Abelianization of free groups

Question

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"?

Answer 1

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) $S\cong T$.

Answer 2

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.

Answer 3

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.

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I hope this helps ^_^