Free Groups and Presentations

In the last section we've taken groups, found sets of elements that generate these groups, and constructed a metric graph with a natural isometry corresponding to every group element. We can also take a reverse approach: instead of asking what elements generate a group, we can construct groups by defining them to be generated by some elements.

Free groups

A shift in perspective is in order. Before we looked at groups as either isometries/symmetries of a space or figure or as a bunch of elements with a defined product. Another useful way of looking at groups is as words in some alphabet. For instance, take the alphabet consisting of a single letter: a. Our rule for constructing words is the following. We can use the letter a or its upper case version A. If a word contains the string aA or Aa we can cancel it. Thus, in our language the word aaAa is the same as aa (since we can cancel aA or Aa); we write aaAa=aa.

Activity 1: Words, words, words

The word with no letters is called the empty word, and will be denoted by the letter e. Note that e=aA=Aa.
  1. Suppose we wanted to minimize the number of letters in our words by canceling all aA or Aa. Can we reduce all the words in our language to the form aaa...a (for some number of a's) or AAA...A or the empty word e?
  2. Is the word aaa equal to aaaa? That is, can we transform one to the other by addition or deletion of aA and Aa?
  3. Define the product operation in our language to be concatenation, so that aaAa•AAA=aaAaAAA. Are the words in our language together with this product satisfy the conditions for being a group? (Hint: what is the identity? Does every word have an inverse? Does condition 3 hold?)
  4. Draw a Cayley graph for this group with the generator a. Label all elements with reduced words, those in which all aA and Aa have been canceled.

In the last part of the above activity you should notice that the Cayley graph of the group defined by our one letter language looks exactly the same as the Cayley graph of Z with the generator 1. Both are long "lines". If we simply relabel the letter a in our language as 1 and A as -1, we get the exact same structure as the group of integers Z. If a relabeling of elements produces identical Cayley graphs, we say that the two groups are isomorphic--for our purposes they are the same. Note, however, that just because two Cayley graphs can't be made the same by relabeling does not mean that the groups are not isomorphic. After all, choosing different generators will often lead to two different Cayley graphs of the same group.

We don't have to restrict ourselves to just one letter. We can take, for example the alphabet {a,b}. Again, words are composed of the letters a,A,b,B such that we can cancel aA, Aa, bB, and Bb.

Activity 2: a and b sat on a tree
  1. Which of the following words are equal? aBaAAABbaabA, aBAbBa, aBBa, ab, bABbaa, ABBaAbbaaBbb, BaAbAaabbBBBbBBa, e
  2. Is this group abelian? (Recall that a group G is abelian if for every two elements g,h∈G we have gh=hg; is, for example, ab=ba?)
  3. Draw part of the Cayley graph of this group with generators a and b; again label the vertices with the reduced words. (Hint: another way to draw a Cayley graph is to start with one vertex, the identity, then draw an outward directed edge corresponding to every generator, terminating it with an appropriately labeled vertex, and an inward directed edge corresponding to every inverse of a generator. Repeat recursively for the new vertices just created.)

Your finished product from part 2 above should look something like this. This is just a small part of the infinite Cayley graph comprised of the vertices close to the identity. Recall our convention that that each edge of the Cayley graph is length 1. Thus my picture of the Cayley graph is not true to life, so to speak; it's just more convenient to visually shrink the edges when drawing the graph.

Of course we can have arbitrary many letters in our alphabet, with the only cancellation being that of a lower case and an upper case of the same letter. Such groups are called free groups. We'll denote the free group generated by an n letter alphabet Fn. We've already seen that F1=Z, the group of integers.

Activity 3: Have groups—will product

Given any two groups G and H we can form the direct product G×H by taking as elements pairs all distinct pairs of elements (g,h), g∈G, h∈H and defining the product as (g1,h1)•(g2,h2)=(g1g2,h1h2), where the product on the left is the group product of G and the product on the right is the group product of H.
  1. Check that G×H is in fact a group.
  2. If G has five elements and H six, how many elements does G×H have?
  3. Recall that the group of rotational symmetries of a regular n-gon, denoted Zn is generated by (among other elements) the 360/n rotation. Is Z3×Z6 generated by a single element? How about Z3×Z4?
  4. For which positive integers m and n can Zn×Zm be generated by a single element? (Hint: this is another number theory question in disguise. Look at small cases to notice the pattern.)


A relation in a group is simply a word that's equal to the empty word. In F1, for example, some relations are Aa, aAAa, aA, etc. In D4, relations include L1L1, R90R90R180, and R180L4L2. We can use the Cayley graph to simply read off the relations of a group: trace along a path from the identity vertex e to itself, recording the generators corresponding to the edges you pass along the way. Below is part of the Cayley graph of the group Z×Z (or equivalently, F1×F1) with generators (1,0) and (0,1).

Cayley graph of ZxZ
Activity 4: One word
  1. What are some relations in Z×Z? Write them in terms of the generators.
  2. Recall that the group F2 can be generated by letters a and b (with inverses A and B respectively) and consists of all words in these letters up to cancellation of inverses. Suppose we allow the cancellation of another word in addition to the four above, the word abAB. Call this new group H. Show that ab=ba in H.
  3. Below is the Cayley graph of F2. How would you modify it to get the Cayley graph for H? (Hint: some words (vertices) that are distinct in F2 are equal in H. Which ones? If you're having difficulties, start out by drawing part of the Cayley graph of H around the identity.)
Cayley graph of F_2
  1. How would you relabel the Cayley graph of H to get the the Cayley graph of Z×Z given previously?

As you've see in the activity above, while F1=Z, we've had to add relations to the group F2 to get the group Z×Z. These additional relations were all the words we could reduce to the identity by canceling the word abAB, that is, setting abAB=e. A group presentation is a list of letters which generate a group and a list of words which we cancel, usually written as ⟨letters|words⟩. For instance, F2=⟨a,b| ⟩. The list of canceling words is empty because by convention the words aA, Aa, bB, etc. are implied; they are part of every group presentation since elements in a group must have inverses. As we've seen above, Z×Z=⟨a,b|abAB⟩. Thus a free group is one with no relations other than the implied inverse ones.

Activity 5: A rather presentable group
  1. How many elements does the group G=⟨a|aaaa⟩ have? (Hint: since aaaa=e, aa is its own inverse, so aa=AA.)
  2. Construct a product table for G and draw its Cayley graph. What familiar group is G isomorphic to?
  3. Here's a more challenging problem. Consider the group H=⟨a,b|aa,bbbb,abab⟩. How many elements does it have? Draw its Cayley graph with the generators a and b. Which group does this presentation define?
  4. Write down any presentation. Is the resulting group abelian? How many elements does it have? What are their orders? Draw a Cayley graph.

Next: Subgroups, normality, and quotients