# Abelianization of free groups

I'm reading Hatcher's Algebraic Topology and I have some questions about an argument on Page 42:

The abelianization of a free group is a free abelian group with basis the same set of generators, so since the rank of a free abelian group is well-defined, independent of the choice of basis, the same is true for the rank of a free group.

The first question is: why is the rank of a free non-abelian group not well-defined, independent of the choice of basis?

The second question is: why "the same is true"?

It is not a priori clear that for two free groups $$F(S)$$ and $$F(T)$$ on sets $$S$$ and $$T$$ we have that if $$F(S)\cong F(T)$$ is an isomorphism of groups, there is an isomorphism of the sets $$S\cong T$$. For example a priori it could be the case that there is an isomorphism from the free group on one generator to a free group on two generators.

Luckily a posteriori this is not the case. We know from linear algebra that the rank of a free abelian group is welldefined, ie. that $$\Bbb Z[S]\cong\Bbb Z[T]$$ as abelian groups implies $$S\cong T$$ as sets. Now there is a canonical isomorphism from the abelianization of the free group $$(F(S))^{ab} := F(S)/[F(S),F(S)]$$ to the free abelian group $$\Bbb Z[S]$$. So if we assume $$F(S)\cong F(T)$$ we also get that $$\Bbb Z[S]\cong \Bbb Z[T]$$ and hence (by the aforementioned result) $$S\cong T$$.

why is the rank of a free non-abelian group not well-defined, independent of the choice of basis?

The rank of a free group is well-defined. This is exactly what Hatcher is proving. Here is his proof, spelled out in more detail:

Suppose $$B$$ and $$B'$$ are two bases for a free group $$G$$. Let $$\pi : G \to G_{\text{ab}}$$ be the abelianization map. Then $$\pi(B)$$ and $$\pi(B')$$ are bases for $$G_{\text{ab}}$$, and moreover the restrictions $$\pi|_B$$ and $$\pi|_{B'}$$ are injective. We conclude that $$\lvert B \rvert = \lvert \pi(B) \rvert = \lvert \pi(B') \rvert = \lvert B' \rvert$$, as desired.

For the first question, the rank of a free abelian group $$G$$ is exactly the dimension of the vector space $$\mathbb{Q} \otimes G$$. So if you're convinced that the dimension of a vector space is independent of the choice of basis, then the same must be treu for the rank of a free abelian group.

For the second question, we run a similar argument. The rank of $$G$$ (free abelian) is the same thing as the dimension of $$\mathbb{Q} \otimes G$$, and computationally this is because a basis for $$G$$ becomes a basis for $$\mathbb{Q} \otimes G$$. But what about a free group $$F$$? Well, a "basis" for this free group (by which I mean the generators) becomes the basis for the abelianization $$F^\text{ab}$$!

So if we have our favorite free group with generating set $$X$$, $$F(X)$$, then the abelianization $$F(X)^\text{ab}$$ will be the free abelian group with basis $$X$$, and then the tensor product $$\mathbb{Q} \otimes F(X)^\text{ab}$$ will be a $$\mathbb{Q}$$ vector space with basis $$X$$.

In particular, the rank of a free group (that is, the cardinality $$|X|$$) must be well defined! Since if we can write $$F(X) \cong F(Y)$$ then chasing through the above process we must have $$|X| = |Y|$$, since they become bases for a vector space.

I hope this helps ^_^

• I have always found it strange to use tensor products in proofs about ranks of abelian groups. It seems to be unnecessarily using concepts outside of group theory that not everybody will be familiar with. (Tensor products are rarely covered in undergraduate courses.) Commented Aug 30, 2022 at 18:39
• @DerekHolt -- I see where you're coming from. When I first started doing research, my advisor gave me a paper by Nekrashevych and Sidki that proved a fact about integer lattices by first tensoring with $\mathbb{Q}$, and it looked like magic to me. I was an undergrad at the time, and didn't know much more than the definition of tensor products. Of course, the idea of base change is incredibly useful, so I'm glad that paper started me on the long road to understanding it (which I'm still on, of course). Commented Aug 30, 2022 at 19:46
• Either way, I definitely pitched this answer at the wrong level for OP (it's also far from my most coherently written answer besides that...), so it's a good thing that other answerers came along to write up other approaches ^_^ Commented Aug 30, 2022 at 19:47