If $A$ is a set of generators for the free group $F(X)$ then $|A| \geqslant |X|$.

Question

I'm currently revising my course Geometric Group Theory - my notes say that if $A$ is a set of generators for the free group $F(X)$ then $|A| \geqslant |X|$ because otherwise there are at most $A^{|A|}$ homomorphisms from $F(X)$ to $\mathbb{Z}_2$, a contradiction. I don't quite understand what $\mathbb{Z}_2$ has to do with it - is it to do with if each of the elements of $A$ are in an element or something? If somebody could clarify that would be great, thanks!

Answer 1

The reasoning goes like this: Suppose that $A$ is a generating set of $F(X)$. Then any homomorphism $\phi: F(X) \to \mathbb{Z}_2$ is completely determined by where it sends each element of $A$. The number of functions from $A$ to $\mathbb{Z}_2$ is clearly $2^{|A|}$, so there are at most $2^{|A|}$ homomorphisms from $F(X)$ to $\mathbb{Z}_2$.

On the other hand, we know that there are precisely $2^{|X|}$ homomorphisms from $F(X)$ to $\mathbb{Z}_2$ since $F(X)$ is free on the set $X$. This therefore implies that $|A| \geq |X|$.

Answer 2

Here is a related approach that works even for free groups of infinite rank. One uses that the abelianization of $F(X)$ is the direct sum of $|X|$ copies of $\mathbb{Z}$. So, the abelianization tensored with $\mathbb{Z}/2$ is the direct sum of $|X|$ copies of $\mathbb{Z}/2$, which is a vector space over $\mathbb{Z}/2$ of dimension $|X|$; I'll call this the "$\mathbb{Z}/2$ abelianization" of $F(X)$. But if $F(X)$ were generated by fewer than $|X|$ elements it would follow that the $\mathbb{Z}/2$ abelianization is spanned by a number of elements less than the dimension, contradiction.