As a prelude to the investigation of maps between groups, in this section we will look at substructures within groups and their properties.

When looking at the multiplication table of
the dihedral group D_{4} you probably noticed that the upper
left quarter consists solely of rotation. You've also seen that
rotation symmetries of regular n-gons are themselves groups. Thus for
every regular n-gon we have its group of rotation sitting inside its
symmetry group D_{n}. A *subgroup* of a group G is a
subset of elements of G which is itself a group under the group
operation of G. Note that a subgroup must always contain the identity
element of G. The usual notation for H being a subgroup of G is
H<G.

- Given a group G, is the subset consisting of only the identity element a subgroup? (Hint: use the conditions Luke!)
- One way of finding a subgroup is to simply take one or more elements in a group and see what group those elements generate. Use this method to find three subgroups of Z×Z (that are not just (0,0) and the whole of Z×Z itself).
- In the group D
_{4}, does the set of all the reflections plus the identity element form a subgroup? - Find three different subgroups of D
_{4}. - Show that for groups K, H, and G if K<H and H<G, then K<G.

We can also find and investigate subgroups using the Cayley graph of a group. The procedure for constructing subgroups using the Cayley graph is as follows:

- mark the identity vertex,
- pick one or more other vertices and mark them too,
- record a directed path from the identity to those vertices you picked; these paths and their inverses will be the patterns you'll now use
- mark, recursively, all vertices that can be reached from those you've already marked by one of the patterns recorded in the step above (starting at the marked vertices, not the identity).

Here's an example, using the group Z×Z.

- Denote the subgroup of Z×Z in the example above H. What are the elements of H?
- Within Z×Z, this subgroup is generated by ab and bb (where a=(1,0) and b=(0,1)). Draw a Cayley graph of H, using the generators ab and bb (you can call them x and y, for instance).
- Relabeling this Cayley graph produces the Cayley graph of a familiar group. What is it?

Congratulation, you've now found a subgroup of Z×Z isomorphic
to... Z×Z. While this may seem strange at first, it actually not
that surprising. The even integers is a subgroup of the integers which
is isomorphic to the integers. What is more strange is that a free
group on two elements, F_{2}, has subgroups isomorphic to
F_{n} for every n. Let's construct one!

- Consider a subgroup of F
_{2}generated by the three elements ab, abb, and abbb. Draw part of a Cayley graph of F_{2}and mark some elements in this subgroup. - Is the element b in the subgroup? (Hint: can you write b as a product of ab,abb,abbb and their inverses? Why not?)
- How do we know that this subgroup is in fact a free group? Because
next we'll prove the very neat result that
*every*subgroup of a free group is free! - Thus this subgroup is not the whole group (by part 2) and it's free.
So we have F
_{3}<F_{2}. Can you find an F_{5}<F_{2}? How about F_{1}<F_{2}?

Note that if a group has non-trivial relations (i.e. not just
inverses), then it's Cayley graph must have a non-trivial loop. In the
group Z×Z for instance, we have a loop corresponding to abAB
and in the Cayley graph of D_{4} one corresponding to
L_{1}L_{1}. Thus the isometry induced by a relation,
like the element corresponding to abAB in Z×Z is always the
identity isometry.

To show this we'll need to resort to the following fact from geometry: Suppose S is a group of isometries of an (infinite) tree having no fixed points. That is, if the isometry f is in S, then f(x)≠x for every point x in the tree. Then S is a free group.

- The Cayley graph of any
F
_{n}=⟨a_{1},...,a_{n}| ⟩ with respect to the generators a_{1},...,a_{n}is a tree. Why? (Hint: see discussion above.) - We've seen that any group is a subgroup of the isometry group of its
Cayley graph. Do the isometries of the Cayley graph of F
_{2}corresponding to the elements of F_{2}have fixed points? Does the same argument work for F_{n}? (Hint: draw the graph, work out a few examples.) - Thus any subgroup H<F
_{n}is itself a group of fixed point free isometries of a tree, and hence is free.

We saw previously that when every two elements in a group commute,
when gh=hg for every g and h in G, then the group is said to be abelian.
A subgroup H<G is *normal* if it behaves like a
commuting element, that is, if gH=Hg for every g∈G. By gH we mean
the set of elements {gh_{1},gh_{2},...} if
H={h_{1},h_{2},...}. Similarly,
Hg={h_{1}g,h_{2}g,...}. The subgroup of even integers
in Z is a normal subgroup since the whole group Z is abelian. A more
interesting example of normality and non-normality can be found in
D_{4}.

- Is the subgroup of D
_{4}consisting of rotations normal? - How about the subgroup generated by a in F
_{2}?

- Suppose H<G and g∈G is
not an element of H. Show that H and the coset gH have no common
elements. (Hint: if they do, then h
_{1}=gh_{2}for some h_{1},h_{2}∈H.) - Let H be a subgroup of G. Is HH=H (does the set product of a subgroup with itself produce new elements)?
- Let N<G be a normal subgroup. Show that (aN)(bN)=abN, as subsets of G. (Hint: use the fact that (aN)(bN)=aNbN, normality, and the previous problem.)

Given a normal subgroup N of G, we can look at the set of all cosets
gN, for all g∈G. The miraculous fact (which we'll shortly prove)
is that this set is itself a group! It's called a *quotient
group* of G by N and denoted G/N.

- Suppose N<G is normal. Is there a coset of N that acts as the identity in G/N? (Hint: try the identity coset eN=N.)
- What is the coset inverse to gN? (Hint: use part 5 from the last activity.)
- Is condition 3 satisfied? (Hint: use the definition of set product given in the last activity; this is just a simple check.)

Next: Homomorphisms