I’ll begin my second (!) blog post about free groups by recalling a proposition that cuts right to the heart of what a free group is.

**Universal property.** Let be a generating set. Then is the unique froups os that for any gnerated by there exists a unique homomorphism such that the diagram below commutes.

This tells us that every group on some generating set is the surjected image of a free group, and hence we can think of the group as a quotient of that free group. This exactly what a group presentation is: in the set of relators we specify all the elements which normally generate (i.e. form a normally generated subgroup which is) the kernel of the map from .

I’ll restate a puzzle which has to do with this fact: Prove that a group has a surjection to if and only if has a presentation in which the sum of the exponents on every relator is . I won’t spoil the puzzle for anyone who’s working on it but for a slightly cryptic hint you can read the following sentence backwards: STNEMELE HCUS GNINIAMER EHT YB TUO DOM.

Next up we address the following postponed theorem:

**Theorem: **For free groups given by generating sets and , we have if and only if .

The proof of this theorem relies on the observation that much of the difficulty in working with free groups comes from the lack of commutativity. A surprisingly effective way of dealing with this problem is to create an abelian quotient group by forcing elements which would be zero if the generators *did* commute into the kernel of a quotient map. To see this in rigorous terms we need a few definitions.

For elements the element is called a commutator. The subgroup of G generated by the set of commutators is called the commutator subgroup and is denoted . To see that this subgroup is normal, let , where the are commutators. We must show that for all , we have . To do so we need only show which we can do by writing this element as a product of commutators: . The resulting group is called the ablianization of , and it is, as its name suggests, abelian.

We can now return to the question of settling the isomorphism classes of free groups. On the one hand, if , there exists a bijection between and which can be used to induce an isomorphism . For the harder direction, we note that taking if , then their abelianizations must be isomorphic, , and those are much easier to understand. Namely,

And hence we find that under our assumptions . Which is only true if . We can illustrate this fact in the finite case by noting that contains an isomorphic copy of , the quotient with which is . Performing successive mods of these subgroups will reduce to the identity after exactly steps, a fact which uniquely identifies this group with the integer , and prevents it from being isomorphic to for . The details are left to be sorted out.

We move now to the surprising fact that subgroups of finitely generated groups are not necessarily finitely generated. The proof is constructive, and it works by offering a subgroup of which is not finitely generated, namely

To see that is not finitely generated, we must show that no freely reduced word in is the identity. Towards a contradiction, suppose that

For the above to be true, there must be some sequence of cancellations that reduces the right hand expression to the identity, which means there is a pair of and which are immediately adjacent except for the powers of which cancel with each other, and hence must be part of an expression of the form which in turn looks like in the original word, meaning that it was not freely reduced with respect to our generating set, contradicting our assumption that there is such a freely reduced word which equals the identity. We have shown that , which is the free group on countably many generators.

We only need to modify this argument slightly to conclude that has subgroups isomorphic to for every integer . Indeed take the generating set .

To contrast with this surprising fact, we note that every quotient group of is finitely generated, since we can use the image of its generators under the natural surjection , where the same idea holds in any finitely generated group. In practice this looks like using the “same” generators for the quotient group.

Since we’ve started down the path of analyzing subgroups of , and I’ve mentioned a surprising fact followed but an underwhelming fact, I’ll conclude the post by mentioning a surprising plan of attack on a somewhat reasonable fact. In choosing a subgroup of the free group , we have the freedom to choose its generators, but not to introduce any relators, this makes it seem that the resulting group, whatever the cardinality of a minimal generating set happens to be, should probably be free.

**Theorem:** If then is a free group.

This fact is true, but the proof is surprisingly difficult, it turns out that the key idea is to use a seemingly more difficult theorem that invokes the geometric structure of and its actions.

**Theorem: **A group acts freely on a tree without edge inversions if and only if that group is free.

The proof of the first theorem now becomes simple. We simply state that acts on the tree (without edge inversion), and therefore each of its subgroups act on , and indeed by the above theorem this makes free.

This is a beautiful result of the Cayley graph machinery we’ve seen so far. It raises another question about what it means when we find , but the graph has degree four at every vertex whereas the the graph has degree six at every vertex. There certainly can’t be such a tree contained as a sub-tree of , but we do have a geometric object which is essentially a disjoint union of paths which we can treat as such a “sub-tree.” An illustration of this tree in red and orange is given below.

We take the subgroup generators .

Here we can imagine that the tendrils drawn in blue are symmetrically extending in an identical manner at each vertex at the ends of the blue baths. The symmetries of this set of paths are exactly the symmetries of the tree of degree .

Nice post, Horace!

I first have a comment on your puzzle since I would like to elaborate on the solution a bit more. Let me remind everyone of the problem: Prove that a group G has a surjection to \mathbb{Z} if and only if G has a presentation in which the sum of the exponents on every relator is 0.

Going back to the universal property of free groups, then, there is a surjection from to if and only if the concatenated diagram (gluing the surjection of to at the bottom of Horace’s diagram) is commutative. We would like to get the information about the kernel to say the information about the relators. Indeed, we note that , where is a map from to so that all but one generators are in the kernel. Backtrack a bit, since has to send all generators of to 1, the map needs to have the kernel of all relators summing up to 0 to make the map compatible (in other words, canceling all generators but one).

Nice post, Horace!

I first have a comment on your puzzle since I would like to elaborate on the solution a bit more. Let me remind everyone of the problem: Prove that a group G has a surjection to \mathbb{Z} if and only if G has a presentation in which the sum of the exponents on every relator is 0.

Going back to the universal property of free groups, then, there is a surjection from to if and only if the concatenated diagram (gluing the surjection of to at the bottom of Horace’s diagram) is commutative. We would like to get the information about the kernel to say the information about the relators. Indeed, we note that , where is a map from to so that all but one generators are in the kernel. Backtrack a bit, since has to send all generators of to 1, the map needs to have the kernel of all relators summing up to 0 to make the map compatible (in other words, canceling all generators but one).

Free groups (and infinite groups more generally) really make one question what it means for groups to be “smaller” than other groups. For example, we can think of as being “one-fifth” of . Even though both are infinite, we can fit five copies of the subgroup in the larger group. However, this flies out the window with free groups since for any . Using the same logic as before, this would imply that is both smaller than and larger than , which appears to be a contradiction. I wonder how many cosets has in and if that might help me wrap my head around this.

Great post Horace! One thing I wonder about is whether there are free subgroups of non-free groups. Is it a requirement for a group to be free and have free subgroups or do things get more complicated? Could we also use the free subgroup theorem for this?

I think things get more complicated; isn’t a free group, but it’s got as a subgroup, which is free.

(We also see this in light of more recent developments in class about free products!)