# Show that if $G$ is a free group with $n$ free generators, then any system of free generators for $G$ has $n$ elements.

This is the Corollary 69.5 of Munkres section 69, which states as follows:

If $$G$$ is a free group with $$n$$ free generators, then any system of free generators for $$G$$ has $$n$$ elements.

He defined free group and system of free generators as follows:

[Definition]

Let $$\{a_{\alpha}\}$$ be a family of elements of a group $$G$$. Suppose each $$a_{\alpha}$$ generates an infinite cyclic subgroup $$G_{\alpha}$$ of $$G$$. If $$G$$ is the free product of the groups $$\{G_{\alpha}\}$$, then $$G$$ is said to be a free group, and the family $$\{a_{\alpha}\}$$ is called a system of free generators for $$G$$.

To prove this corollary, he firstly proved a theorem, stated as follows:

[Theorem 69.4]

If $$G$$ is a free group with free generators $$a_{\alpha}$$, then $$G/[G,G]$$ is a free abelian group with basis $$[a_{\alpha}]$$, where $$[a_{\alpha}]$$ denotes the coset of $$a_{\alpha}$$ in $$G/[G,G]$$.

Then the proof of this corollary is one-line:

Proof:

The free abelian group $$G/[G,G]$$ has rank $$n$$.

This confused me. By his definition, the system of free generators is just set of the generator of each infinite cyclic subgroup $$G_{\alpha}$$ in the free product, so where does the "any system" come from?

Then by his definition, if $$G$$ has $$n$$ free generators, does it by definition imply that the system of generators have $$n$$ elements?

Why did he take a detour and use the rank of free abelian group?

I think I maybe got mixed up about the definition of system of generators, but the definition I quoted above is the only definition about system of generators in the book.

Any idea? Thank you!

• A common idea in algebra is to use techniques to reduce the question to an easier one. To show that a commutative ring has invariant basis number we quotient by maximal ideals to make it a problem about vector spaces. Once we know this we know that the rank of a free abelian group is well defined. Now we want to reduce our problem about free groups to this one, so we quotient about by the commutator to get a free abelian group. Which free abelian group it is depends on the number of free generators of the free group, so this must be constant. Commented Nov 14, 2019 at 17:15

The point is that given a free group $$G$$, you can always choose different system of free generators (just like you can choose different bases for a vector space). In linear algebra any vector space is free, but when we define the dimension, we have to make sure that all bases have the same number of elements.
Here it's the same: for instance, if $$G$$ is free on $$\{a,b\}$$, it is also free on $$\{a,ab\}$$, or on $$\{ab,aba\}$$. Now all those sets have the same number of elements (and the theorem precisely says that they have to), but there is no reason why it should be true at first sight.
Especially since non-commutative free groups are complicated and can have weird behavior: for instance, a free group on $$n$$ generators can have a free subgroup on $$m$$ generators with $$m>n$$ (actually, a free group of rank $$2$$ has free subgroups of any finite rank, and even of infinite countable rank).
Now as a comment pointed out, the idea of the proof is to reduce to a case we already know: the case of abelian groups, where it is much easier to see that a free abelian group (so a free $$\mathbb{Z}$$-module) has a well-defined rank, ie any system of free generators has the same length. It turns out that a system of free generators of a free group given after abelianization a system of free generators of the corresponding free abelian group, so they all have the same size.