,i+1} + q_{m,i+1} − q_{k,i} | < 2^{−i+1}.

The definitions of p_{n,i+1} and p_{m,i+1} are p_{n,i+1} a ≤ 2^{i+1}g_{n} < (p_{n,i+1} + 1) a

p_{m,i+1} a ≤ 2^{i+1}g_{m} < (p_{n,i+1} + 1) a.

Adding these two equations together yields

(p_{n,i+1} + p_{m,i+1}) a ≤ 2^{i+1}g_{k} < (p_{n,i+1} + p_{m,i+1} + 2) a.

Thus p_{k,i+1} is either p_{n,i+1} + p_{m,i+1} or p_{n,i+1} + p_{m,i+1} + 1. In either case, q_{k,i+1} − q_{n,i+1} − q_{m,i+1} ≤ 2^{−i−1} and we have

| q_{n,i+1} + q_{m,i+1} − q_{k,i} | ≤ | q_{n,i+1} + q_{m,i+1} − q_{k,i+1} | + | q_{k,i+1} − q_{k,i} |
≤ 2^{−i−1} + 2^{−i}
< 2^{−i+1}.

The map that sends g_{n} to r_{n} is onto by definition. The following claim implies that it is one-to-one. Claim. If n ≠ m, then r_{n} ≠ r_{m}.

To establish r_{n} ≠ r_{m}, we need to find an i such that

| q_{n,i} − q_{m,i} | > 2^{−i+1}.

Equivalently, we can find an i such that

| p_{n,i} − p_{m,i} | > 2.

Because n ≠ m implies g_{n} ≠ g_{m} assume without loss of generality that g_{n} < g_{m}. Split into four cases. Case. g_{n} < 1_{G} ≤ g_{m}

By the Archimedean property of G there is an i such that

2^{i} g_{n} < −3a < 1_{G} ≤ 2^{i} g_{m}.

It follows that p_{n,i} < −3 and p_{m,i} ≥ 0. Hence | p_{n,i} − p_{m,i} | ≥ 3.

Case. g_{n} = 1_{G} < g_{m} There is an i such that

1_{G} < 3a < 2^{i} g_{m}.

It follows that p_{n,i} = 0 while p_{m,i} ≥ 3. Case. 1_{G} < g_{n} < g_{m}

Since 1_{G} < g_{m} − g_{n}, there is an i which yields the following equations:

1_{G} < a < 2^{i}(g_{m} − g_{n}) = 2^{i} g_{m} − 2^{i} g_{n}
2^{i} g_{n} < a + 2^{i} g_{n} < 2^{i} g_{m}
2^{i+2} g_{n} < 4a + 2^{i+2} g_{n} < 2^{i+2} g_{m}.

There is an m such that

ma ≤ 2^{i+2} g_{n} < (m + 1)a.

Combining these equations

ma ≤ 2^{i+1} g_{n} < (m + 4)a ≤ 4a + 2^{i+2} g_{n} < 2^{i+2} g_{m}.

It follows that p_{n,i+2} = m and p_{m,i+2} ≥ m + 4. Case. g_{n} < g_{m} < 1_{G}

In this case, 1_{G} < g_{m} − g_{n} and so the same argument works as in the previous case. This case completes the proof of the claim and shows that the map is one-to-one.

The claims show that A is a subgroup of (ℝ, +) and that A is isomorphic to G by the map that sends g_{n} ↦ r_{n}. Finally, to show that g_{n} < g_{m} implies that r_{n} < r_{m}, notice that from the construction, if g_{n} < g_{m} then q_{n,i} ≤ q_{m,i} for every i. Thus, r_{n} ≤ r_{m}. But, since g_{n} ≠ g_{m} implies r_{n} ≠ r_{m}, we have r_{n} < r_{m}.

6.3 Strong Divisible Closures

Definition 6.20. (RCA_{0}) Let A be an abelian group. A strong divisible closure of A is a divisible closure h : A → D such that h is an isomorphism of A onto a subgroup of D. If A is a f.o. group, D is fully ordered and h is order preserving, then we call h : A → D an f.o. strong divisible closure.

Recall from Theorem 6.14 that RCA_{0} suffices to prove the uniqueness of the divisible closure for f.o. abelian groups, but that ACA_{0} is required to prove the uniqueness for abelian groups in general. Given these results, it is reasonable to hope that proving the existence of a strong divisible closure would be easier for f.o. abelian group than for abelian groups. The next theorem shows this is not the case.

Theorem 6.21. (RCA_{0}) The following are equivalent:

1. ACA_{0}

2. Every abelian group has a strong divisible closure.

3. Every fully ordered Archimedean group has an f.o. strong divisible closure.

The idea of proving (3) implies (1) is fairly simple. Let p_{k} be an enumeration of the primes in increasing order. Given a one-to-one function f, let G be the subgroup of ℚ generated by 1 and p^{−k} for each k in the range of f. This group has an Archimedean full order. If D is a strong divisible closure of G, then the image h(G) exists and the range of f can be recovered by

range(f) = \{ k | \frac{h(1)}{p_{k}} ∈ h(G) \}.

Lemma 6.22. (RCA_{0}) Let p_{k} enumerate the primes in increasing order. If k ∈ ℤ, j ∈ ℕ and ∀i ≤ j (0 ≤ m_{i} < p_{i}) then ∑_{i≤j} m_{i}/p_{i} = k implies that k = 0 and m_{i} = 0 for all i ≤ j. Proof. For a contradiction, suppose that some m_{i} ≠ 0. Notice that k > 0 since all m_{i} ≥ 0. Let p̂ be the product of p_{0}, …, p_{j} and p̂_{i} be p̂/p_{i}. If we multiply the sum by p̂ we obtain

∑_{i≤j} m_{i}p̂_{i} = kp̂.

This equation must hold modulo p_{i} for all i ≤ j.

(∑_{i≤j} m_{i}p̂_{i} = kp̂) mod p_{i}

However, if u ≠ i, then (m_{u}p̂_{u} = 0) mod p_{i} because p_{i} divides p̂_{u}. Therefore, we have

(∑_{i≤j} m_{i}p̂_{i} = m_{i}p̂_{i}) mod p_{i}.

Also, (kp̂ = 0) mod p_{i} and so we have

(m_{i}p̂_{i} = 0) mod p_{i}.

It follows that p_{i} divides m_{i}. Because 0 ≤ m_{i} < p_{i}, m_{i} must be 0. This argument holds for all i ≤ j.

Using Lemma 6.22, we can give a proof of Theorem 6.21.

Proof. (1) implies (2) because ACA_{0} is strong enough to show both that the divisible closure exists and that the image of h exists. (2) implies (3) because a full order on D can be defined which makes h order preserving.

P (D) = { d ∈ D | ∃n > 0∃g ∈ P (G)(nd = h(g)) }
= { d ∈ D | ∀n > 0∀g ∈ P (G)(nd ≠ h(g)) }

Because P (D) has a Δ_{1}^{0} definition, RCA_{0} suffices to prove it exists and to verify that it is a full order on D.

To show (3) implies (1), let f be a one-to-one function and let p_{k} be an enumeration of the primes in increasing order. It suffices to show that the range of f exists. Let G be the group given by the generators a, x_{i} for i ∈ ℕ and the relations p_{f(i)}x_{i} = a. The intuition is that G is isomorphic to a subgroup of ℚ with a ↦ 1 and x_{i} ↦ p_{f(i)}^{−1}. In RCA_{0} we can represent elements of G by finite sums:

ka + ∑_{i≤j} m_{i} x_{i}.

where k ∈ ℤ, 0 ≤ m_{i} < p_{f(i)} and m_{j} ≠ 0. Using the relation equations, any element of G can be reduced to one of these finite sums. We need to show that no two of these finite sums represent the same element of G. Claim. If ka + ∑_{i≤j} m_{i} x_{i} = k̃a + ∑_{i≤j̃} m̃_{i}x_{i} then k = k̃, j = j̃ and ∀i ≤ j (m_{i} = m̃_{i}).

First notice that 1_{G} has a unique representation as the finite sum 0a. Indeed, if

ka + ∑_{i≤j} m_{i} x_{i} = 1_{G} = 0a

then using the relations, we obtain

∑_{i≤j} m_{i} \frac{a}{p_{f(i)}} = −ka.

Because G is torsion free, this equation implies

∑_{i≤j} \frac{m_{i}}{p_{f(i)}} = −k.

By Lemma 6.22 k = 0 and m_{i} = 0. To show that j must equal j̃ as in the claim, suppose that j < j̃ and

ka + ∑_{i≤j} m_{i} x_{i} = k̃a + ∑_{i≤j̃} m̃_{i}x_{i}.

Reducing (k − k̃) a + ∑_{i≤j}(m_{i} − m̃_{i})x_{i} to k′a + ∑_{i≤j′} m′_{i}x_{i}, we obtain

k′a + ∑_{i≤j′} m′_{i}x_{i} + ∑_{j<i<j̃} m̃_{i}x_{i} = 1_{G}.

Thus, m̃j̃ = 0 which gives the desired contradiction. Hence j = j̃.

A similar argument shows that m_{i} = m̃_{i} for all i ≤ j. Suppose there is an i ≤ j such that m_{i} ≠ m̃_{i}. Since we can always subtract off equal terms, we can assume without loss of generality that m_{j} ≠ m̃_{j}. If

(k̃ − k)a + ∑_{i≤j−1}(m̃_{i} − m_{i})x_{i}

reduces to the normal form

k′a + ∑_{i≤j′} m′_{i}x_{i}

then

(k̃ − k)a + ∑_{i≤j}(m̃_{i} − m_{i})x_{i}

reduces to the normal form

k′a + ∑_{i≤j′} m′_{i}x_{i} + (m̃_{j} − m_{j})x_{j} = 0_{G}.

By the uniqueness of the normal form for 0_{G}, we have that m̃_{j} − m_{j} = 0, which is a contradiction. Therefore, m̃_{i} = m_{i} for all i ≤ j. Our equation reduces to ka = k̃a which implies that k = k̃. Claim. G is fully orderable.

Define the positive cone P (G) by

ka + ∑_{i≤j} m_{i} x_{i} ∈ P (G) ↔ k + ∑_{i≤j} \frac{m_{i}}{p_{f(i)}} ≥ 0.

P (G) is normal because G is abelian. To verify the other properties, notice that if there are two finite sums, not necessarily in normal form, that are equivalent under the group relations:

ka + ∑_{i≤j} m_{i} x_{i} and k̃a + ∑_{i≤j̃} m̃_{i}x_{i}

then:

k + ∑_{i≤j} \frac{m_{i}}{p_{f(i)}} = k̃ + ∑_{i≤j̃} \frac{m̃_{i}}{p_{f(i)}} .

This property is proved by induction on the number of applications of relation equations it takes to transform one sum into the other. This property immediately yields that P (G) is a pure, full semigroup with identity. Furthermore, it shows that G is Archimedean under this order because ℚ is Archimedean.

Applying condition (3) from the theorem, we have a divisible closure h : G → D and the image h(G) exists.

X = \{k | \frac{h(a)}{p_{k}} ∈ h(G) \}

\frac{h(a)}{p_{k}} ∈ h(G) ↔ p_{k} divides a in G

↔ ∃i (p_{k}x_{i} = a)

↔ ∃i (f(i) = k)

Thus X is the range of f.

The last issue to discuss before leaving divisible closures is the relationship between the complexity of the full orders on a torsion free abelian group and on its divisible closure. We have already mentioned the following theorem. Theorem 6.23 (Smith (1981)). Every computable abelian group has a computable divisible closure. Lemma 6.24. If D is a divisible closure of G, then D is fully orderable if and only if G is fully orderable. Furthermore, each order on G extends uniquely to a full order on D. Proof. To prove the first statement, notice that D has torsion elements if and only if G has torsion elements.

Suppose h : G → D is as in the definition of a divisible closure and P is the positive cone of a full order on G. The second statement claims that h(P) extends uniquely to a full order on D. Consider d ∈ D \ 1_{D}. There exists n > 0 and g ∈ G \ 1_{G} such that nd = h(g). Therefore, if g ∈ P, then nd ∈ h(P) and so d must be positive. If g ∉ P, then g^{−1} ∈ P and so nd^{−1} ∈ h(P) and d must be negative. Therefore, the unique extension of h(P) to the positive cone of a full order on D is

{d ∈ D|d = 1_{D} ∨ (d ≠ 1_{D} ∧ ∃n > 0∃g ∈ P \ 1_{G}(nd = h(g)))}.

Proposition 6.25. Let G be a computable torsion free abelian group and D a computable divisible closure of G. There is a Turing degree preserving bijection ψ from the space of full orders on G to the space of full orders on D.

ψ : 𝕏(G) → 𝕏(D)

Proof. By Lemma 6.24, 𝕏(G) can be mapped bijectively to 𝕏(D) by sending P to its unique extension on D. An examination of the definition of this extension shows that the two orders have the same degree.

Order Types

One of the fundamental problems in the theory of ordered groups is to classify all possible orders for various classes of O-groups. In general, this problem is extremely difficult to solve. An easier problem is to classify the possible order types for countable fully ordered groups. Mal’tsev (1949) proved that the order type of a countable f.o. group is ℤ^{α}ℚ^{ϵ} where ℤ denotes the order type of the integers, ℚ denotes the order type of the rationals, α is a countable ordinal, and ϵ is either 0 or 1. The goal of this chapter is to prove that this theorem is equivalent over RCA_{0} to Π_{1}^{1} − CA_{0}.

7.1 Order Type of a Group

The definitions for a linear order and a well order were given in Chapter 1. The definition of a well order X says there are no infinite descending chains in X. In keeping with the notation of set theory, we will use the letters α, β, γ to stand for well orders. If X is a linear order, it is useful to talk about the largest initial segment which is well ordered.

Definition 7.1. (RCA_{0}) The well ordered initial segment of X is defined by

W (X) = {x ∈ X | ¬∃f : ℕ → X(f(0) = x ∧ ∀i(f(i + 1) < f(i)))}.

Notice that W (X) need not exist in systems like RCA_{0}. From the definition it may require Π_{1}^{1} − CA_{0}, but the only place we will use it is inside Π_{1}^{1} − CA_{0}. It is clear from the definition that if (W (X), ≤) does exist, then it is a well order and that if y ∈ W (X) and z ≤ y then z ∈ W (X).

Definition 7.2. (RCA_{0}) Let (X, ≤_{X}) and (Y, ≤_{Y}) be linear orders. The product X · Y is the linear order (Z, ≤_{Z}) where

Z = { ⟨x, y⟩ | x ∈ X ∧ y ∈ Y }
⟨x_{1}, y_{1}⟩ ≤_{Z} ⟨x_{2}, y_{2}⟩ ↔ y_{1} <_{Y} y_{2} ∨ (y_{1} = y_{2} ∧ x_{1} ≤_{X} x_{2}).

X · Y is frequently written XY.

We also need a definition of ℤ^{X} for a well order X. In set theoretic terms, ℤ^{X} is given by the set of functions f : X → ℤ with finite support. If f ≠ g, then f < g if and only if f(x) <_{ℤ} g(x) where x is the maximum value of X on which f and g disagree. To represent ℤ^{X} in second order arithmetic, we use finite sequences of pairs ⟨x, z⟩ with x ∈ X and z ∈ ℤ \ 0. To give a normal form for the sequences, we require that the X-components in each sequence be in decreasing order. By convention, ℤ^{∅} is the single element linear order. Recall that π_{1} and π_{2} are the projection functions for pairs.

Definition 7.3. (RCA_{0}) Let X be a non-empty linear order and Y = X ×(ℤ\{0}). ℤ^{X} is given by

{ x | x ∈ Fin_{Y} ∧ ∀i < (lh(x) − 1) (π_{1}(x(i)) >_{X} π_{1}(x(i + 1))) }.

Two elements x, y are equal if and only if they are identical as sequences. If x ≠ y and lh(x) ≤_{ℕ} lh(y), then there are two cases to consider.

1. If x ⊂ y then x < y ↔ π_{2}(y(lh(x))) > 0

2. If x ⊄ y, let i be the least number such that x(i) ≠ y(i) and suppose x(i) = ⟨x_{i}, u_{i}⟩ and y(i) = ⟨y_{i}, v_{i}⟩.

(a) If x_{i} <_{X} y_{i} then x < y ↔ v_{i} > 0

(b) If x_{i} >_{X} y_{i} then x < y ↔ u_{i} < 0

(c) If x_{i} = y_{i} then x < y ↔ u_{i} < v_{i}

To see why this definition captures the set theoretic notion, think of each sequence x ∈ ℤ^{X} as representing the function that sends π_{1}(x(i)) to π_{2}(x(i)) for all i < lh(x) and sends all other values in X to 0. In case 1 of the definition, π_{1}(y(lh(x))) represents the largest value of X on which the functions associated to x and y differ. The function for x sends this element to 0, so x < y if and only if y maps this element to something greater than 0. The other cases have similar explanations.

Definition 7.4. (RCA_{0}) If G is an f.o. group and X is a linear order, then X is the order type of G if there is an order preserving bijection f : G → X.

If f : G → X and g : G → Y are two order types of G, then the map g ∘ f^{−1} : X → Y is an order preserving bijection between X and Y. So, in RCA_{0} the order type is unique up to order preserving bijection.

To clear up a possibly confusing point of terminology, an order preserving bijection is a bijection between linearly ordered structures that preserves the order, but ignores any other structure they might have. On the other hand, an order isomorphism, or o-isomorphism, is a group isomorphism that preserves order. We can now state the main theorem of this chapter. Theorem 7.5. (RCA_{0}) The following are equivalent:

1. Π_{1}^{1} − CA_{0}

2. Let G be a countable f.o. group. There is a well order α and ϵ = 0 or 1 such that ℤ^{α}ℚ^{ϵ} is the order type of G.

3. Let G be an abelian countable f.o. group. There is a well order α and ϵ = 0 or 1 such that ℤ^{α}ℚ^{ϵ} is the order type of G.

In this section, we prove that (1) implies (2). The idea of the proof is that if G is an f.o. group, then either G has a least strictly positive element or it does not. If G does not have such an element, then it has order type ℚ. If G does have a least strictly positive element a, then the order type of G is the product of ℤ and G/⟨a⟩, where ⟨a⟩ is the convex normal subgroup generated by a. This process is repeated with G/⟨a⟩ and continues to be repeated until we have either used up all of G or found a quotient of G which has order type ℚ. The recursion can be done in AT R_{0}, but Π_{1}^{1} − CA_{0} is required to prove that the process eventually terminates.

The implication from (2) to (3) is trivial. In the next section, we show that ACA_{0} proves that (3) implies (1). First, we prove that RCA_{0} plus statement (3) suffices to prove that the well ordered initial segment of every linear order exists. This fact is used together with properties of the Kleene-Brouwer order on trees (which require ACA_{0}) to prove Π_{1}^{1} − CA_{0} in the form given in Theorem 1.19.

In the last section, we prove that the reversal can be done over RCA_{0} instead of ACA_{0}. We show that any model of RCA_{0} and statement (3) is closed under the Turing jump and hence is a model of ACA_{0}. By Gödel’s Completeness Theorem, RCA_{0} and statement (3) prove the axioms of ACA_{0}, which in turn prove Π_{1}^{1} −CA_{0}.

Lemma 7.6. (RCA_{0}) Let G be an f.o. group and H a convex normal subgroup. If X is the order type of H and Y is the order type of the induced order on G/H, then XY is the order type of G.

Proof. Since G/H is a set of representatives for the cosets, each element of G can be uniquely written as ah where a ∈ G/H and h ∈ H. If g_{1} ≠ g_{2}, g_{1} = a_{1}h_{1} and g_{2} = a_{2}h_{2}, then by the definition of the induced order g_{1} <_{G} g_{2} if and only if a_{1} <_{G/H} a_{2} or a_{1} = a_{2} and h_{1} <_{G} h_{2}. Suppose f_{H} : H → X and f_{G/H} : G/H → Y are the order preserving bijections. Define

f : G → XY
g ↦ ⟨f_{H}(h), f_{G/H}(a)⟩

where g = ah is the decomposition of g given above. This map is the desired order preserving bijection.

Definition 7.7. (RCA_{0}) Let G be an f.o. group. The set Arch(G) is a set of unique representatives of the Archimedean classes of G.

Arch(G) = {g ∈ G | ∀h ∈ G (h <_{ℕ} g → ¬(h ≈ g))}

In general, RCA_{0} is not strong enough to prove the existence of Arch(G). Arch(G) is ordered by taking x < y if and only if x ≪ y. If Arch(G) exists, then we can define a function a : G → Arch(G) which assigns to each g the element in Arch(G) to which it is Archimedean equivalent.

a(g) = d ⇔ d ∈ Arch(G) ∧ d ≈ g
⇔ d ∈ Arch(G) ∧ ∃n∃m (|g^{m}| > |d| ∧ |d^{n}| > |g|)
⇔ d ∈ Arch(G) ∧ ∀y <_{ℕ} g∀n∀m (y ≠ d → (|y^{m}| < |g| ∨ |g^{n}| < |y|))

Since a(g) has a Δ_{1}^{0} definition with Arch(G) as a parameter, it is definable in RCA_{0} if Arch(G) exists.

Lemma 7.8. (RCA_{0}) Let G be an f.o. group. Suppose Arch(G), Y = W (Arch(G)) and H = { g ∈ G | ∃y ∈ Y (g ≪ y ∨ g ≈ y) } exist and Y ≠ Arch(G). Then H is a convex normal subgroup of G and G/H has order type ℚ.

Proof. H is clearly a convex subgroup. To show H is normal, suppose h ∈ H, g ∈ G and ghg^{−1} ∉ H. Let a ∈ Arch(G) be such that a ≈ ghg^{−1}. Because ghg^{−1} ∉ H, it follows that a ∉ Y and there is an infinite descending chain f : ℕ → Arch(G) below a. By Lemma 6.17, f(n + 1) ≪ f(n) implies g^{−1}f(n + 1)g ≪ g^{−1}f(n)g. Define f̃ : ℕ → Arch(G) by setting f̃(n) to be the element of Arch(G) which is Archimedean equivalent to g^{−1}f(n)g. f̃ is an infinite descending chain below f̃(0), and so f̃(0) is not in Y. However, f̃(0) = h which contradicts the fact that h ∈ H.

Because G is fully ordered, RCA_{0} suffices to form G/H with the induced order. To finish the proof, it suffices to show that this order is dense with no endpoints. The key fact is that for any b ∈ Arch(G) \ Y there is a c ∈ Arch(G) \ Y such that c ≪ b. For example, if 1_{G}H < g_{1}H < g_{2}H, then there are b, c ∈ Arch(G) \ Y such that g_{1} ≈ b and c ≪ g_{1}. Since h ≪ c for all h ∈ H and g_{1} + |c| ≪ g_{2}, it follows that (g_{1} + |c|)H is strictly between g_{1}H and g_{2}H. The other cases showing that G/H is dense and has no endpoints are similar.

Lemma 7.9. (ACA_{0}) Let G be an f.o. group. If G has a least strictly positive element x then ∀g ∈ G(gx = xg) and the subgroup generated by x is convex.

Proof. Let P^{+} be the set of strictly positive elements of G. Suppose there is a g with gx ≠ xg. Without loss of generality assume gx < xg. It follows that gxg^{−1} < x. But, x ∈ P^{+} implies gxg^{−1} ∈ P^{+} which contradicts the fact that x is the least strictly positive element. The subgroup generated by x has a Σ_{1}^{0} definition, so its existence can be proved in ACA_{0}. The elements of this subgroups have the form x^{n} for n ∈ ℤ. Suppose there is an n ∈ ℤ and a c ∈ G such that x^{n} < c < x^{n+1}. It follows that 1_{G} < cx^{−n} < x which contradicts the hypothesis.

Lemma 7.10. (RCA_{0}) Let G be an f.o. group. If G contains elements a_{1}, a_{2} such that a_{1} < a_{2} and ∀g(a_{1} ≤ g ≤ a_{2} → (a_{1} = g ∨ a_{2} = g)), then G has a least strictly positive element.

Proof. Let x = a_{2}a_{1}^{−1}. 1_{G} < x and if 1_{G} < b < x, then a_{1} < ba_{1} < a_{2} which contradicts the hypothesis. Lemma 7.11. (Π_{1}^{1} −CA_{0}) For any linear order X, the well ordered initial segment W (X) exists. Proof. For each a ∈ X, let

T_{a} = { σ ∈ Fin_{X} | σ(0) = a ∧ ∀i < (lh(σ) − 1) (σ(i + 1) < σ(i)) }.

T_{a} has a path if and only if there is an infinite descending chain below a. Hence

W (X) = { a | T_{a} has no path }.

By Theorem 1.19, Π_{1}^{1} − CA_{0} proves that this set exists.

We are ready to prove that (1) implies (2) in Theorem 7.5.

Proof. Let G be an f.o. group. By Lemma 7.11, Π_{1}^{1} − CA_{0} suffices to prove that W (Arch(G)) exists. Let X = W (Arch(G)) if W (Arch(G) has a greatest element and otherwise let X be the well ordered obtain by adding a greatest element onto W (Arch(G)). We use ordinal notation for elements of X: 0 denotes the least element of X, β + 1 denotes the successor of β, and γ is a limit if γ has no immediate predecessor. For any β ∈ X, let

β̂ = { y ∈ X | y < β }.

The strategy is to use AT R_{0} to construct a chain of convex normal subgroups, A_{β} ⊆ G for β ∈ X. At each step, we prove the order type of A_{β} is ℤβ̂ and that unless β is the maximal element of X, A_{β} is strictly contained in G. If we reach a step where A_{β} cannot be extended to A_{β+1}, the construction terminates early.

Construction: Define A_{0} = {1_{G}}. Successor Step: Assume A_{β} is a convex normal subgroup, A_{β} ≠ G and the order type of A_{β} is ℤβ̂. G/A_{β} is an f.o. group with the induced order. There are two cases to consider:

1. If G/A_{β} has no least strictly positive element, then terminate the construction early at β. In this case, G/A_{β} has order type ℚ.

2. If G/A_{β} has a least strictly positive element, let a_{β+1} ∈ G represent this least

positive coset. Define A_{β+1} to be the subgroup generated by A_{β} and a_{β+1}. Limit Step: If λ is a limit ordinal in X, A_{λ} = ⋃_{β<λ} A_{β}. End of Construction

We need to verify that at each step of the construction, A_{β} is a convex normal subgroup with order type ℤβ̂. Consider the successor step β + 1. Since a_{β+1}A_{β} is the least positive element of G/A_{β}, Lemma 7.9 says that a_{β+1}A_{β} is in the center of G/A_{β} and the subgroup it generates is convex. This fact means that a_{β+1} commutes with elements of G modulo A_{β}. That is, for every g there is an a ∈ A_{β} such that ga_{β+1} = a_{β+1}ga. Thus any element of A_{β+1} can be written in the form a_{β+1}^{n}b for some n ∈ ℤ and b ∈ A_{β}. Also, since A_{β} is convex and a_{β+1} ∉ A_{β}, a_{β+1} is Archimedean greater than all the elements of A_{β}. We can now verify the following facts:

1. A_{β+1} is normal: Let x = a_{β+1}^{n}b. Because a_{β+1} commutes with elements of G

modulo A_{β}, there is a b̃ ∈ A_{β} such that

gxg^{−1} = ga_{β+1}^{n}bg^{−1} = a_{β+1}^{n}gb̃bg^{−1}.

The fact that A_{β} is normal implies that gb̃bg^{−1} ∈ A_{β} and therefore that gxg^{−1} ∈ A_{β+1}.

2. A_{β+1} is convex: If a_{β+1}^{n}b < z < a_{β+1}^{m}b̃ then a_{β+1}^{n}A_{β} ≤ zA_{β} ≤ a_{β+1}^{m}A_{β}. Since the subgroup of G/A_{β} generated by a_{β+1}A_{β} is convex, zA_{β} = a_{β+1}^{p}A_{β} for some p. It follows that z = a_{β+1}^{p}c for some c ∈ A_{β}, so z ∈ A_{β+1}.

3. The order type of A_{β+1}/A_{β} is ℤ: Elements of A_{β+1}/A_{β} are of the form a_{β+1}^{n}A_{β}. Since b ≪ a_{β+1} for all b ∈ A_{β}, it follows that a_{β+1}^{n} ≠ a_{β+1}^{m} modulo A_{β} if n ≠ m.

4. For all b ∈ A_{β+1}, either b ≪ a_{β+1} or b ≈ a_{β+1}: If a_{β+1} ≪ b, then a ≪ b for all a ∈ A_{β} and so b is not in the subgroup generated by a_{β+1} and A_{β}.

(4) shows that unless β + 1 is the maximum element of X, A_{β+1} ≠ G. By Lemma 7.6 and the induction hypothesis, (3) shows that the order type of A_{β+1} is ℤβ̂+1.

To check the properties at a limit step, assume λ is a limit in X. From the construction it is clear that A_{λ} is a convex normal subgroup and that unless λ is the maximum element of X, there are elements of Arch(G) above A_{λ}, and so A_{λ} ≠ G. For β < γ assume f_{β} : A_{β} → ℤβ̂ is an order preserving bijection. Define

f_{λ} : A_{λ} → ℤλ̂
a ↦ f_{β}(a)

where β is the least element of X such that a ∈ A_{β}. Notice that ℤβ̂ ⊂ ℤλ̂, so we can view f_{β}(a) as an element of ℤλ̂. This map is an order preserving bijection, so A_{λ} has the desired order type.

Since the construction may have terminated early and W (Arch(G)) may or may not be Arch(G), there are four cases to consider to finish the proof.

1. If W (Arch(G)) = Arch(G)

(a) If the construction terminates early at β, then A_{β} has order type ℤβ̂ and G/A_{β} has order type ℚ, so G has order type ℤβ̂ℚ.

(b) If the construction completes and β is the maximum element of X, then G = A_{β} and so G has order type ℤβ̂.

2. If Arch(G) is not well ordered:

(a) If the construction terminates early at β, then as in the first case, G has order type ℤβ̂ℚ

(b) If the construction is completed and β is the maximum element of X, then G/A_{β} has order type ℚ by Lemma 7.8 and G has order type ℤβ̂ℚ.

7.2 The Reversal

The goal of this section is to show that ACA_{0} suffices to prove that (3) implies (1) in Theorem 7.5. The proof takes place in two steps. First, we show that RCA_{0} plus statement (3) in Theorem 7.5 suffices to prove the well ordered initial segments of every linear order exists. Second, we use this fact plus some properties of the Kleene-Brouwer order on trees to prove in ACA_{0} that (3) implies (1).

Definition 7.12. (RCA_{0}) For a linear order X, U ⊆ X is dense if U has at least two elements and for every u, v ∈ U, if u <_{X} v then there is a w ∈ U such that u <_{X} w <_{X} v. Lemma 7.13. (RCA_{0}) Let X be a well order and U be a dense subset of ℤ^{X}. There are sequences of elements of U, u_{0}, u_{1}, … and v_{0}, v_{1}, … such that for each n ∈ ℕ

1. u_{n} < u_{n+1} < v_{n+1} < v_{n}

2. lh(u_{n}) > n and lh(v_{n}) > n

3. u_{n}(0) = v_{n}(0), u_{n}(1) = v_{n}(1), …, u_{n}(n) = v_{n}(n) Proof. If X = ∅, then ℤ^{X} has only one element and hence has no dense subsets. Assume that X ≠ ∅ and U ⊆ ℤ^{X} is dense. We define the sequences by induction starting with u_{0} and v_{0}. Claim. There are u ≠ v such that π_{1}(u(0)) = π_{1}(v(0)).

Suppose there are no such u and v. We will produce a contradiction to the fact that X is a well order. Since U is infinite, we can pick u and v such that either π_{2}(u(0)) and π_{2}(v(0)) are both positive or are both negative. Without loss of generality, assume they are both positive and u <_{ℤ^{X}} v. Define a function g : ℕ → ℤ^{X} such that g(0) = v and g(i + 1) is the ℕ-least element of U strictly between u and g(i). The density of U insures that g(i + 1) is defined. For any i ∈ ℕ we have

u <_{ℤ^{X}} g(i + 1) <_{ℤ^{X}} g(i) <_{ℤ^{X}} v.

We verify that π_{1}(u(0)) <_{X} π_{1}(g(i + 1)(0)). Assume that this inequality does not hold. By assumption, π_{1}(u(0)) ≠ π_{1}(g(i + 1)(0)), so we must have π_{1}(g(i + 1)(0)) <_{X} π_{1}(u(0)). However, by the definition of ≤_{ℤ^{X}} and because π_{2}(u(0)) > 0, this inequality implies that g(i + 1) <_{ℤ^{X}} u, which is a contradiction.

Because π_{1}(u(0)) <_{X} π_{1}(g(i + 1)(0)) and u <_{ℤ^{X}} g(i + 1), the definition of ≤_{ℤ^{X}} implies that π_{2}(g(i + 1)(0)) > 0. Therefore, we can apply the reasoning of the previous paragraph to g(i + 1) <_{ℤ^{X}} g(i) and conclude that π_{1}(g(i + 1)(0)) <_{X} π_{1}(g(i)(0)). Define the function h : ℕ → X by h(i) = π_{1}(g(i)(0)). The properties of g imply that h(i + 1) <_{X} h(i) for all i, which contradicts the fact that X is a well order and proves the claim.

Let u, v ∈ U be such that u < v and π_{1}(u(0)) = π_{1}(v(0)). Let U_{1} = {x ∈ U|u ≤ x ≤ v}. U_{1} is also a dense subset of X and for any x ∈ U_{1}, π_{1}(x(0)) = π_{1}(u(0)). To finish the n = 0 case, it suffices to find r, s ∈ U_{1} such that r ≠ s and π_{2}(r(0)) = π_{2}(s(0)). Suppose there are no such elements. If r ≠ s ∈ U_{1}, then

r <_{ℤ^{X}} s ⇔ π_{2}(r(0)) < π_{2}(s(0)).

However, if r ∈ U_{1}, then π_{2}(r(0)) is between π_{2}(u(0)) and π_{2}(v(0)). Thus there are a finite number of elements in U_{1}, which contradicts the density of U_{1}. Let ⟨u_{0}, v_{0}⟩ be the ℕ-least pair of elements of U such that u_{0} < v_{0} and u_{0}(0) = v_{0}(0).

The argument for the induction step is similar. Assume we have u_{n} and v_{n}. Consider the set V of elements x ∈ U with u_{n} ≤_{ℤ^{X}} x ≤_{ℤ^{X}} v_{n}. For any x ∈ V, x(i) = u_{n}(i) for 0 ≤ i ≤ n. By a notationally cumbersome, but similar argument, we can find r, s ∈ V such that r ≠ s and r(n + 1) = s(n + 1). Let ⟨u_{n+1}, v_{n+1}⟩ be the ℕ-least pair in V such that u_{n+1} < v_{n+1} and u_{n+1}(n + 1) = v_{n+1}(n + 1).

Lemma 7.14. (RCA_{0}) If X is a well order, then there are no dense subsets of ℤ^{X}.

Proof. Suppose X is a well order and U is a dense subset of ℤ^{X}. Let u_{0}, u_{1}, … and v_{0}, v_{1}, … be the sequences from Lemma 7.13. Define F : ℕ → X by

F (n) = π_{1}(u_{n}(n)).

F is an infinite descending chain which contradicts the fact that X is a well order.

Proposition 7.15. (RCA_{0}) (1) implies (2) where

1. For any countable abelian f.o. group A, there is a well order α and ϵ = 0 or 1 such that ℤ^{α}ℚ^{ϵ} is the order type of A.

2. The well ordered initial segment W (X) exists for all linear orders X.

The proof of Proposition 7.15 follows from the next two lemmas. Let X = {x_{0}, x_{1}, … } be an infinite linear order and G be the free abelian group on the generators {a_{0}, a_{1}, … }. Elements of G are represented by finite sums,

∑_{i∈I} r_{i}a_{i}

where I is a finite set and r_{i} ∈ ℤ \ {0}. For our purposes, it is more convenient to represent the elements of ℤ^{X} as finite sums rather than as sequences

∑_{i∈I} r_{i}x_{i}

where I is a finite set and r_{i} ∈ ℤ \ {0}. When G and ℤ^{X} are presented this way, there is a natural bijection between them that sends ∑_{i∈I} r_{i}a_{i} to ∑_{i∈I} r_{i}x_{i}.

X is used to define a full order on G. To compare two distinct elements, ∑_{i∈I} r_{i}a_{i} and ∑_{j∈J} s_{j}a_{j}, let K = I ∪ J, let r_{k} = 0 for k ∈ J \ I and let s_{k} = 0 for k ∈ I \ J.

Let n be such that x_{n} is X-maximal in {x_{k}|k ∈ K ∧ r_{k} ≠ s_{k}}. The order is defined by

∑_{i∈I} r_{i}a_{i} < ∑_{j∈J} s_{j}a_{j} ↔ r_{n} < s_{n}.

This definition yields a full order on G. Furthermore, under this order, the bijection from G to ℤ^{X} is order preserving. Statement (1) in Proposition 7.15 gives an order preserving bijection from G to ℤ^{α}ℚ^{ϵ} for some well order α and ϵ = 0 or 1. Lemma 7.16. X is a well order if and only if ϵ = 0. Proof. Case. (⇒)

Suppose X is a well order. Because there is an order preserving bijection between G and ℤ^{X} and between G and ℤ^{α}ℚ^{ϵ}, it follows that there is an order preserving bijection between ℤ^{X} and ℤ^{α}ℚ^{ϵ}. If ϵ = 1, then the set {⟨ ⟨⟩, q ⟩ | q ∈ ℚ} is dense in ℤ^{α}ℚ. Because RCA_{0} proves that the image of any subset of the domain of a bijection exists, the image of this set exists and is dense in ℤ^{X}. This statement contradicts Lemma 7.14. Case. (⇐)

Suppose X is not a well ordering and g : ℕ → X is an infinite descending chain in X. There is an infinite descending chain of generators in G,

… ≪ a_{g(2)} ≪ a_{g(1)} ≪ a_{g(0)}.

Let H be the subgroup of G generated by p_{n}a_{g(n)} where p_{n} is the n^{th} prime.

∑_{i∈I} r_{i}a_{i} ∈ H ↔ ∀i ∈ I ∃p_{n} ≤ |r_{i}| (g(n) = i ∧ p_{n} divides r_{i})

Because the quantification is bounded, this condition is Σ_{0}^{0}. It suffices to show that H is dense, for in that case, the image of H in ℤ^{α}ℚ^{ϵ} is dense and by Lemma 7.14, ϵ must be 1.

To show H is dense, consider two elements of H,

∑_{i∈I} r_{i}a_{i} < ∑_{j∈J} s_{j}a_{j}.

Define the coefficients in the sums for all the elements of I ∪ J by setting r_{j} = 0 for j ∈ J \ I and s_{i} = 0 for i ∈ I \ J. Let n be such that x_{n} is X-maximal in {x_{k}|k ∈ K ∧ s_{k} ≠ r_{k}}. n is in the range of g, so there is a k with g(k) = n. The element ∑_{i∈I} r_{i}a_{i} + p_{k+1}a_{g(k+1)} lies in H and is strictly between the two elements given above.

If X is not well ordered, then by Lemma 7.16 G has order type ℤ^{α}ℚ. Let f : G → ℤ^{α}ℚ be the order preserving bijection. For any x ∈ X, there is an associated generator a_{x} in G. f(a_{x}) has two components, one from ℤ^{α} and one from ℚ. The second component is the key to defining W (X).

Lemma 7.17. If X is not well ordered, then x ∈ X is in W (X) if and only if π_{2}(f(a_{x})) = π_{2}(f(1_{G})).

Proof. Case. (⇒)

Suppose x ∈ W (X). Let H be the subgroup generated by a_{y} for y ≤_{X} x. H exists since its elements are exactly those of the form ∑_{i∈I} r_{i}a_{i} where ∀i ∈ I(x_{i} ≤_{X} x). Claim. H is convex.

Let ∑_{i∈I} r_{i}a_{i} <_{G} ∑_{j∈J} s_{j}a_{j} be two elements in H and let g ∈ G lie strictly between them. ∑_{i∈I} r_{i}a_{i} <_{G} g implies that for all y >_{X} x, the coefficient of a_{y} in g is greater than or equal to 0. On the other hand, g <_{G} ∑_{j∈J} s_{j}a_{j} implies that for all y >_{X} x, the coefficient of a_{y} in g is less than or equal to 0. Hence, g ∈ H as required.

f(H) exists because f is a bijection and f(H) is convex because f is order preserving. Suppose π_{2}(f(a_{x})) ≠ π_{2}(f(1_{G})). The contradiction we derive is that f(H) has a dense subordering while H does not.

Define the well order X̂ by

X̂ = { y ∈ X | y ≤ x }.

Since H is the torsion free abelian group on the generators a_{y} with y ∈ X̂, it follows from the definition of ≤_{G} that the order type of H is ℤX̂. Lemma 7.14 shows that H has no dense suborderings.

As for f(H), since π_{2}(f(a_{x})) ≠ π_{2}(f(1_{G})) and 1_{G} ≤_{G} a_{x}, it follows that π_{2}(f(1_{G})) <_{ℚ} π_{2}(f(a_{x})). Thus, for any q ∈ ℚ strictly between these values, ⟨ ⟨⟩, q ⟩ is strictly between f(1_{G}) and f(a_{x}). Since f(H) is convex, the set of such points is in f(H). Thus f(H) has a dense subordering. Case. (⇐)

This case is similar to the proof of Lemma 7.16. Suppose x is not in the well ordered initial segment of X. Let g : ℕ → X be an infinite descending chain below x. This function also gives a descending chain of generators.

… ≪ a_{g(2)} ≪ a_{g(1)} ≪ a_{g(0)} Let H be the subgroup of G which is generated by the elements of the form p_{n}a_{g(n)} for n ≥ 1. Let P be the positive cone of G. As in the proof of Lemma 7.16, P ∩ H exists and is a dense subordering of G. Now, suppose that π_{2}(f(a_{x})) = π_{2}(f(1_{G})). To complete the proof, we show that f(P ∩ H) does not have a dense subordering. Claim. For any y ∈ f(H ∩ P), π_{2}(y) = π_{2}(f(1_{G})).

Suppose not. If π_{2}(y) <_{ℚ} π_{2}(f(1_{G})) then y <_{ℤ^{α}ℚ} f(1_{G}) and f^{−1}(y) <_{G} 1_{G}. This contradicts the fact that 1_{G} is the least element of f(H ∩P). If π_{2}(f(1_{G})) <_{ℚ} π_{2}(y), then by the similar reasoning and the fact that π_{2}(f(1_{G})) = π_{2}(f(a_{x})) we have that a_{x} <_{G} f^{−1}(y). However, since H is generated by p_{n}a_{g(n)} for n ≥ 1, any element of H is below a_{x}.

To show that f(H ∩ P) has no dense suborderings, let f̃: f(H ∩ P) → ℤ^{α} be the map that takes y to π_{1}(y). f̃ is order preserving and one-to-one, but is not necessarily a bijection. However, the range of f̃ is

{ z ∈ ℤ^{α} | ⟨z, π_{2}(f(1_{G}))⟩ ∈ f(H ∩ P) }.

This condition is Σ_{0}^{0}, so the range exists. Also, if U ⊆ f(H ∩ P) then

f̃(U) = { z ∈ ℤ^{α} | ⟨z, π_{2}(f(1_{G}))⟩ ∈ U }.

If U ⊆ f(H ∩ P) is dense, so is f̃(U) ⊆ ℤ^{α}. Hence, by Lemma 7.14, there are no dense subsets of f(H ∩ P)

We can now give a proof of Proposition 7.15.

Proof. If X is well ordered, then W (X) = X. Otherwise, given the definitions above

W (X) = {x ∈ X | π_{2}(f(a_{x})) = π_{2}(f(1_{G}))}.

We now define the Kleene-Brouwer order on a tree and use properties of this order to prove the reversal of Theorem 7.5 in ACA_{0}.

Definition 7.18. (RCA_{0}) The Kleene-Brouwer order, KB, on Fin_{ℕ} is given by: σ ≤_{KB} τ if and only if σ ⊇ τ or there is a j < min(lh(σ), lh(τ)) with σ(j) < τ(j) and σ(i) = τ(i) for all i < j. If T is a tree, then KB(T), the Kleene-Brouwer order of T, is KB ∩ (T × T).

Lemma 7.19. (ACA_{0}) A tree T has a path if and only if KB(T) is not a well ordering.

Lemma 7.20. (ACA_{0}) In Theorem 7.5, (3) implies (1).

Proof. Assume (3) and let ⟨T_{i} | i ∈ ℕ⟩ be a sequence of trees. By Theorem 1.19, it suffices to construct the set

{ i | T_{i} has a path }.

If none of the T_{i}’s has a path, then RCA_{0} can form X. Therefore, assume at least one T_{i} has a path.

First we define a tree T that contains each T_{i} as a subtree.

T = { ⟨⟩ } ∪ { i^{⌢}τ | τ ∈ T_{i} }

T is ordered by the Kleene-Brouwer order, KB(T). Since at least one T_{i} has a path, T has a path and KB(T) is not a well order. Define A_{i} by

A_{i} = { τ ∈ T | ⟨i − 1⟩ <_{KB} τ ≤_{KB} ⟨i⟩ }.

The map which sends τ ∈ T_{i} to i^{⌢}τ ∈ T is one-one, preserves the tree structure and has image A_{i}. Thus, T_{i} and A_{i} are isomorphic as trees, and KB(T_{i}) and KB(T) ∩ (A_{i} × A_{i}) are isomorphic as linear orders.

By Lemma 7.19, T_{i} has a path if and only if KB(T_{i}) is not well ordered, which holds if and only if KB(T) ∩ (A_{i} × A_{i}) is not well ordered. Thus it suffices to form the set:

{ i | KB(T) ∩ (A_{i} × A_{i}) is not well ordered }.

Let G be the torsion free group on generators the a_{τ}, τ ∈ T. Order the generators by a_{τ} ≪ a_{γ} if and only if τ <_{KB} γ. As above, this order of the generators determines a full order on G. By statement (3) in Theorem 7.5, there is an order preserving bijection f : G → ℤ^{α}ℚ^{ϵ}. Because KB(T) is not a well order, it follows that ϵ = 1. Claim. If Y ⊆ T and KB(T) ∩ (Y × Y) is well ordered, then there is an order preserving bijection between ℤ^{Y} and the subgroup generated by a_{τ} with τ ∈ Y, denoted ⟨a_{τ}|τ ∈ Y ⟩.

As shown above, when the elements of ⟨a_{τ}|τ ∈ Y ⟩ and ℤ^{Y} are represented as finite sums, there is a natural order preserving bijection between them. Claim. For each i, ⟨a_{τ}|τ ∈ A_{i}⟩ has a dense subordering if and only if A_{i} is not well ordered by KB(T) ∩ (A_{i} × A_{i}).

If A_{i} is well ordered, then by the first claim ⟨a_{τ}|τ ∈ A_{i}⟩ has no dense suborderings. If A_{i} is not well ordered, then we can use a descending chain to build a dense subordering as in Lemma 7.16. Claim. For each i > 0, ⟨a_{τ}|τ ∈ A_{i}⟩ has a dense subordering if and only if π_{2}(f(a_{⟨i−1⟩})) ≠ π_{2}(f(a_{⟨i⟩})).

Suppose π_{2}(f(a_{⟨i−1⟩})) ≠ π_{2}(f(a_{⟨i⟩})). Then

U = { γ ∈ T | π_{2}(f(a_{⟨i−1⟩})) < π_{2}(f(a_{γ})) < π_{2}(f(a_{⟨i⟩})) }

is contained in A_{i} by the definition of the Kleene-Brouwer ordering. Since the elements of T are all actually elements of ℕ we can trim this set down father.

V = { τ ∈ U | ∀γ ∈ U(γ <_{ℕ} τ → π_{2}(f(a_{γ})) ≠ π_{2}(f(a_{τ})) }

V ⊆ A_{i} is dense so by the second claim ⟨a_{τ}|τ ∈ A_{i}⟩ has a dense subordering. To prove the other direction, suppose π_{2}(f(a_{⟨i−1⟩})) = π_{2}(f(a_{⟨i⟩})) = q. The image of ⟨a_{τ}|τ ∈ A_{i}⟩ in ℤ^{α}ℚ is contained in ℤ^{α} × {q} which has no dense suborderings.

Similarly, we can show ⟨ a_{τ}|τ ≤ ⟨0⟩ ⟩ has a dense suborder if and only if π_{2}(f(a_{⟨0⟩})) ≠ π_{2}(f(1_{G})).

For i > 0 we have shown KB(T) ∩ (A_{i} × A_{i}) is not well ordered if and only if π_{2}(f(a_{⟨i−1⟩})) ≠ π_{2}(f(a_{⟨i⟩})). For i = 0, KB(T) ∩ (A_{0} × A_{0}) is not well ordered if and only if π_{2}(f(a_{0})) ≠ π_{2}(f(1_{G})). These equivalences give us a Σ_{0}^{0} condition to form the set

{ i | KB(T) ∩ (A_{i} × A_{i}) is not well ordered }.

7.3 Coding Turing Jumps

This section shows that RCA_{0} can prove the implication from (3) to (1) in Theorem 7.5. The idea is that for any given set Y, RCA_{0} can produce a linear order X such that the well ordered initial segment of X codes the Turing jump of Y. Thus, by Proposition 7.15, any model of RCA_{0} and condition (3) from Theorem 7.5 is closed under Turing jumps and hence is a model for ACA_{0}. By Gödel’s Completeness Theorem, RCA_{0} plus condition (3) suffice to prove ACA_{0}, which in turn proves Π_{1}^{1} − CA_{0}. Once we have proved Lemma 7.23, the proof of Theorem 7.5 will be complete. Definition 7.21. (RCA_{0}) Let π(e, r, X) be a Π_{1}^{0} formula with exactly the displayed variables free. π is universal lightface Π_{1}^{0} if for all Π_{1}^{0} formulas ψ(e, r, X) with the same free variables

RCA_{0} ⊢ ∀e∃e′∀r∀X (π(e′, r, X) ↔ ψ(e, r, X)).

Definition 7.22. (RCA_{0}) Let π(e, r, Y) be a fixed universal lightface Π_{1}^{0} formula. The Turing jump of Y, T J(Y), is given by

T J(Y) = { ⟨e, r⟩ | π(e, r, Y) }.

Notice that in contrast to the usual definition of the Turing jump as a complete Σ_{1}^{0} set, this definition makes it a complete Π_{1}^{0} set. Lemma 7.23. (RCA_{0}) (1) implies (2):

1. Let G be a countable abelian f.o. group. There is a well order α and ϵ = 0 or 1 such that ℤ^{α}ℚ^{ϵ} is the order type of G.

2. ACA_{0}

The strategy for building X so that W (X) codes T J(Y) is to use a marker construction. The function m(n, t) gives the marker for n at stage t. At stage t we will only have markers for n ≤ t, so m(n, t) = 0 for all n > t. For this reason, X will order ℕ^{+}. There is also a function b(t) such that at the end of stage t, X_{t} will order 1, … b(t). While the markers may change during the construction, the limit of m(n, t) always exists and

n ∈ T J(Y) ↔ lim_{t→∞} m(n, t) ∈ W (X).

To build X we keep track of which elements we know are not in T J(Y) and which elements we guess may be in T J(Y). Because the formula π(e, r, Y) introduced in the definition of the Turing jump is Π_{1}^{0}, it can be written as ∀xφ(e, r, x, Y). If we find a number x such that ¬φ(e, r, x, Y) then we know ⟨e, r⟩ ∉ T J(Y). Once we find such a witness, we begin to build an infinite descending chain below the marker for ⟨e, r⟩ so that this marker will not be in W (X).

To keep track of this knowledge, at stage t each number n < b(t) is labeled either not fixed or possibly fixed. A number labeled not fixed will have an infinite descending chain below it when the construction of X is completed. Once a number is labeled not fixed, it remains not fixed forever. However, a possibly fixed number may be changed at a later stage to not fixed. However, as long as a number remains possibly fixed, nothings enters X below it. Thus, at each stage the possibly fixed numbers form an initial segment of the part of X defined so far.

At stage t + 1 of the construction, we check if ¬φ(e, r, x, Y) holds where t = ⟨e, r, x⟩. If it does not hold, then we do not yet have a witness for ⟨e, r⟩ ∉ T J(Y). In this case, we want to extend the construction of the infinite descending chain below the not fixed elements and otherwise leave X as it is.

However, if ¬φ(e, r, x, Y) and m(⟨e, r⟩, t) is possibly fixed, then we have found a witness to ⟨e, r⟩ ∉ T J(Y) and we did not already know this fact. In this case we label m(⟨e, r⟩, t) not fixed and begin to build a descending chain below m(⟨e, r⟩, t). The complication is that there may be other numbers m(n, t) which are possibly fixed and such that m(⟨e, r⟩, t) <_{X} m(n, t). We do not want to build a descending chain below m(n, t). To handle this problem, we define the marker m(n, t + 1) to be a new large number, place this number below m(⟨e, r⟩, t) in X, label the number m(n, t) not fixed since we no longer care about it, and label m(n, t + 1) possibly fixed. This redefinition of markers will happen only finitely often for each n.

We now give the formal construction. X_{t} denotes the portion of X defined by the end of stage t. Construction: Stage 0: Define m(n, 0) = 0 for all n > 0, m(0, 0) = 1, and b(0) = 1. X_{0} consists of a single element 1 which is labeled possibly fixed. Stage t+1: By induction, assume the following facts hold at the end of stage t.

1. Each l ≤ b(t) except 0 is labeled either not fixed or possibly fixed.

2. The numbers labeled possibly fixed form an initial segment of X_{t}.

3. If l is not fixed, then either l ≠ m(n, t) for any n ≤ t or l = m(n, t), n = ⟨e, r⟩ and ∃k ≤ t(¬φ(e, r, k, Y)).

4. If l is possibly fixed then there is an n ≤ t such that l = m(n, t).

5. If m(n, t) and m(ñ, t) are both possibly fixed, then m(n, t) <_{X} m(ñ, t) if and only if n <_{ℕ} ñ.

6. b(t) is the least not fixed element in X_{t}.

There are two cases to consider in the construction. Assume t = ⟨e, r, x⟩. Case. ¬φ(e, r, x, Y) and m(⟨e, r⟩, t) is possibly fixed.

We have a new witness to the fact that ⟨e, r⟩ ∉ T J(Y). Let n = ⟨e, r⟩ and suppose X_{t} looks like

\underbrace{m(b_{l}, t) < … < m(b_{1}, t) < m(n, t) < m(a_{1}, t) < … < m(a_{k}, t)}_{possibly fixed} \underbrace{< b(t) < …}_{not fixed} .

For all c, define m(c, t + 1) as follows. If c > t + 1, m(c, t) is not fixed, or m(c, t) ≤_{X} m(n, t), then m(c, t + 1) = m(c, t). For the elements m(a_{1}, t) through m(a_{k}, t), define m(a_{i}, t + 1) = b(t) + i and define m(t + 1, t + 1) = b(t) + k + 1. The numbers m(a_{i}, t) and m(n, t) = m(n, t + 1) have their labels changed to not fixed. The new elements m(a_{i}, t+1) are labeled possibly fixed and are placed below m(n, t+1), above m(b_{1}, t+1) and in ≤_{ℕ}-order on the a_{i}’s. The new element b(t)+k+1 = m(t+1, t+1) is labeled possibly fixed and inserted as the largest possibly fixed element. Finally, the number b(t) + k + 2 is labeled not fixed and added to X as the least not fixed element. b(t + 1) is set to b(t) + k + 2. X_{t+1} looks like

m(b_{l}, t + 1) < … m(b_{1}, t + 1) < m(a_{1}, t + 1) < …

possibly fixed

\underbrace{… < m(a_{k}, t + 1) < m(t + 1, t + 1)}_{possibly fixed} < \underbrace{b(t + 1) < m(n, t + 1) < …}_{not fixed} .

Case. Either φ(e, r, x, Y) or m(⟨e, r⟩, t) is labeled not fixed.

Define m(c, t+1) = m(c, t) for all c except c = t+1, define m(t+1, t+1) = b(t)+1 and define b(t + 1) = b(t) + 2. The labels on n ≤ b(t) remain the same, b(t) + 1 is labeled possibly fixed and b(t) + 2 = b(t + 1) is labeled not fixed. b(t) + 1 is added to X as the greatest possibly fixed element and b(t + 1) is added as the least not fixed element. If X_{t} had the numbers n_{1} and n_{2} as the greatest possibly fixed and the least not fixed elements respectively, then X_{t+1} looks like

\underbrace{… < n_{1} < m(t + 1, t + 1)}_{possibly fixed} < \underbrace{b(t + 1) < n_{2} < …}_{not fixed} .

Notice that in either case, the inductive assumptions are satisfied at the end of stage t + 1. End of Construction

The following facts are clear from the construction.

1. If m(n, t) is not fixed, then ∀t′ ≥ t (m(n, t′) = m(n, t)).

2. If m(n, t) and m(n′, t) are possibly fixed, then m(n, t) <_{X} m(n′, t) if and only if n <_{ℕ} n′.

3. If n ∉ T J(Y) and n = ⟨e, r⟩, then there is a least t = ⟨e, r, x⟩ such that ¬φ(e, r, x, Y). The number m(n, t + 1) is labeled not fixed at stage t + 1.

4. If n ∈ T J(Y) and n = ⟨e, r⟩ then for all t = ⟨e, r, x⟩, we have φ(e, r, x, Y) and hence m(n, t) is never labeled not fixed at stage t + 1. If the number m(n, t) is later labeled not fixed at stage t′ > t then m(n, t′) ≠ m(n, t).

5. If m(n, t) is labeled possibly fixed and m(a, t) <_{X} m(n, t) with m(a, t + 1) ≠ m(a, t), then m(n, t + 1) ≠ m(n, t).

It remains to show that lim_{t→∞} m(n, t) exists for all n, that there is an infinite descending chain below lim_{t→∞} m(n, t) in X if and only if n ∉ T J(Y), and that this fact can be used to give a Δ_{1}^{0} definition of T J(Y).

Lemma 7.24. (RCA_{0}) For all n, there exists t such that for all t′ > t, m(n, t′) = m(n, t). That is, lim_{t→∞} m(n, t) exists.

Proof. This lemma cannot be proved by induction in RCA_{0} because it would require Σ_{2}^{0} induction. We fix n and prove the lemma for this specific n. Case. n ∉ T J(Y) and n = ⟨e, r⟩

There is a t = ⟨e, r, x⟩ with ¬φ(e, r, x, Y). At stage t + 1, m(n, t + 1) is labeled not fixed and ∀t′ > t (m(n, t′) = m(n, t + 1)). Case. n ∈ T J(Y)

Notice that m(n, t + 1) ≠ m(n, t) if and only if t = ⟨e, r, x⟩, ¬φ(e, r, x, Y), m(⟨e, r⟩, t) is labeled possibly fixed, and m(⟨e, r⟩, t) <_{X} m(n, t). Since m(n, t) is labeled possibly fixed at stage t, it follows from m(⟨e, r⟩, t) <_{X} m(n, t) that ⟨e, r⟩ <_{ℕ} n. Thus we have

m(n, t) ≠ m(n, t + 1)

if and only if

∃⟨e, r⟩ ≤ n∃x ≤ t∀k ≤ x ( t = ⟨e, r, x⟩ ∧ ¬φ(e, r, x, Y) ∧ φ(e, r, k, Y) ).

Therefore the set

{ t | m(n, t) ≠ m(n, t + 1) }

exists. It is finite since for each ⟨e, r⟩ ≤ n there is at most one x such that ¬φ(e, r, x, Y) and φ(e, r, k, Y) for all k ≤ x. Hence lim_{t→∞} m(n, t) exists.

Lemma 7.25. (RCA_{0}) If n ∉ T J(Y) then there is a t ≥ n such that m(n, t) is labeled not fixed at stage t.

Proof. Let t = ⟨e, r, x⟩ where ¬φ(e, r, x, Y) and ∀k ≤ t(φ(e, r, k, Y)).

Lemma 7.26. (RCA_{0}) If n ∉ T J(Y) and m(n, t̃) is labeled not fixed at stage t̃, then m(n, t̃) = lim_{t→∞} m(n, t) and there is an infinite descending chain below m(n, t̃) in X.

Proof. The function f : ℕ → ℕ given by k ↦ b(t̃+k) is an infinite descending chain below m(n, t̃).

In fact it is clear that if n is any number labeled not fixed at stage t̃ then the function f in the proof of the previous lemma gives an infinite descending chain below n.

Lemma 7.27. (RCA_{0}) If n ∈ T J(Y) and lim_{t→∞} m(n, t) = ñ then there is a finite number of elements below ñ in X.

Proof. Let t̃ be such that m(n, t̃) = ñ. Since m(n, t) is an increasing function, m(n, t) never changes after t̃ and since n ∈ T J(Y), ñ is labeled possibly fixed at all stages after t̃. Because ñ is labeled possibly fixed, nothing is ever placed below ñ in X after stage t̃. Since the size of Xt̃ is bounded by b(t̃), there are only a finite number of elements below ñ in X.

Lemma 7.28. (RCA_{0}) n ∈ T J(Y) if and only if ∃t(m(n, t) ∈ W (X)).

Proof. We have shown that any number labeled not fixed has an infinite descending chain below it and that n ∈ T J(Y) if and only if lim_{t→∞} m(n, t) ∈ W (X). Notice that if m(n, t+1) ≠ m(n, t), then the number m(n, t) is labeled not fixed and hence m(n, t) ∉ W (X). Thus, a number k is in W (X) if and only if k = lim_{t→∞} m(n, t) for some n and n ∈ T J(Y). However, if m(n, t̃) ∈ W (X), then the label on m(n, t̃) never changes to not fixed. Hence, m(n, t̃) = lim_{t→∞} m(n, t) and so n ∈ T J(Y). If m(n, t) ∉ W (X) for all t, then in particular lim_{t→∞} m(n, t) ∉ W (X) and so n ∉ T J(Y).

To complete the proof of Lemma 7.23, notice that the original definition of T J(Y) was given by a Π_{1}^{0} formula. Lemma 7.28 gives a definition of T J(Y) that is Σ_{1}^{0} in W (X). However, by Proposition 7.15, we know that RCA_{0} plus statement (1) of Lemma 7.23 imply that W (X) exists. Hence, assuming statement (1), T J(Y) is defined by a Δ_{1}^{0} condition. This step completes the proof of Lemma 7.23, which in turn completes the proof of Theorem 7.5.

Spaces of Orders and Computable

Presentations

In this chapter, we turn our attention to two questions from Downey and Kurtz (1986). First, is every c.b. Π_{1}^{0} class representable up to Turing degree by the space of orders on some computable torsion free abelian group? Second, is every fully orderable computable group classically isomorphic to a computably fully orderable computable group? Since we will only be concerned with full orders, the terms order and full order will be used interchangeably.

8.1 Π_{1}^{0} Classes and Spaces of Orders

Recall that for an orderable group G, 𝕏(G) denotes the space of positive cones of full orders on G. We proved the following connection between 𝕏(G) and c.b. Π_{1}^{0} classes in Chapter 3.

Theorem 8.1. Let G be a fully orderable computable group. There is a c.b. Π_{1}^{0} class C and a Turing degree preserving bijection φ : 𝕏(G) → C.

Metakides and Nerode (1979) proved a similar result for ordered fields. Let 𝕏(F) denote the set of positive cones of orders on an orderable field F. As with groups, there is a set of algebraic conditions that determines if a subset of F is the positive cone of a full order.

Theorem 8.2 (Metakides and Nerode (1979)). Let F be a computable field. There is a c.b. Π_{1}^{0} class C and a Turing degree preserving bijection φ : 𝕏(F) → C.

Metakides and Nerode (1979), however, carried this argument one step farther. They proved that every c.b. Π_{1}^{0} class can be represented by 𝕏(F) for some computable field F.

Theorem 8.3 (Metakides and Nerode (1979)). Let C be a c.b. Π_{1}^{0} class. There is a computable field F and a degree preserving bijection φ : 𝕏(F) → C.

Downey and Kurtz (1986) asked if there is a result similar to Theorem 8.3 for computable torsion free abelian groups. Unfortunately, it is not possible to have a result this strong. For groups, if P is the positive cone of a full order and

P^{−1} = { g^{−1} | g ∈ P }.

then P^{−1} is also the positive cone of a full order. Notice that P ≡_{T} P^{−1}, so any degree which contains a member of 𝕏(G) contains at least two members of 𝕏(G). However, there is a c.b. Π_{1}^{0} class C such that for any f, g ∈ C, if f ≠ g then deg(f) ≠ deg(g). Therefore, there cannot be a degree preserving bijection between C and 𝕏(G) for any fully orderable computable group G.

There are two natural modifications of the question of representing c.b. Π_{1}^{0} classes. Given a c.b. Π_{1}^{0} class C:

1. Is there a computable torsion free abelian group G such that

{ deg(f)|f ∈ C } = { deg(P)|P ∈ 𝕏(G) } ?

2. Is there a fully orderable computable group G and a two-to-one degree preserving map φ : 𝕏(G) → C?

The answer to both questions turns out to be no. Notice that if the answer to (1) is no, then the answer to (2) is also no. Before answering (1), we need some abelian group theory.

Definition 8.4. Let G be an abelian group. The elements g_{0}, …, g_{n} ∈ G \ 1_{G} are linearly independent if and only if for any α_{0}, …, α_{n} ∈ ℤ the equality

α_{0}g_{0} + α_{1}g_{1} + ⋯ + α_{n}g_{n} = 0_{G}

implies α_{i}g_{i} = 0_{G} for all i. If G is torsion free, this condition means that α_{i} = 0 for all i. An infinite set B ⊆ G \ 1_{G} is linearly independent if every finite subset of B is linearly independent. A maximal set of linearly independent elements is called a basis for G and the cardinality of any basis is called the rank of G.

For example, G has rank 1 if and only if G is isomorphic to a subgroup of ℚ. If G is a torsion free divisible abelian group, then G can be viewed as a vector space over ℚ. In this case, the definitions of linear independence, basis and rank for G as a group agree exactly with the definitions of the same terms for G as a vector space.

Our study of computable torsion free abelian groups breaks into three categories: groups of rank 1, groups of finite rank > 1, and groups of infinite rank. Recall that if G is a computable torsion free abelian group and D is a computable divisible closure of G, then there is a degree preserving bijection φ : 𝕏(G) → 𝕏(D). Therefore, when studying the cardinality of 𝕏(G) or the degrees of members of 𝕏(G), we can assume without loss of generality that G is divisible. Lemma 8.5. If G is a computable torsion free abelian group of rank 1, then G has exactly two orders, both of which are computable. Proof. Any divisible closure of G is computably isomorphic to ℚ. ℚ has exactly two orders both of which are computable, so G has exactly two orders both of which are computable. Lemma 8.6. If G is a computable torsion free abelian group with finite rank strictly greater than 1, then G has 2^{ω} orders and has orders of every Turing degree. Proof. We assume that G is divisible. Since G has finite rank, we can also assume that we have a basis for G. For simplicity, we consider the case when G has rank 2 and later prove the lemma for larger ranks.

Let {a, b} be a basis for G. Any element of G can be written as p_{1}a + p_{2}b for some p_{1}, p_{2} ∈ ℚ. The lexicographic order on G is computable:

p_{1}a + p_{2}b ∈ P ⇔ p_{1} > 0 ∨ (p_{1} = 0 ∧ p_{2} ≥ 0).

Therefore, G has a computable order. Let a be an arbitrary noncomputable degree and r ∈ a be a real such that 0 < r < 1. Notice that r must be irrational since 0 <_{T} r. We define the map f_{r} : G → ℝ that sends p_{1}a + p_{2}b to p_{1} + p_{2}r. Because r is irrational, this map is a monomorphism and defines an isomorphism between G and a subgroup of ℝ. We use the structure of ℝ to define a full order ≤_{r} on G. For g, h ∈ G

g ≤_{r} h ⇔ f_{r}(g) ≤_{ℝ} f_{r}(h).

It remains to show that deg(r) = deg(≤_{r}), for then we have deg(≤_{r}) = a. Claim. deg(r) ≤_{T} deg(≤_{r})

The idea is to use ≤_{r} to compute a binary expansion for r. We need to determine the coefficients a_{i} ∈ {0, 1} in

r = ∑_{i=1}^{ω} \frac{a_{i}}{2^{i}} .

To compute a_{1}, notice that

r >_{ℝ} \frac{1}{2} ⇔ 2r >_{ℝ} 1 ⇔ 2b >_{r} a

r <_{ℝ} \frac{1}{2} ⇔ 2r <_{ℝ} 1 ⇔ 2b <_{r} a.

Using ≤_{r} we compute:

2b >_{r} a ⇒ a_{1} = 1
2b <_{r} a ⇒ a_{1} = 0.

To compute a_{n+1}, we assume by induction that we know a_{1}, …, a_{n} and that

∑_{i=1}^{n} \frac{a_{i}}{2^{i}} <_{ℝ} r <_{ℝ} ∑_{i=1}^{n} \frac{a_{i}}{2^{i}} + \frac{1}{2^{n}} .

To find a_{n+1}, we need to know if

r <_{ℝ} ∑_{i=1}^{n} \frac{a_{i}}{2^{i}} + \frac{1}{2^{n+1}} .

If this inequality holds, then a_{n+1} = 0 and otherwise a_{n+1} = 1. By the definition of ≤_{r}, we have that

r <_{ℝ} ∑_{i=1}^{n} \frac{a_{i}}{2^{i}} + \frac{1}{2^{n+1}} ⇔ b <_{r} ∑_{i=1}^{n} \frac{a_{i}}{2^{i}} a + \frac{a}{2^{n+1}} .

Note that the required induction hypothesis holds once we set a_{n+1} to the correct value. The claim is proved. Claim. deg(≤_{r}) ≤_{T} deg(r)

Assume we have coefficients a_{i} ∈ {0, 1} for the sum

r = ∑_{i=1}^{ω} \frac{a_{i}}{2^{i}} .

We need to be able to compare elements of the form

\frac{p}{q}a + \frac{p′}{q′} b and \frac{m}{n}a + \frac{m′}{n′} b

where the numbers in the fractions are from ℤ and q, q′, n, n′ > 0. If we multiply by qq′nn′ we see that it suffices to be able to compare elements of the form

n_{1}a + m_{1}b and n_{2}a + m_{2}b.

However, by collecting terms we have

n_{1}a + m_{1}b ≤_{r} n_{2}a + m_{2}b ⇔ (n_{1} − n_{2})a ≤_{r} (m_{2} − m_{1})b.

Therefore, it suffices to be able to compare na and mb for n, m ∈ ℤ. We will further assume that n, m ∈ ℕ\0 since the other cases are either easy (i.e. n = 0 and m > 0) or reduce to this case (i.e. n < 0 and m < 0).

The map f_{r} sends na ↦ n and mb ↦ mr. To compare n and mr as elements of ℝ, we need to compute an approximation to r to an accuracy such that there are no integers within the error bounds for m times the approximation. For any k we know that

m · ∑_{i=1}^{k} \frac{a_{i}}{2^{i}} <_{ℝ} m · r <_{ℝ} m · (∑_{i=1}^{k} \frac{a_{i}}{2^{i}} + \frac{1}{2^{k}} ) .

We need to find a k such that there are no integers in the interval

[ m · ∑_{i=1}^{k} \frac{a_{i}}{2^{i}} , m · (∑_{i=1}^{k} \frac{a_{i}}{2^{i}} + \frac{1}{2^{k}} ) ] .

Once we know such a k exists, we can find it by searching. Because mr is irrational, it must sit strictly between two integers. Let d be the distance from mr to the nearest integer and let k be such that m/2^{k} < d. It follows that

m · (r − ∑_{i=1}^{k} \frac{a_{i}}{2^{i}} ) < \frac{m}{2^{k}} < d

and

m · (∑_{i=1}^{k} \frac{a_{i}}{2^{i}} + \frac{1}{2^{k}} − r) < \frac{m}{2^{k}} < d.

Thus, there is a k as required for the approximation. We can now compare na and mb as follows:

na <_{r} mb ⇔ n <_{ℝ} mr ⇔ n <_{ℝ} m · ∑_{i=1}^{k} \frac{a_{i}}{2^{i}} .

The claims show that deg(r) = deg(≤_{R}) and hence G has orders of every degree. Since there are 2^{ω} degrees, G has 2^{ω} distinct orders. This completes the proof for groups of rank 2.

Assume G is a computable torsion free abelian group of finite rank n > 2. We assume that G is divisible and that we have a basis {g_{1}, …, g_{n}} for G. Any element of G can be written as

q_{1}g_{1} + ⋯ + q_{n}g_{n}

for some q_{1}, …, q_{n} ∈ ℚ. G is the direct product of the computable subgroups generated by {g_{1}, g_{2}} and {g_{3}, …, g_{n}}, so we can lexicographically order G using the orders on these subgroups as in Theorem 3.21.

For any degree a, the subgroup generated by {g_{1}, g_{2}} has an order of degree a. The subgroup generated by {g_{3}, …, g_{n}} is computably orderable and therefore, the direct product has an order of degree a. Lemma 8.7. If G is a computable torsion free abelian group with infinite rank, then G has 2^{ω} distinct orders and has orders of every degree a ≥_{T} 0′ Proof. Unlike in the proof of Lemma 8.6, we cannot assume that we have a basis for G. However, there is a basis computable in 0′. Define

A = { ⟨g_{1}, …, g_{n+1}⟩ | ∃⟨q_{1}, …, q_{n}⟩ (g_{n+1} = q_{1}g_{1} + ⋯ + q_{n}g_{n}) }.

A is defined by a Σ_{1}^{0} statement, so A ≤_{T} 0′. To obtain a basis for G, define a sequence of finite sequences

⟨e_{1}⟩, ⟨e_{1}, e_{2}⟩, …, ⟨e_{1}, …, e_{n}⟩, …

by: e_{1} is the least nonidentity element of G and e_{n+1} is the ℕ-least element of G such that ⟨e_{1}, …, e_{n}, e_{n+1}⟩ ∉ A. A basis for G is given by

B = { g | g appears in a sequence ⟨e_{1}, …, e_{n}⟩ }.

B is computable from A, so B ≤_{T} 0′.

Once we have a basis e_{1}, e_{2}, … for G, we can write each element of G as

q_{1}e_{1} + ⋯ + q_{n}e_{n}

for some q_{1}, … q_{n} ∈ ℚ with q_{n} ≠ 0. We split G into the direct product of the subgroup generated by {e_{1}, e_{2}} and the subgroup generated by {e_{3}, e_{4}, … }. The lexicographic order on the subgroup generated by {e_{3}, e_{4}, … } is computable from the basis B.

Suppose a is any degree above 0′. The basis B is computable from a and so the subgroup generated by {e_{3}, e_{4}, … } has an order computable in a. By Lemma 8.6, the subgroup generated by {e_{1}, e_{2}} has an order exactly of degree a. Therefore, the lexicographic product of these orders has degree exactly a.

Theorem 8.8. There is a c.b. Π_{1}^{0} class C such that for any computable torsion free abelian group G

{ deg(f) | f ∈ C } ≠ { deg(p) | P ∈ 𝕏(G) }.

Proof. Recall from Theorem 1.9 that there is an infinite c.b. Π_{1}^{0} class C such that for any f, g ∈ C, if f ≠ g then f and g are Turing incomparable. Let C be such a c.b. Π_{1}^{0} class. Let G be any computable torsion free abelian group. By Lemmas 8.5, 8.6 and 8.7 we know that G has either only computable orders, orders of every degree or orders of every degree above 0′. In any of these cases, it is impossible for the set of degrees of elements of 𝕏(G) to be equal to the set of degrees of elements of C.

Theorem 8.8 shows that the spaces of orders on computable torsion free abelian groups do not suffice to represent all c.b. Π_{1}^{0} classes even in a weak sense. The next theorem shows that they cannot represent even the restricted class of Π_{1}^{0} classes of separating sets.

Theorem 8.9. There is a Π_{1}^{0} class of separating sets C such that for any computable torsion free abelian group G

{ deg(f) | f ∈ C } ≠ { deg(p) | P ∈ 𝕏(G) }.

Proof. The proof is exactly the same as the proof of Theorem 8.8 except that it relies on a different result about Π_{1}^{0} classes of separating sets. From Theorem 1.10 we know that there is a Π_{1}^{0} class of separating sets C such that for any A, B ∈ C either A ≡_{T} B or A and B are Turing incomparable.

8.2 Extension to Nilpotent Groups

Recall from Theorem 3.34 that the space of orders of any orderable computable group can be represented by a c.b. Π_{1}^{0} class. So far in this chapter we have only looked at torsion free abelian groups. Perhaps we need to widen our view to the class of all fully orderable groups if we want to represent all c.b. Π_{1}^{0} classes. This problem of whether all orderable computable groups suffice to represent c.b. Π_{1}^{0} classes is still open. We can, however, extend the negative results from abelian groups to nilpotent groups. The goal for this section is to show that there is a c.b. Π_{1}^{0} class C such that for any computable torsion free nilpotent group G we have

{ deg(f) | f ∈ C } ≠ { deg(p) | P ∈ 𝕏(G) }.

Before proving results about the number and complexity of full orders on nilpotent groups, we need a description of a general method for building these full orders. Recall that if N is a normal subgroup of an O-group G, then a full order ≤_{N} on N is called a G-order if for any a, b ∈ N and g ∈ G

a ≤_{N} b ⇒ gag^{−1} ≤_{G} gbg^{−1}.

Let P (N) be the positive cone of a full G-order on N and ≤_{G/N} be a full order on the quotient group. These orders induce a full order on G defined by

g ≤_{G} h ⇔ aN <_{G/N} bN ∨ (aN = bN ∧ a^{−1}b ∈ P (N)).

Under ≤_{G}, N is convex and the induced orders on N and G/N are the same as the ones used to build ≤_{G}.

Let G be a torsion free nilpotent group. Recall that this means G has a finite upper central series

⟨1_{G}⟩ = ζ_{0}(G) ≤ ζ_{1}(G) ≤ ⋯ ≤ ζ_{n}(G) = G

where ζ_{1}(G) is the center of G and for each 0 ≤ i < n, ζ_{i+1}(G)/ζ_{i}(G) is the center of G/ζ_{i}(G). By Lemma 3.17, we know that each ζ_{i+1}(G)/ζ_{i}(G) is a torsion free abelian group and therefore is fully orderable. We will use this structure to build orders on G under which the terms in the upper central series are convex.

To order G, start by ordering the torsion free abelian subgroup ζ_{1}(G). Because ζ_{1}(G) is the center of G, any order we put on it will be a G-order. Next, consider ζ_{2}(G)/ζ_{1}(G). This quotient group is torsion free and abelian, so it is orderable. Because ≤_{ζ_{1}(G)} is a G-order, and hence a ζ_{2}(G)-order, we have an induced order, ≤_{ζ_{2}(G)}, on ζ_{2}(G).

The key fact is that the order ≤_{ζ_{2}(G)} is a G-order. Suppose a <_{ζ_{2}(G)} b. There are two cases to consider. If aζ_{1}(G) ≠ bζ_{1}(G), then it must be that aζ_{1}(G) < bζ_{1}(G). Because ζ_{2}(G)/ζ_{1}(G) is the center of G/ζ_{1}(G) we have that

gag^{−1}ζ_{1}(G) = aζ_{1}(G) < bζ_{1}(G) = gbg^{−1}ζ_{1}(G).

Therefore, by the definition of the induced order, gag^{−1} <_{ζ_{2}(G)} gbg^{−1}. The second case is when aζ_{1}(G) = bζ_{1}(G). In this case, gag^{−1}ζ_{1}(G) = gbg^{−1}ζ_{1}(G) and the order on these elements is determined by ≤_{ζ_{1}(G)}, which we already know is a G-order.

Now that we have ordered ζ_{2}(G), we proceed up the upper central series ordering each term from the order on the previous term and the order on the quotient. This method does not construct all the possible orders on G, but it will give enough for our purposes. In particular, it is possible to have orders under which the terms in the upper central series are not convex and this procedure will never yield an order like that.

Proposition 8.10. Let G be a properly 2 step torsion free nilpotent group and C be the center of G. The rank of G/C is greater than or equal to 2.

Proof. For a contradiction, suppose that the rank of G/C is 1. This statement is equivalent to saying that G/C is isomorphic to a subgroup of ℚ. Since G is properly 2 step nilpotent, and hence not abelian, there must be elements a, b ∈ G such that ab ≠ ba. Thus, neither a nor b is in C and so aC ≠ 1_{G}C and bC ≠ 1_{G}C.

Since G/C is a subgroup of ℚ, there must be integers p, q ≠ 0 such that

a^{p}C = b^{q}C.

This equality implies that there is a c ∈ C such that a^{p} = b^{q}c and we get

a^{p}b = b^{q}cb = b^{q}bc = bb^{q}c = ba^{p}.

Thus a^{p} commutes with b and so [a^{p}, b] = 1_{G}. By commutator identities we have that

[x^{2}, y] = [x, y] · [[x, y], x] · [x, y] = [x, y]^{2}.

The second inequality follows because in a 2 step nilpotent group the commutators commute with all elements of the group. By induction, we have that

[a^{p}, b] = [a, b]^{p}

for any p ≥ 0. In a 2 step nilpotent group [x^{−1}, y] = [y, x], so [a^{p}, b] = [b, a^{−p}] for p < 0. Applying another commutator identity for 2 step nilpotent groups, [x, y^{n}] = [x, y]^{n}, we have that for p < 0

[a^{p}, b] = [b, a^{−p}] = [b, a]^{−p} = [a, b]^{p}.

So, for any p ∈ ℤ, [a^{p}, b] = [a, b]^{p}. Therefore, since [a^{p}, b] = 1_{G}, we know that [a, b]^{p} = 1_{G}. But, G is torsion free, so [a, b] = 1_{G}. This fact contradicts our original choice of a and b as noncommuting elements.

Corollary 8.11. If G is a countable properly 2 step torsion free nilpotent group, then G has 2^{ω} distinct full orders.

Proof. For any O-group there is always a full order under which the center is convex (see Kokorin and Kopytov (1974)). In fact, our general method for constructing full orders on nilpotent groups makes the center convex. Let ≤ be such an order on G and let C be the center of G. Notice ≤ |_{C} is a full G-order. Therefore, if we have any order on G/C, we can combine it lexicographically with ≤ |_{C} to get a full order on G. Furthermore, if we take two distinct orders on G/C, this process yields distinct orders on G.

By Proposition 8.10, G/C has rank ≥ 2 and so by Lemma 8.6, G/C has 2^{ω} distinct orders. Therefore, G has 2^{ω} orders.

Lemma 8.12 (Robinson (1982)). If m ≥ 3 and G is a properly m step nilpotent group, then G/ζ_{m−2}(G) is a properly 2 step nilpotent group.

Corollary 8.13. If m ≥ 2 and G is a torsion free properly m step nilpotent group, then G has 2^{ω} orders.

Proof. The case for m = 2 was handled in Corollary 8.11. If m ≥ 3, then by Lemma 8.12, G/ζ_{m−2}(G) is a properly 2 step nilpotent and hence has 2^{ω} orders.

Let ≤ be an order on G under which the terms in the upper central series are convex as in our general method. As in the proof of Corollary 8.11, the restriction ≤ |_{ζ_{m−2}(G)} is a full G-order on ζ_{m−2}(G). Therefore, ≤ |_{ζ_{m−2}(G)} can be combined lexicographically with any order on G/ζ_{m−2}(G) to give an order of G. Thus, G has 2^{ω} orders.

We need to know not only about the number of orders, but also about the degrees of orders of computable torsion free nilpotent groups.

Lemma 8.14. A computable torsion free properly 2 step nilpotent group has orders of every degree above 0′′.

Proof. Let G be a computable torsion free properly 2 step nilpotent group. The center C is computable in 0′, so G/C is a torsion free abelian group computable in 0′ with rank > 1. By the relativized versions of Lemmas 8.6 and 8.7, G/C has orders of every degree above 0′′.

Since C is a torsion free abelian group computable in 0′, it has an order computable in 0′′ (in fact, it has an order which is low over 0′). Fix such an order. Because C is the center of G, any order on C is a G-order. Let a be any degree above 0′′. An order on G/C of degree a can be lexicographically combined with the order on C to produce an order on G which has degree a.

We want to use similar ideas to handle nilpotent groups with longer upper central series. Notice that ζ_{1}(G) is computable in 0′, so G/ζ_{1}(G) is computable in 0′ and ζ_{1}(G) has a G-order computable in 0′′. The center of G/ζ_{1}(G) is computable in 0′′, so ζ_{2}(G) and G/ζ_{2}(G) are both computable in 0′′. Therefore, ζ_{2}(G)/ζ_{1}(G) has a G-order computable in 0^{(3)} and there is an induced order on ζ_{2}(G) computable in 0^{(3)}. Continuing this process, it is clear that ζ_{n}(G) will be computable in 0^{(n)} and have a G-order computable in 0^{(n+1)}.

Lemma 8.15. Let n > 1 and let G be a computable torsion free properly n step nilpotent group. For every degree a ≥_{T} 0^{(n)}, G has an order of degree a.

Proof. The case for n = 2 was done in Lemma 8.14, so assume that n > 2. By Lemma 8.12 and the comments above, we know that G/ζ_{n−2}(G) is a torsion free properly 2 step nilpotent group computable in 0^{(n−2)}. By the relativized version of Lemma 8.14, G/ζ_{n−2}(G) has orders of every degree above 0^{(n)}. By the comments made above, ζ_{n−2}(G) is computable in 0^{(n−2)} and has a G-order computable in 0^{(n−1)}.

Let a be any degree above 0^{(n)}. An order of degree a on G/ζ_{n−2}(G) together with a G-order of degree 0^{(n−1)} on ζ_{n−2}(G) induces an order of degree a on G.

Theorem 8.16. There is a c.b. Π_{1}^{0} class of separating sets C such that for any torsion free nilpotent group G

{ deg(f) | f ∈ C } ≠ { deg(p) | P ∈ 𝕏(G) }.

Proof. If G is abelian, the theorem follows from Theorem 8.9. Otherwise, G must be properly n step nilpotent for some n > 1. The theorem now follows from Lemma 8.15 and the proof of Theorem 8.9.

8.3 Computable Presentations

In this section, we turn our attention to the question of whether every orderable computable group is classically isomorphic to a computably orderable computable group. Downey and Kurtz (1986) asked this question because their example of a computable torsion free abelian group with no computable order was isomorphic to Π_{i=1}^{ω}ℤ_{i}, which with the right presentation is computably orderable. The answer to this question for general groups is still unknown, but we will answer it for abelian groups and for finitely generated nilpotent groups.

In Section 8.1, we made the simplifying assumption that our abelian groups were divisible because it did not change the change the number or the complexity of the orders. Here, we cannot make such an assumption because it changes the structure of the group. The next lemma shows that we can reduce the problem of finding computable orders to the problem of finding a computable basis.

Lemma 8.17. Let G be a computable torsion free abelian group. If B is a basis for G, then G has a full order that is computable from B.

Proof. Let e_{1}, e_{2}, …, e_{n}, … be a (possible finite) list of the elements of B. We order the elements of the basis by

e_{1} ≪ e_{2} ≪ ⋯ ≪ e_{n} ≪ · · · .

This order of the basis elements induces an order on all of G. To compare g and h ∈ G with g ≠ h, first write each element in the form

ng = ∑_{i∈I} p_{i}e_{i}

mh = ∑_{j∈J} q_{j}e_{j}

where I, J are finite sets and n, m, p_{i} and q_{j} ∈ ℤ \ {0}. Let K be the maximum of I ∪ J and set p_{k} = 0 if k ∉ I and q_{k} = 0 if k ∉ J. The order on g and h is given by

g < h ⇔ \frac{p_{k}}{n} <_{ℚ} \frac{q_{k}}{m}.

Because the summation forms for each element are computable from B, this order is computable from B.

Theorem 8.18. Every computable torsion free abelian group of finite rank has a computable order.

Proof. This theorem follows immediately from Lemma 8.17 because every finite set is computable.

We know that Theorem 8.18 is not true for groups of infinite rank because of the Downey and Kurtz (1986) example and the fact that W KL_{0} is required to prove that every torsion free abelian group is orderable. However, we can use the following result from the study of computable abelian groups.

Theorem 8.19 (Dobritsa (1983)). Let G be a computable torsion free abelian group. There is a computable group H which is classically isomorphic to G and has a computable basis.

This theorem was originally stated and proved using the terminology of the Russian school’s approach to computable mathematics (i.e. using the language of constructivizations). The proof below is written using the language of the Western approach.

Definition 8.20. Let G be a torsion free abelian group. The elements a_{0}, …, a_{k} are t-dependent if there are integers m_{0}, …, m_{k} such that each |m_{i}| ≤ t, at least one m_{i} ≠ 0 and

m_{0}a_{0} + m_{1}a_{1} + ⋯ + m_{k}a_{k} = 0_{G}.

If there are no such integers, then a_{0}, … a_{k} are called t-independent.

Proof. Notice that if G has finite rank, then G has a computable basis. Assume that G has infinite rank. The main idea is to build H in stages while approximating a basis for G. During the construction, we define the elements of H and a bijection

ψ : H → G.

Only at the end, when we have verified certain properties of the construction, do we define the group structure on H and verify that ψ is a group isomorphism. Because the approximation of ψ changes frequently, ψ will not be computable, but will be Δ_{2}^{0}.

Before starting the construction, it is worth introducing the notation that is used. c_{i}^{t} is the guess at the i^{th} basis element of G at stage t. There is a computable function f(t) which defines the domain of H at stage t,

H_{t} = {h_{0}, h_{1}, …, h_{f(t)}}.

The first t + 1 basis elements of H are denoted by

{b_{0}, …, b_{t}} ⊆ H_{t}.

If i > 0 and h_{i} enters H at stage t, then h_{i} is assigned a (t + 2)-tuple of integers from [−t, t]. The statement ⟨α, α_{0}, …, α_{t}⟩ is assigned to h_{i} is written as

αh_{i} = α_{0}b_{0} + α_{1}b_{1} + ⋯ + α_{t}b_{t}.

The intuition is that these tuples uniquely determine each element of H in terms of the basis elements. A slight technical point about our use of tuples in this proof is that we only consider tuples up to trailing zeros. That is, the tuples ⟨1, 2⟩, ⟨1, 2, 0⟩ and ⟨1, 2, 0, 0, 0, 0⟩ are considered equal. Also, any tuple which is assigned will have the property that the greatest common divisor of α, α_{0}, …, α_{t} is 1.

The approximation to ψ at stage t is denoted ψ_{t}. The elements

{c_{0}^{t}, …, c_{t}^{t}}

are defined at stage t. We guarantee at stage t that

αh_{i} = α_{0}b_{0} + ⋯ + α_{t}b_{t}

if and only if

αψ_{t}(h_{i}) = α_{0}c_{0}^{t} + ⋯ + α_{t}c_{t}^{t}.

If at a later stage any of the c_{i}^{t} change, we will change the approximation of ψ to guarantee that this equality continues to hold. At stage t, all tuples of the form ⟨1, α_{0}, …, α_{t}⟩ with each |α_{i}| ≤ t will be assigned and possibly some tuples of the form ⟨α, α_{0}, …, α_{t}⟩ where the greatest common divisor of the elements of the tuple is 1. Construction Stage 0: Define c_{0}^{0} to be the ℕ-least nonidentity element of G. Set f(0) = 1, so H_{0} contains 0 and 1, which to avoid confusion are denoted h_{0} and h_{1}. The map ψ_{0} : H_{0} → G sends h_{0} ↦ 0_{G} and h_{1} ↦ c_{0}^{0}. b_{0} is set to h_{1} and the tuples 1 · h_{1} = 1 · b_{0} and 1 · h_{0} = 0 · b_{0} are assigned.

Stage t + 1 : At the end of stage t we have the following objects:

{b_{0}, …, b_{t}} ⊆ {h_{0}, …, h_{f(t)}} = H_{t}
ψ_{t} : H_{t} → G
{c_{0}^{t}, …, c_{t}^{t}} ⊆ G.

Each h_{i}, 0 < i ≤ f(t), has been assigned to a tuple, the length of which depends on the stage at which h_{i} entered H. The following property is satisfied, assuming h_{i} entered H at stage s ≤ t:

αh_{i} = α_{0}b_{0} + ⋯ + α_{s}b_{s} ⇔ αψ_{t}(h_{i}) = α_{0}c_{0}^{t} + ⋯ + α_{s}c_{s}^{t}.

The construction proceeds are follows. Let c_{t+1}^{t} = 0_{G}. Let I be the least number such that c_{0}^{t}, …, c_{I}^{t} are t + 1-dependent. Find the ℕ-least elements c′_{I+j}, 0 ≤ j ≤ t + 1 − I, of G such that

c_{0}^{t}, …, c_{I−1}^{t}, c′_{I}, …, c′_{t+1}

are (t + 1)-independent. Define c_{k}^{t+1} = c_{k}^{t} for k < I and c_{k}^{t+1} = t!c′_{k} + c_{k}^{t} for I ≤ k ≤ t + 1.

Let N be the number of tuples ⟨1, α_{0}, …, α_{t+1}⟩ where |α_{i}| ≤ t + 1 for each i ≤ t + 1 and either α_{t+1} ≠ 0 or |α_{i}| = t + 1 for at least one i ≤ t. Notice that none of these tuples has been assigned at a previous stage. Let M be the number of elements of G which are less than t + 1 and satisfy an equation of the form

αg = α_{0}c_{0}^{t+1} + ⋯ + α_{t+1}c_{t+1}^{t+1}

with α ≠ 0, 1, the greatest common divisor of α, α_{0}, …, α_{t+1} is 1 and the tuple ⟨α, α_{0}, …, α_{t+1}⟩ has not been used before. Recall, that we equate tuples up to trailing zeros, so these tuples could have been used before.

Define f(t + 1) = f(t) + N + M. The definition of ψ_{t+1} splits into cases. For i = 0,

ψ_{t+1}(h_{0}) = 0_{G}.

For 1 ≤ i ≤ f(t), assume that h_{i} was introduced at stage s ≤ t and assigned the tuple

αh_{i} = α_{0}b_{0} + ⋯ + α_{s}b_{s}

with |α|, |α_{0}|, …, |α_{s}| ≤ s. By assumption we know that

α_{0}c_{0}^{t} + ⋯ + α_{s}c_{s}^{t} = αψ_{t}(h_{i}).  (8.1)

We split into two subcases. If s < I, then set ψ_{t+1}(h_{i}) = ψ_{t}(h_{i}). Because c_{0}^{t+1} = c_{0}^{t}, …, c_{s}^{t+1} = c_{s}^{t}, we know that ψ_{t+1}(h_{i}) still satisfies Equation (8.1). If s ≥ I, then

α_{0}c_{0}^{t} + ⋯ + α_{s}c_{s}^{t} ≠ α_{0}c_{0}^{t+1} + ⋯ + α_{s}c_{s}^{t+1}.

However, by the definition of the elements c_{k}^{t+1} for k ≥ I we have

α_{0}c_{0}^{t} + ⋯ + α_{s}c_{s}^{t} = α_{0}c_{0}^{t} + ⋯ + α_{s}c_{s}^{t} + t!(c′_{I} + ⋯ + c′_{s}).

Because α_{0}c_{0}^{t} +⋯+α_{s}c_{s}^{t} appears on one side of Equation 8.1, we know it is divisible by α in G. Notice that if s = t + 1, then this statement is not quite true because α_{t+1}c_{t+1}^{t} does not appear in Equation 8.1. However, we can leave this term out because c_{t+1}^{t} = 0_{G}. Because |α| ≤ s ≤ t, t!(c′_{I} + ⋯ + c′_{s}) is also divisible by α in G. Let ψ_{t+1}(h_{i}) be the ℕ-least solution to

α_{0}c_{0}^{t+1} + ⋯ + α_{s}c_{s}^{t+1} = αx

in G. Notice that we have extended Equation 8.1 to stage t + 1.

For f(t) < i ≤ f(t) + N, let ⟨1, α_{0}, …, α_{t+1}⟩ be the (i − f(t))^{th} tuple satisfying the conditions in the definition of N. Let

ψ_{t+1}(h_{i}) = α_{0}c_{0}^{t+1} + ⋯ + α_{t+1}c_{t+1}^{t+1}

and assign h_{i} the tuple

1 · h_{i} = α_{0}b_{0} + ⋯ + α_{t+1}b_{t+1}.

b_{t+1} is defined to be whichever h_{i} is assigned the tuple

1 · h_{i} = 0b_{0} + ⋯ + 0b_{t} + 1b_{t+1}.

For f(t) + N < i ≤ f(t + 1), let g be the (i − f(t) − N)^{th} element of G satisfying the conditions in the definition of M. Suppose g satisfies the equation

αg = α_{0}c_{0}^{t+1} + ⋯ + α_{t+1}c_{t+1}^{t+1}

with the same restrictions on the coefficients as in the definition of M. Set ψ_{t+1}(h_{i}) = g and assign h_{i} the tuple

αh_{i} = α_{0}b_{0} + ⋯ + α_{t+1}b_{t+1}.

This case completes stage t + 1 of the construction. Notice that we have met the induction requirements assumed at the beginning of state t.

End of Construction

The following lemmas verify the required properties of the construction. Lemma 8.21. For each i, lim_{t→∞} c_{i}^{t} = c_{i} exists and the set of elements c_{i} is independent.

Proof. The lemma is proved by induction on i. Because c_{0}^{0} ≠ 0_{G}, it never changes and so c_{0} = c_{0}^{0}.

Assume s is a stage such that c_{0} = c_{0}^{s}, …, c_{i−1} = c_{i−1}^{s}. Let g be the ℕ-least element of G such that

{c_{0}, …, c_{i−1}, g}

is an independent set. Let t ≥ s be the least stage such that each g̃ ≤_{ℕ} g is (t + 1)-dependent on {c_{0}, …, c_{i−1}}. By definition, at stage t + 1 we set c′_{i} = g and c_{i}^{t+1} = c_{i}^{t} + t!g. Notice that {c_{0}, …, c_{i−1}, c_{i}^{t} + t!g} is independent, so c_{i}^{t+1} never changes again. Therefore, c_{i} = c_{i}^{t+1}.

Lemma 8.22. For each i ∈ ω, lim_{t→∞} ψ_{t}(h_{i}) = ψ(h_{i}) exists.

Proof. Fix i and we show that lim_{t→∞} ψ_{t}(h_{i}) exists. Suppose h_{i} is assigned the tuple

αh_{i} = α_{0}b_{0} + ⋯ + α_{s}b_{s}.

Let t be a stage such that c_{0}^{t} = c_{0}, …, c_{s}^{t} = c_{s}. For any stage t′ ≥ t, when defining ψ_{t′+1}(h_{i}) we will be in the case of 1 ≤ i ≤ f(t′) and s < I. Thus, ψ_{t′+1}(h_{i}) = ψ_{t′}(h_{i}) for every t′ ≥ t.

Lemma 8.23. The domain of ψ is ω and the range of ψ is G.

Proof. The domain of ψ is all of ω because f(t) is a strictly increasing function. To show that the range of ψ is G, let g ∈ G. Let t be a stage such that c_{0}^{t} = c_{0}, …, c_{s}^{t} = c_{s} and αg = α_{0}c_{0} + ⋯ + α_{s}c_{s}, where the greatest common divisor of the coefficients in 1 and α_{s} ≠ 0. Let t′ be the maximum of t, the code for g, and the absolute values of α and the α_{i}’s. If the tuple ⟨α, α_{0}, …, α_{s}⟩ has not been assigned by stage t′, it will be assigned at stage t′. In either case, whichever element h_{i} is assigned this tuple will satisfy

αψ_{t′}(h_{i}) = α_{0}c_{0}^{t′} + ⋯ + α_{s}c_{s}^{t′} = α_{0}c_{0} + ⋯ + α_{s}c_{s} = αg.

Notice that because the approximations to the basis elements of G have stabilized by stage t′, we know that ψ_{t′}(h_{i}) = ψ(h_{i}). Thus, we have

αψ(h_{i}) = αg.

Because G is torsion free, this equation implies that ψ(h_{i}) = g.

We have now shown that ψ is a bijection and that {c_{0}, c_{1}, … } is a basis for G. It remains to show how to define the group structure on H and to check that ψ is a group homomorphism.

To define +_{H}, consider h_{i} +_{H} h_{j}. By possibly adding some zeros to the end of the tuples assigned to h_{i} or h_{j}, we can assume that we have

αψ(h_{i}) = α_{0}c_{0} + ⋯ + α_{t}c_{t}
βψ(h_{j}) = β_{0}c_{0} + ⋯ + β_{t}c_{t}.

Multiplying the top equation by β, the bottom equation by α and adding them together we get

αβ(ψ(h_{i}) + ψ(h_{j})) = (α_{0}β + αβ_{0})c_{0} + ⋯ + (α_{t}β + αβ_{t})c_{t}.

Let n be the greatest common divisor of αβ and the α_{i}β + αβ_{i} terms. We define h_{i} +_{H} h_{j} to be the element h_{k} that is assigned the tuple

⟨ \frac{αβ}{n} , \frac{α_{0}β + αβ_{0}}{n} , …, \frac{α_{t}β + αβ_{t}}{n} ⟩.

Notice that because ψ is onto G, there must be some h_{k} that is assigned this tuple. In fact, h_{k} is exactly the element such that ψ(h_{k}) = ψ(h_{i}) +_{G} ψ(h_{j}). Lemma 8.24. H is a group under this definition of +_{H} and ψ is an isomorphism.

Proof. It is clear that +_{H} is defined to mimic +_{G}. That is

h_{i} +_{H} h_{j} = h_{k} ⇔ ψ(h_{i}) +_{G} ψ(h_{j}) = ψ(h_{k}).

The fact that G is a group implies that H is a group and this equivalence shows that ψ is an isomorphism.

The set of elements {b_{0}, b_{1}, … } is a basis for H. This set is computably enumerable and so is its complement. To test if g is not one of the b_{i}, just look for an equation of the form

αx = α_{0}b_{0} + ⋯ + α_{t}b_{t}

that is satisfied by g. Thus, H has a computable basis, which completes the proof of Theorem 8.19.

Theorem 8.25. Every computable torsion free abelian group is classically isomorphic to a computable group with a computable order.

Proof. Let G be a computable torsion free abelian group. Theorem 8.18 has already handled the case when G has finite rank. If G has infinite rank, then let H be as in Theorem 8.19. By Lemma 8.17, H has a computable order.

As in Section 8.2, we would like to extend this result to the class of nilpotent groups. Unfortunately, for reasons explained below, we are only able to extend it to finitely generated nilpotent groups.

The class of finitely generated nilpotent groups has been extensively studied in computational algebra. These groups has many nice computational properties. For example, every finitely generated nilpotent group is finitely presented, they have the max property, which implies that every subgroup is finitely generated, and they are residually finite which implies that the word problem is solvable. In addition, the conjugacy and isomorphism problems are solvable.

Theorem 8.26 (Baumslag et al. (1991)). Let G be a computable finitely generated nilpotent group. Each term in the upper central series of G is computable.

Theorem 8.27. If G be a computable torsion free finitely generated nilpotent group, then G has a computable order.

Proof. The idea is to build up the computable order using the terms in the upper central series as in Section 8.2. Notice that since each subgroup ζ_{i}(G) is computable, the factors ζ_{i+1}(G)/ζ_{i}(G) are computable. Also, because every subgroup of G is finitely generated, the groups ζ_{i+1}(G)/ζ_{i}(G) are finitely generated torsion free abelian groups. In particular, they have finite rank and therefore, by Theorem 8.18, they are computably orderable.

All that remains is to put together the computable orders on the finite number of fact groups. Because ζ_{1}(G) is a subgroup of G, ζ_{1}(G) is finitely generated and so computably orderable. Combining this order with a computable order on ζ_{2}(G)/ζ_{1}(G) induces a computable order on ζ_{2}(G). We continue ordering the terms of the upper central series until we reach G.

The proof of Theorem 8.27 does not work for infinitely generated nilpotent groups because the terms in the upper central series are not necessarily computable. In Theorem 3.28, we saw that even for 2 step nilpotent groups, the center need not be computable.

Appendix A

Miscellaneous Definitions and

Proofs

In this appendix, we present some of the background definitions and technical proofs required in Chapter 5. Section A.1 contains the definitions and basic facts about free groups and free products. Section A.2 contains the proofs of the basic formulas for the class of triangular matrices. Section A.3 contains the proofs of the formulas used in the embedding of the free product A∗B into the class of triangular matrices.

A.1 Free Groups and Free Products

Let A ⊆ ℕ. For the purposes of defining the free group on the set of generators A, it is convenient to think of the elements of A as distinct symbols in some alphabet. Let a^{1} stand for the pair ⟨a, 1⟩ and a^{−1} stand for the pair ⟨a, −1⟩. In this section ϵ will always denote either 1 or −1, and hence a^{ϵ} is either ⟨a, 1⟩ or ⟨a, −1⟩.

Definition A.1. (RCA_{0}) If A ⊆ ℕ, then the set of words over A, denoted by Word_{A}, is the set of finite sequences of pairs ⟨a, ϵ⟩ where a ∈ A and ϵ = ±1. In our notation,

Word_{A} = FinÃ

where Ã = {a^{ϵ} | a ∈ A ∧ ϵ = ±1}. The empty sequence in Word_{A} is denoted by 1_{A}.

In keeping with standard mathematical notation, we write a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}} for the sequence σ ∈ Word_{A} with σ(i) = a_{i}^{ϵ_{i}} for 1 ≤ i ≤ k. We also write w_{1} · w_{2} for the concatenation of the sequences w_{1}, w_{2} in Word_{A}.

A sequence x ∈ Word_{A} is called reduced if there is no place in the sequence where a^{1} and a^{−1} appear next to each other for any a ∈ A. This notion is defined formally below. Definition A.2. (RCA_{0}) The set of reduced words over A, denoted by Red_{A}, is the subset of Word_{A} such that x ∈ Red_{A} if and only if x ∈ Word_{A} and

∀i < (lh(x) − 1) ( π_{1}(x(i)) ≠ π_{1}(x(i + 1)) ∨ π_{2}(x(i)) = π_{2}(x(i + 1)) )

where π_{1} and π_{2} are the standard projection functions on pairs.

Both Word_{A} and Red_{A} have Σ_{0}^{0} definitions, so RCA_{0} proves they exist. We next want to define an equivalence relation on Word_{A} such that each equivalence class contains exactly one element of Red_{A}. This equivalence relation is used to put a group structure on Red_{A}. Two words are 1 step equivalent if either they are the same sequence or one results from the other by deleting a pair a^{1}, a^{−1} that appear next to each other.

Definition A.3. (RCA_{0}) Two words x, y ∈ Word_{A} are 1 step equivalent, x ∼_{1} y, if and only if one of the following conditions holds:

1. x = y

2. lh(x) = lh(y) + 1 and

∃i < lh(x) ( ∀j < i (x(j) = y(j)) ∧

∧ ∀j ≥ i ( j < lh(y) → y(j) = x(j + 2) ) ∧

∧ π_{1}(x(i + 1)) = π_{1}(x(i)) ∧ π_{2}(x(i + 1)) + π_{2}(x(i)) = 0)

3. Same as 2 with the roles of x and y switched. The elements x(i) and x(i + 1) are said to be cancelled in x or inserted in y.

The conditions in this definition are Σ_{0}^{0} so RCA_{0} proves the existence of the set of all pairs ⟨x, y⟩ with x ∼_{1} y. Definition A.4. (RCA_{0}) Two words x, y ∈ Word_{A} are freely equivalent, x ∼ y, if there is a finite sequence σ of elements of Word_{A} such that

1. σ(0) = x

2. σ(lh(σ) − 1) = y

3. σ(i) ∼_{1} σ(i + 1) for all i < lh(σ) − 1.

This defines an equivalence relation. Notice that the condition in this definition is Σ_{1}^{0}, so we have to work harder to prove the existence of the set of pairs ⟨x, y⟩ with x ∼ y in RCA_{0}. In order to form this set, we define a function ρ by recursion.

ρ : Word_{A} → Red_{A}

ρ(1_{A}) = 1_{A}
ρ(a^{ϵ}) = a^{ϵ} for a ∈ A, ϵ = ±1

If ρ(U) = a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}} then

ρ(U · a^{ϵ}) = \{ \begin{matrix}a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}}a^{ϵ} & if a ≠ a_{k} or a = a_{k} ∧ ϵ_{k} + ϵ ≠ 0 \\ a_{i}^{ϵ_{1}} ⋯ a_{k−1}^{ϵ_{k−1}} & if a = a_{k} ∧ ϵ_{k} + ϵ = 0 \end{matrix}

Lemma A.5. (RCA_{0}) The following properties hold of ρ for all words W, W_{1} and W_{2} in Word_{A} and all a ∈ A.

1. ρ(W) ∈ Red_{A}

2. ρ(W) ∼ W

3. W ∈ Red_{A} → ρ(W) = W

4. ρ(W_{1} · W_{2}) = ρ(ρ(W_{1}) · W_{2})

5. ρ(W · a^{ϵ} · a^{−ϵ}) = ρ(W)

6. ρ(W_{1} · a^{ϵ} · a^{−ϵ} · W_{2}) = ρ(W_{1} · W_{2}) Proof. The proofs are all by induction either on the length of W or on the length of W_{2}. To prove that ρ(W) ∈ Red_{A}, we prove ∀nφ(n) by induction where φ(n) is

lh(W) = n → ρ(W) ∈ Red_{A}.

The only element of Word_{A} with length 0 is 1_{A}. Since ρ(1_{A}) = 1_{A}, we have that φ(0) holds. If lh(W) = 1, then W = a^{ϵ} for some a ∈ A. By the definition of ρ, ρ(a^{ϵ}) = a^{ϵ} and so φ(1) holds. In the case when lh(W) > 1, we write W as the concatenation W = U · a^{ϵ}. By the induction hypothesis, ρ(U) ∈ Red_{A}. Assume ρ(U) = a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}} and split into two cases.

If a_{k} ≠ a or a_{k} = a but ϵ_{k} + ϵ ≠ 0, then by definition ρ(W) = a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}}a^{ϵ} and ρ(W) ∈ Red_{A}. If a_{k} = a and ϵ_{k} + ϵ = 0, then ρ(W) = a_{1}^{ϵ_{1}} ⋯ a_{k−1}^{ϵ_{k−1}}. Again, since ρ(U) ∈ Red_{A} we have ρ(W) ∈ Red_{A}. This proves property (1).

To prove ρ(W) ∼ W, we use Σ_{1}^{0} induction on lh(W). If lh(W) = 0 or lh(W) = 1, then the argument is the same as for property 1. Assume lh(W) > 1 and W = U ·a^{ϵ} with U ∼ ρ(U) = a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}}. Let σ be the sequence which shows the free equivalence of U and ρ(U). Split into the same two cases as in the proof of property 1. If ρ(W) = a_{1}^{ϵ_{1}} ⋯ a_{k}^{ϵ_{k}}a^{ϵ} then σ̃ gives the free equivalence of W and ρ(W) where σ̃ is defined from σ by

σ̃(i) = σ(i)a^{ϵ}.

If ρ(W) = a_{1}^{ϵ_{1}} ⋯ a_{k−1}^{ϵ_{k−1}} then σ̃ gives the free equivalence of W and ρ(W) where σ̃ is defined by

∀i < lh(σ)(σ̃(i) = σ(i) · a^{ϵ})
σ̃(lh(σ)) = ρ(W)).

To verify that this proof is indeed Σ_{1}^{0} induction, notice that we proved ∀nφ(n) where

φ(n) ≡ (W ∈ Word_{A} ∧ lh(W) = n) → ρ(W) ∼ W.

φ(n) is Σ_{1}^{0}.

The proofs of the remaining properties involve similar case analysis, except for the last property. Property 6 is a direct consequence of Properties 3 and 5. For more details, see Magnus et al. (1965)

Lemma A.6. (RCA_{0}) If x ∼ y then ρ(x) = ρ(y).

Proof. From the definition of 1 step equivalence and from Property 6 of Lemma A.5, it follows that if x ∼_{1} y, then ρ(x) = ρ(y). Assume x ∼ y and let σ be the sequence that shows x ∼ y. Since σ(i) ∼_{1} σ(i + 1) for all i < (lh(σ) − 1), we have ρ(σ(i)) = ρ(σ(i + 1)). Thus, ρ(σ(0)) = ρ(σ(lh(σ) − 1)) and so ρ(x) = ρ(y).

Proposition A.7. (RCA_{0}) For every x ∈ Word_{A} there is a unique y ∈ Red_{A} such that x ∼ y.

Proof. Since ρ(x) ∈ Red_{A} and x ∼ ρ(x), we know that there is at least one y ∈ Red_{A} such that x ∼ y. It remains to show that if x ∼ y and y ∈ Red_{A} then y = ρ(x). Because x ∼ y implies that ρ(x) = ρ(y) and y ∈ Red_{A} implies ρ(y) = y, we have ρ(x) = y as required.

Because free equivalence is an equivalence relation, it follows that if ρ(x) = ρ(y) then x ∼ y. Together with Lemma A.6 this shows that x ∼ y if and only if ρ(x) = ρ(y). The set of pairs ⟨x, y⟩ such that x ∼ y can be formed by Σ_{0}^{0} comprehension.

{⟨x, y⟩ | x ∼ y} = {⟨x, y⟩ | ρ(x) = ρ(y)}

Using this set, we can give the formal definition of the free group on the set of generators A.

Definition A.8. (RCA_{0}) Let A ⊆ ℕ. The set of elements of the free group on the set of generators A is Red_{A}. The empty sequence 1_{A} is the identity element and multiplication is defined by

x · y = ρ(x · y)

where ρ(x · y) is ρ applied to the concatenation of the strings x and y.

The definition of the free product of two groups A∗B is similar to this definition of free groups. Instead of using sequences of generators and inverses as elements, we will use sequences whose elements alternate between A and B. For example, if a_{i} ∈ A and b_{i} ∈ B then strings such as

⟨a_{1}, b_{3}, a_{2}⟩ and ⟨b_{2}, a_{1}⟩

are in A ∗ B. To form this group, we start with the set of finite strings over A ∪ B. Strings are reduced by removing occurrences of 1_{A} and 1_{B} and by multiplying elements of the same group which appear next to each other in the string. For example,

⟨a_{1}, 1_{A}, b_{2}, b_{3}⟩ ↦ ⟨a_{1}, b_{2} ·_{B} b_{3}⟩.

The definitions and lemmas for free products parallel those given for free groups. Definition A.9. (RCA_{0}) If A, B are groups then Word_{A∗B} is the set of finite sequences of elements of A ∪ B.

Word_{A∗B} = Fin_{A∪B}

1_{A∗B} denotes the empty sequence.

Definition A.10. (RCA_{0}) The set of reduced words, Red_{A∗B}, is the subset of Word_{A∗B} such that x ∈ Red_{A∗B} if and only if x ∈ Word_{A∗B} and one of the following conditions holds:

1. x = 1_{A∗B}

2. For all i < lh(x), σ(i) is not 1_{A} or 1_{B}, and for all i < (lh(x) − 1), if σ(i) ∈ A

then σ(i + 1) ∈ B and if σ(i) ∈ B then σ(i + 1) ∈ A. Definition A.11. (RCA_{0}) Two words x, y ∈ Word_{A∗B} are 1 step equivalent, x ∼_{1} y, if and only if one of the following conditions holds:

1. x = y

2. lh(x) = lh(y) + 1 and the sequence y is the same as x except one occurrence of 1_{A} or 1_{B} is removed.

3. lh(x) = lh(y) + 1 and the sequence y is the same as x except there is an i < lh(y) such that either x(i), x(i + 1) ∈ A and y(i) = x(i) ·_{A} x(i + 1) or x(i), x(i + 1) ∈ B and y(i) = x(i) ·_{B} x(i + 1).

4. Switch the roles of x and y in condition 2.

5. Switch the roles of x and y in condition 3. Definition A.12. (RCA_{0}) Two words x, y ∈ Word_{A∗B} are freely equivalent, x ∼ y, if there exists a finite sequence σ of elements of Word_{A∗B} such that

1. σ(0) = x 2. σ(lh(σ) − 1) = y

3. ∀i < (lh(σ) − 1) (σ(i) ∼_{1} σ(i + 1)).

As in the case of free groups, this is a Σ_{1}^{0} condition and so we have to use a function ρ : Word_{A∗B} → Red_{A∗B} to help form the set of pairs ⟨x, y⟩ with x ∼ y. Unlike the case of free groups, we will retain the sequence notation to make the definition clearer.

ρ(1_{A∗B}) = 1_{A∗B}

ρ(⟨g⟩) = \{ \begin{matrix}1_{A∗B} & if g = 1_{A} ∨ g = 1_{B} \\ ⟨g⟩ & otherwise \end{matrix}

If ρ(U) = ⟨h_{1}, h_{2}, …, h_{r}⟩ then

ρ(U^{⌢}⟨g⟩) = \{ \begin{matrix}⟨h_{1}, …, h_{r}⟩ & if g = 1_{A} ∨ g = 1_{B} \\ ⟨h_{1}, …, h_{r−1}⟩ & if g = h_{r}^{−1} \\ ⟨h_{1}, …, h_{r}, g⟩ & if (g ∈ A \ 1_{A} ∧ h_{r} ∈ B) \\ & ∨ (g ∈ B \ 1_{B} ∧ h_{r} ∈ A) \\ ⟨h_{1}, …, h_{r−1}, h_{r} ·_{A} g⟩ & if g ∈ A \ 1_{A} ∧ h_{r} ∈ A \ g^{−1} \\ ⟨h_{1}, … h_{r−1}, h_{r} ·_{B} g⟩ & if g ∈ B \ 1_{B} ∧ h_{r} ∈ B \ g^{−1} \end{matrix}

As in the free group case, we want to show that each word in Word_{A∗B} is freely equivalent to a unique reduced word. To prove this fact, we prove various properties of ρ.

Lemma A.13. (RCA_{0}) The following properties of ρ hold for all W_{1}, W_{2}, W in Word_{A∗B}.

1. ρ(W) ∈ Red_{A∗B}

2. ρ(W) ∼ W

3. W ∈ Red_{A∗B} → ρ(W) = W

4. ρ(W_{1}^{⌢}W_{2}) = ρ(ρ(W_{1})^{⌢}W_{2})

5. ρ(W^{⌢}⟨1_{A}⟩) = ρ(W^{⌢}⟨1_{B}⟩) = ρ(W)

6. ρ(W_{1}^{⌢}⟨1_{A}⟩^{⌢}W_{2}) = ρ(W_{1}^{⌢}⟨1_{B}⟩^{⌢}W_{2}) = ρ(W_{1}^{⌢}W_{2})

7. If g, h ∈ A then ρ(W^{⌢}⟨g, h⟩) = ρ(W^{⌢}⟨g ·_{A} h⟩)

8. If g, h ∈ B then ρ(W^{⌢}⟨g, h⟩) = ρ(W^{⌢}⟨g ·_{B} h⟩)

9. If g, h ∈ A then ρ(W_{1}^{⌢}⟨g, h⟩^{⌢}W_{2}) = ρ(W_{1}^{⌢}⟨g ·_{A} h⟩^{⌢}W_{2})

10. If g, h ∈ B then ρ(W_{1}^{⌢}⟨g, h⟩^{⌢}W_{2}) = ρ(W_{1}^{⌢}⟨g ·_{B} h⟩^{⌢}W_{2}) The proof of this lemma is a series of inductions as in Lemma A.5. For more

details, see Magnus et al. (1965). As in the free group case, we use this lemma to show that x ∼ y if and only if ρ(x) = ρ(y). The proof of the following proposition is also the same as in the free group case. Proposition A.14. (RCA_{0}) For every x ∈ Word_{A∗B} there is a unique y ∈ Red_{A∗B} such that x ∼ y.

From this proposition, we obtain {⟨x, y⟩ | x ∼ y} = {⟨x, y⟩ | ρ(x) = ρ(y)}.

Thus, RCA_{0} can form the set of pair ⟨x, y⟩ such that x ∼ y. Using this set we give the formal definition of the free product A ∗ B. Definition A.15. (RCA_{0}) The set of elements of the free product of the groups A and B, denoted A ∗ B, is Red_{A∗B}. The empty sequence 1_{A∗B} is the identity element and multiplication is given by

x · y = ρ(x^{⌢}y).

Unraveling the definitions, we can find a connection between free products of ℤ and free groups. Proposition A.16. (RCA_{0}) The free product ℤ ∗ ℤ is isomorphic to F_{2}, the free group on two generators. Proof. Let a, b denote the generators of F_{2}. A typical element of F_{2} has the form ⟨a, a, b^{−1}, a, b, b⟩. For notational convenience, let a also denote 1 in the first copy of ℤ and b denote 1 in the second copy of ℤ. A typical element of ℤ ∗ ℤ has the form ⟨2a, −b, a, 2b⟩. The isomorphism is built by expanding elements resembling na in sequences in ℤ ∗ ℤ to n-tuple ⟨a, a, …, a⟩. For example

⟨2a, −b, a, 2b⟩ ↦ ⟨a, a, b^{−1}, a, b, b⟩.

Writing this map formally, we obtain the isomorphism.

A.2 Proofs for Tri_{ℚ[C]}

In this section, we prove some of the technical results about Tri_{K} that were left out of Chapter 5. We are looking at the class of infinite upper triangular matrices with entries from an f.o. ring K such that the elements along the main diagonal are positive and invertible. Definition A.17. (RCA_{0}) Let (K, ≤) be a fully ordered ring with positive cone P. The function f : ℕ^{+} × ℕ^{+} → K is in the class Tri_{K} if and only if it satisfies the following conditions:

1. For all i > j, f(i, j) = 0_{K}.

2. For all i, f(i, i) ∈ P and ∃x ∈ K(f(i, i) · x = 1_{K}).

We use the notation f ∈ Tri_{K} to mean that f is a function satisfying the conditions of this definition. The product of f and g is defined to be the function

f · g : ℕ^{+} × ℕ^{+} → K

f · g(i, j) = ∑_{n=i}^{j} f(i, n)g(n, j).

The identity function I ∈ Tri_{K} is given by I(i, i) = 1_{K} and I(i, j) = 0_{K} for i ≠ j.

Given f, g ∈ Tri_{K}, we say that f < g if and only if for some pair ⟨i, j⟩ ∈ ℕ^{+} ×ℕ^{+} with i ≤ j the following two conditions hold:

1. f(i, j) <_{K} g(i, j)

2. f(k, k + s) = g(k, k + s) for all k, s such that i + s < j or i + s = j and k < i. A pair ⟨i, j⟩ for which these conditions hold is called a witness for f < g. Writing this condition in terms of a positive cone, we say that f ∈ P (Tri_{K}) if and only if f = I or I < f in the order given above. If f ≠ I this is equivalent to either

∃i [f(i, i) > 1 ∧ ∀j < i(f(j, j) = 1)]

or

∀i (f(i, i) = 1) ∧ ∃i, j [ i < j ∧ f(i, j) > 0 ∧
∧ ∀k ∀s > 0 ((i + s < j ∨ (i + s = j ∧ k < i)) → f(k, k + s) = 0) ].

We want to verify that the elements of Tri_{K} satisfy the axioms of a f.o. group. In Chapter 5, we proved in RCA_{0} that the multiplication is associative. Here, we verify that there are inverse elements and that P (Tri_{K}) is normal, pure and closed under multiplication. As pointed out in Chapter 5, RCA_{0} is not strong enough to show that P (Tri_{K}) is full. Lemma A.18. (RCA_{0}) If f ∈ Tri_{K}, then f has an inverse g ∈ Tri_{K}, in the sense that f · g = g · f = I, given by:

g(i, j) = \{ \begin{matrix}0 & j < i \\ f(i, j)^{−1} & i = j \\ −\frac{f(i,j)}{f(i,i)f(j,j)} + ∑_{i<k_{1}<j} \frac{f(i,k_{1})f(k_{1},j)}{f(i,i)f(k_{1},k_{1}),f(j,j)} − \\ − ∑_{i<k_{1}<k_{2}<j} \frac{f(i,k_{1})f(k_{1},k_{2})f(k_{2},j)}{f(i,i)f(k_{1},k_{1})f(k_{2},k_{2})f(j,j)} + ⋯ & i < j \\ ⋯ + (−1)^{j−i} \frac{f(i,i+1)···f(j−1,j)}{f(i,i)f(i+1,i+1)···f(j,j)} \end{matrix}

Since f(n, n) is invertible, we write it in the denominator of a fraction as shorthand for f(n, n)^{−1}.

Proof. We verify that f · g(i, j) = I(i, j) by splitting into the cases of i = j, i < j and i > j. If i = j then

f · g(i, j) = f(i, i)f(i, i)^{−1} = 1.

If i > j then we have already noted that for f, g ∈ Tri_{K}

f · g(i, j) = 0.

The case for i < j is more complicated. For each fixed j, this case is proved by induction on j − i. The base case is when j − i = 0 and is given by our calculation above. If j − i = l, then the induction hypothesis is that the formula is correct for g(j − k, j) for all 0 ≤ k < l. We need to show that f · g(j − l, j) = 0. To prove this, we start by assuming that f · g = I and work backwards to derive g(j − l, j). If f · g(j − l, j) = 0 then we must have

f(j − l, j − l)g(j − l, j) + f(j − l, j − l + 1)g(j − l + 1, j) + ⋯

⋯ + f(j − l, j)g(j, j) = 0.

Solving this equation for g(j − l, j) we have

g(j − l, j) = ∑_{n=0}^{l−1} − \frac{f(j − l, j − n)}{f(j − l, j − l)}g(j − n, j).  (A.1)

To finish the proof, we need to show that this sum is equal to

− \frac{f(j − l, j)}{f(j − l, j − l)f(j, j)} + ∑_{k_{1}=1}^{l−1} \frac{f(j − l, j − k_{1})f(j − k_{1}, j)}{f(j − l, j − l)f(j − k_{1}, j − k_{1})f(j, j)} −

− ∑_{k_{1}=1}^{l−2} ∑_{k_{2}=k_{1}+1}^{l−1} \frac{f(j − l, j − k_{2})f(j − k_{2}, j − k_{1})f(j − k_{1}, j)}{f(j − l, j − l)f(j − k_{1}, j − k_{1})f(j − k_{2}, j − k_{2})f(j, j)} + ⋯

⋯ + (−1)^{l} \frac{f(j − l, j − l + 1)f(j − l + 1, j − l + 2) ⋯ f(j − 1, j)}{f(j − l, j − l)f(j − l + 1, j − l + 1) ⋯ f(j, j)}. (A.2)

To prove this equality, we will split equation (A.2) into a sum with summands of the form

− \frac{f(j − l, j − n)}{f(j − l, j − l)} · X

and show that X = g(j − n, j). For the case n = 0 we see that

− \frac{f(j − l, j)}{f(j − l, j − l)}

appears only in the first summand of equation (A.2). That is, it only appears in

− \frac{f(j − l, j)}{f(j − l, j − l)f(j, j)}.

In this case X = f(j, j)^{−1} = g(j, j) as required.

For the case of n = 1

− \frac{f(j − l, j − 1)}{f(j − l, j − l)}

appears only in the second sum of equation (A.2) and only when k_{1} = 1. Thus in this case, we have

X = − \frac{f(j − 1, j)}{f(j − 1, j − 1)f(j, j)}.

By the bounded induction hypothesis, this is exactly g(j − 1, j).

In general, for 1 ≤ n < l, we have

− \frac{f(j − l, j − n)}{f(j − l, j − l)}

does not appear in the first summand, but does appear in each of the other summands of equation (A.2) up to the (n + 1)^{st} one. If we examine how it appears in each of these summands, we see that in the 2^{nd} term, we get

− \frac{f(j − n, j)}{f(j − n, j − n)f(j, j)}

which is exactly the first term in g(j − n, j). From the 3^{rd} term, we have something every time k_{2} = n and hence we get

∑_{k_{1}=1}^{n−1} \frac{f(j − n, j − k_{1})f(j − k_{1}, j)}{f(j − n, j − n)f(j − k_{1}, j − k_{1})f(j, j)}.

This sum is exactly the second term in g(j − n, j). This process continues until we reach the (n + 1)^{st} term. From this term, we only get something when k_{1} = 1, k_{2} = 2, … k_{n} = n. This give us

(−1)^{n} \frac{f(j − n, j − n + 1) ⋯ f(j − 1, j)}{f(j − n, j − n) ⋯ f(j, j)}.

This product is exactly the last term of g(j − n, j) and shows X = g(j − n, j) as required. We have now shown that f · g = I. From here, we have that g · f · g = g. By a simpler induction, it can be shown that if h · g = g then h = I. Hence g · f = I as well.

Lemma A.19. (RCA_{0})

1. If f, g ∈ P (Tri_{K}) then f · g ∈ P (Tri_{K})

2. If f ∈ P (Tri_{K}) and f ≠ I then f^{−1} ∉ P (Tri_{K})

3. If f ∈ P (Tri_{K}) and g ∈ Tri_{K} then gfg^{−1} ∈ P (Tri_{K})

Proof. To prove the first statement of the lemma, assume f, g ∈ P (Tri_{K}). For notational purposes, let h = f · g. We need to show h ∈ P (Tri_{K}). Without loss of generality, assume that f, g, h ≠ I. There are two cases to consider. Case. ∃i(f(i, i) ≠ 1 ∨ g(i, i) ≠ 1)

Let i be the least such number. Then, h(i, i) = f(i, i)g(i, i) > 1 and for all j < i, h(j, j) = 1. Thus h ∈ P (Tri_{K}). Case. ∀i(f(i, i) = 1 ∧ g(i, i) = 1)

Let the pair ⟨i, j⟩ be a witness for f > I. Without loss of generality, assume that g(i, j) ≥ 0 and that g(k, k + s) = 0 for all k and s > 0 such that i + s < j or i + s = j and k < i. That is, assume that the witness to g > I comes later in the order on the diagonals than the witness for f. We need to show that h(i, j) > 0, that h(k, k + s) = 0 for k, s as above and that h(n, n) = 1 for all n.

Since f(n, n) = g(n, n) = 1, it is clear that h(n, n) = 1. To show that h(i, j) > 0, we examine

h(i, j) = ∑_{n=i}^{j} f(i, n)g(n, j).

By the assumptions made above on f and g, we have that f(i, i) = g(j, j) = 1 and f(i, i + 1) though f(i, j − 1) are all 0. Thus, this sum reduces to g(i, j) + f(i, j). Since g(i, j) ≥ 0 and f(i, j) > 0, we have that h(i, j) > 0 are required.

Suppose s > 0, i+s < j or i+s = j and k < i. We have the following equalities:

h(k, k + s) = ∑_{n=k}^{k+s} f(k, n)g(n, k + s)

= ∑_{n=0}^{s} f(k, k + n)g(k + n, k + s)

= f(k, k)g(k, k + s) + f(k, k + s)g(k + s, k + s) +

+ ∑_{n=1}^{s−1} f(k, k + n)g(k + n, k + s).

The first term in the last equation is 0 because g(k, k + s) = 0. The second term is 0 because f(k, k + s) = 0. For the third term, since i + s ≤ j we have that i + n < j for all n in the sum. Thus f(k, k + n) = 0 and the third term is 0. This shows that h(k, k + s) = 0 and finishes the proof of the first statement of the lemma.

To prove the second assertion of the lemma, assume that f ≠ I, f ∈ P (Tri_{K}) and g = f^{−1}. We need to show that g ∉ P (Tri_{K}). There are two cases to consider. Case. ∃i(f(i, i) ≠ 1)

Let i be the least such number. Since f ∈ P (Tri_{K}) we know f(i, i) > 1. By the formula for inverses, g(i, i) = f(i, i)^{−1} < 1. Hence, g ∉ P (Tri_{K}). Case. ∀i(f(i, i) = 1)

It follows from the formula for inverses that g(i, i) = 1 for all i as well. Let the pair ⟨i, j⟩ be a witness for f > I. Using the formula for inverses and the fact that f(i, i) = 1 for all i, we have

g(i, j) = −f(i, j) + ∑_{i<k_{1}<j} f(i, k_{1})f(k_{1}, j) − ⋯

⋯ (−1)^{j−i}(f(i, i + 1)f(i + 1, i + 2) ⋯ f(j − 1, j)).

For any subscripted k appearing in this sum we have i < k < j and we can set s = k − i. It follows that i + s = k < j and so by the assumptions on f

f(i, k) = f(i, i + s) = 0.

All terms in g(i, j) vanish except for the first one. This computation shows that g(i, j) = −f(i, j) < 0. It remains to show that g(k, k + s) = 0 for k and s such that s > 0 and either i + s < j or i + s = j and k < i. Using the formula for inverses again and the fact that f(i, i) = 1 we have

g(k, k + s) = −f(k, k + s) + ∑_{0<n_{1}<s} f(k, k + n_{1})f(k + n_{1}, k + s) − ⋯

⋯ + (−1)^{j−i}(f(k, k + 1) ⋯ f(k + s − 1, k + s)).

Again, by the restrictions on f, this sum is 0. Hence, the pair ⟨i, j⟩ is the witness for g ∉ P (Tri_{K}).

To verify the third statement of the lemma, let f ∈ P (Tri_{K}) and g ∈ Tri_{K}. For notational simplicity, let h = gfg^{−1}. We need to show that h ∈ P (Tri_{K}). Again, there are two cases to consider. Case. ∃i(f(i, i) > 1)

In this case we have:

h(i, i) = g(i, i)f(i, i)g(i, i)^{−1} = f(i, i).

Thus, h(i, i) > 1 and for all j < i, h(j, j) = 1 as required. Case. ∀i(f(i, i) = 1)

Let ⟨i, j⟩ be the witness for f ∈ P (Tri_{K}). It is clear that h(i, i) = 1 for all i. We need to show that the pair ⟨i, j⟩ is a witness for h ∈ P (Tri_{K}). First, we show that h(i, j) > 0. We have

h(i, j) = ∑_{n=i}^{j}(g · f)(i, n)g^{−1}(n, j)

(g · f)(i, n) = ∑_{m=i}^{n} g(i, m)f(m, n).

Consider f(m, n) as it appears in (g · f)(i, n) for n < j. When i ≤ m < n < j, we have by the assumptions on f and ⟨i, j⟩ that f(m, n) = 0. Also, f(n, n) = 1. Hence, we have

n < j ⇒ (g · f)(i, n) = g(i, n)f(n, n) = g(i, n)
n = j ⇒ (g · f)(i, n) = g(i, i)f(i, j) + g(i, j).

Putting this back into the formula for h(i, j) we have:

h(i, j) = ∑_{n=i}^{j}(g · f)(i, n)g^{−1}(n, j)

= ∑_{n=i}^{j−1}(g · f)(i, n)g^{−1}(n, j) + (g · f)(i, j)g^{−1}(j, j)

= ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j) + (g(i, i)f(i, j) + g(i, j))g^{−1}(j, j)

= ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j) + g(i, i)f(i, j)g^{−1}(j, j) + g(i, j)g^{−1}(j, j).

Using the facts that I = g · g^{−1} and I(i, j) = 0 we have

I(i, j) = ∑_{n=i}^{j} g(i, n)g^{−1}(n, j)
0 = ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j) + g(i, j)g^{−1}(j, j)
− ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j) = g(i, j)g^{−1}(j, j).

Plugging this into the formula above, we obtain

h(i, j) = ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j) + g(i, i)f(i, j)g^{−1}(j, j) − ∑_{n=i}^{j−1} g(i, n)g^{−1}(n, j)
= g(i, i)f(i, j)g^{−1}(j, j).

Since each of g(i, i), f(i, j), and g^{−1}(j, j) are strictly positive, we have that h(i, j) > 0. It remains to show that h(k, k + s) for the appropriate k and s.

h(k, k + s) = ∑_{n=0}^{s}(g · f)(k, k + n)g^{−1}(k + n, k + s)

(g · f)(k, k + n) = ∑_{m=0}^{n} g(k, k + m)f(k + m, k + n)

Recall that the conditions on k and s are that s > 0 and either s+i < j or s+i = j and k < i. To examine the bottom equation, let t = n − m. From the ranges of the indices on the sums, it follows that t ≤ s and t = s only if n = s and m = 0. We can use t to write f(k + m, k + n) as f(k + m, k + m + t). Since ⟨i, j⟩ is the witness for f ∈ P (Tri_{K}), we know that if t > 0, then f(k + m, k + m + t) = 0 if t + i < j or t + i = j and k + m < i. We now look at lots of cases. We first split into the cases when t > 0 and t = 0 and then subdivide further from there.

If t > 0 and s + i < j then, since t ≤ s, we have that t + i < j and so f(k + m, k + m + t) = f(k + m, k + n) = 0. Still assuming t > 0, suppose that s + i = j and m + t = n < s. It follows from n ≠ s that t < s. Hence, t + i < j and f(k + m, k + m + t) = 0. This still leaves the case when s + i = j and n = s. We will consider this case last.

Next assume that t = 0. This implies that m = n and hence f(k + m, k + n) = f(k + n, k + n) = 1. Our calculations so far show that if n < s, then all the terms in the sum for (g · f)(k, k + n) drop out except for the g(k, k + n)f(k + n, k + n) term which is just g(k, k + n).

We return to the case when s + i = j and n = s. The sum looks like

(g · f)(k, k + s) = ∑_{m=0}^{s} g(k, k + m)f(k + m, k + s).

If 0 < m < s then f(k + m, k + s) = 0 since i + s = j and so i + s − m < j. Also, f(k, k + s) = 0 and f(k + s, k + s) = 1. Thus

(g · f)(k, k + s) = g(k, k)f(k, k + s) + g(k, k + s)f(k + s, k + s)
= g(k, k + s).

Altogether, we now have that for 0 ≤ n ≤ s, g · f(k, k + n) = g(k, k + n). Putting this back into the formula for h(k, k + s) we obtain

h(k, k + s) = ∑_{n=0}^{s}(g · f)(k, k + n)g^{−1}(k + n, k + s)

= ∑_{n=0}^{s} g(k, k + n)g^{−1}(k + n, k + s).

As above, we apply the facts that I = g · g^{−1} and I(k, k + s) = 0 for s > 0 to get

0 = ∑_{n=0}^{s} g(k, k + n)g^{−1}(k + n, k + s).

Thus, h(k, k + s) = 0 as required.

A.3 Proofs for Free Product Embedding

In this section, we prove the properties of the embedding of A ∗ B into Tri_{ℚ[C]}. Recall that A and B are f.o. groups and C is the restricted direct product

C = A × B × ∏_{i,j=1}^{∞}⟨x_{ij}⟩ × ∏_{i,j=1}^{∞}⟨y_{ij}⟩ × ∏_{i=1}^{∞}⟨u_{i}⟩ × ∏_{i=1}^{∞}⟨v_{i}⟩

ℚ[C] is the group ring of C over ℚ. The elements are the finite formal sums ∑ α_{i}c_{i} with α_{i} ∈ ℚ \ {0}, c_{i} ∈ C and all the c_{i} distinct. Addition is defined by

∑_{i∈I} α_{i}c_{i} + ∑_{j∈J} β_{j}c_{j} =

∑_{i∈I\J} α_{i}c_{i} + ∑_{j∈J\I} β_{j}c_{j} + ∑_{i∈I∩J}(α_{i} + β_{i})c_{i}

with the stipulation that any terms in the third sum with α_{i} + β_{i} = 0 are removed. Multiplication is defined by

(∑_{i∈I} α_{i}c_{i}) (∑_{j∈J} β_{j}c_{j}) = ∑_{i∈I} ∑_{j∈J}(α_{i}β_{j})c_{i}c_{j}

where the terms with the same value from C in this finite sum are collected and any term with coefficient 0 is dropped. The additive identity here is the empty sum I = ∅, and the multiplicative identity is the sum with one element 1_{ℚ}1_{C}. The sum ∑_{i∈I} α_{i}c_{i} is in P (ℚ[C]) if and only if I = ∅ or α_{j} >_{ℚ} 0 where j is such that c_{j} is the ≤_{C}-least element among the c_{i} with i ∈ I.

See Chapter 5 for the definitions of the elements X, Y, U, V ∈ Tri_{ℚ[C]} and the maps

α, α′, α′′ : A → Tri_{ℚ[C]}
β, β′, β′′ : B → Tri_{ℚ[C]}
γ : A ∗ B → Tri_{ℚ[C]}.

In Chapter 5, we proved that γ satisfies the properties of a group homomorphism. The proof that γ is one-to-one required the explicit formulas for α′(a)(i, j) and β′(b)(i, j) which are proved below.

Lemma A.20. (RCA_{0})

α′(a)(i, i) = \{ \begin{matrix}1 & i is odd \\ a & i is even \end{matrix}

Proof. This comes directly from the formulas for f and g. Lemma A.21. (RCA_{0}) If i < j and i, j are both even, then

α′(a)(i, j) = (1 − a) ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) −

− ∑_{i<k_{1}<k_{2}<n}(x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}n}x_{nj}) + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj})

Proof. The proof consists of grinding through the calculations one step at a time, and breaking the sum up into pieces.

α′(a)(i, j) = ∑_{n=i}^{j} g(i, n)f(n, j)

= \underbrace{g(i, i)f(i, j)}_{(I)} + \underbrace{g(i, j)f(j, j)}_{(II)} + \underbrace{∑_{n=i+1}^{j−1} g(i, n)f(n, j)}_{(III)}

Since i is even, (I) is ax_{ij}. Since j is even, f(j, j) = a, and so (II) equals

a (−x_{ij} + ∑_{i<k_{1}<j}(x_{ik_{1}}x_{k_{1}j}) − ⋯ + (−1)^{j−i}(x_{i(i+1)} ⋯ x_{(j−1)j}))

(III) breaks into two cases: when n is even and when n is odd.

\underbrace{∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (g(i, n) · x_{nj})}_{(IV)} + \underbrace{∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (g(i, n) · ax_{nj})}_{(V)} =

∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) − ⋯ + (−1)^{n−i}(x_{i(i+1)} ⋯ x_{nj})) +

a · ∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) − ⋯ + (−1)^{n−i}(x_{i(i+1)} ⋯ x_{nj}))

There are a couple of important observations. First, (I) cancels with the first term in (II). Second, all of the terms in (V) appear in and cancel with terms in (II). Third, since j is even, it follows that j − 1 is odd and so the last term in (II) does not cancel. Performing the cancelations, we are left with

a · ( ∑_{\begin{matrix}i<k_{1}<j \\ k_{1} odd \end{matrix}}(x_{ik_{1}}x_{k_{1}j)} − ∑_{\begin{matrix}i<k_{1}<k_{2}<j \\ k_{2} odd \end{matrix}}(x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}j}) +

+ ⋯ + (−1)^{j−i}(x_{i(i+1)} ⋯ x_{(j−1)j})) +

+ ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) − ⋯ + (−1)^{n−i}(x_{i(i+1)} ⋯ x_{nj})) .

This equation yields the formula in the statement of the lemma once the following general rewriting principles are applied.

∑_{\begin{matrix}i<k_{1}<j \\ k_{1} odd \end{matrix}} x_{ik_{1}}x_{k_{1}j} ⇒ − ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} −x_{in}x_{nj}

− ∑_{\begin{matrix}i<k_{1}<k_{2}<j \\ k_{2} odd \end{matrix}} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}j} ⇒ − ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} ( ∑_{i<k_{1}<n} x_{ik_{1}}x_{k_{1}n}x_{nj})

These principles continue for longer linear sequences of subscripted k’s. For example, a similar rewriting rule can be applied to the sum over i < k_{1} < k_{2} < k_{3} < j with k_{3} odd.

Lemma A.22. (RCA_{0}) If i < j, i is even, and j is odd then

α′(a)(i, j) = (1 − a)(−x_{ij}) + (1 − a) ∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (x_{in}x_{nj} − ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) +

+ ∑_{i<k_{1}<k_{2}<n} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{n}n}x_{nj} − ⋯ + (−1)^{n−i}x_{ii+1} ⋯ x_{n−1n}x_{nj}) .

Proof. This proof proceeds as the last one. I will outline it and point out what needs to be changed from the last proof. As before we have

α′(a)(i, j) = ∑_{n=i}^{j} g(i, n)f(n, j)

= \underbrace{g(i, i)f(i, j)}_{(I)} + \underbrace{g(i, j)f(j, j)}_{(II)} + \underbrace{∑_{n=i+1}^{j−1} g(i, n)f(n, j)}_{(III)} .

Since i is still even, (I) remains ax_{ij}. However, since j is odd, f(j, j) = 1 and so (II) is

−x_{ij} + ∑_{i<k_{1}<j}(x_{ik_{1}}x_{k_{1}j}) − ⋯ + (−1)^{j−i}(x_{i(i+1)} ⋯ x_{(j−1)j}).

(III) still breaks into two pieces.

\underbrace{∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1}(g(i, n) · x_{nj})}_{(IV)} + \underbrace{∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1}(g(i, n) · ax_{nj})}_{(V)}

This time, (I) does not cancel with the first term of (II), and instead of the terms of (V) appearing in (II), the terms in (IV) appear there. Since j is odd, it follows that j − 1 is even, and so the last term in (II) does not cancel. The lemma follows after cancelling and rewriting as above.

Using similar methods, we can prove the next two lemmas as well.

Lemma A.23. (RCA_{0}) If i < j and both i, j are odd then

α′(a)(i, j) = (1 − a) ∑_{\begin{matrix}n=i+1 \\ n even \end{matrix}}^{j−1} (x_{in}x_{nj} − ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) +

+ ∑_{i<k_{1}<k_{2}<n}(x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{2}n}x_{nj}) + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj}) .

Lemma A.24. (RCA_{0}) If i < j, i is odd, and j is even then

α′(a)(i, j) = (1 − a)(x_{ij}) + (1 − a) ∑_{\begin{matrix}n=i+1 \\ n odd \end{matrix}}^{j−1} (−x_{in}x_{nj} + ∑_{i<k_{1}<n}(x_{ik_{1}}x_{k_{1}n}x_{nj}) −

− ∑_{i<k_{1}<k_{2}<n} x_{ik_{1}}x_{k_{1}k_{2}}x_{k_{n}n}x_{nj} + ⋯ + (−1)^{n−i}x_{i(i+1)} ⋯ x_{(n−1)n}x_{nj}) .

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