# Finitely generated free group has finite rank

Initially I was trying to prove that the commutator subgroup of $F_2$ (=free group of rank $2$) is not finitely generated. It seems possible to prove that it is indeed free of infinite rank. To get to a contradiction I have to prove that if a free group is finitely generated then it has finite rank but I can't find a way, despite it seems trivial.

• If it is generated by $n$ elements then it has at most $2^n$ homomorphisms to the group of order $2$, so its rank must be at most $n$ (since a free group of rank $m$ has exactly $2^m$ such homomorphisms). – Derek Holt May 20 '17 at 10:09
• By the way it very unclear what you are asking, and you should probably define what you mean by rank, since it is one of the most overused words in mathematics. – Derek Holt May 20 '17 at 10:12
• @DerekHolt I will try to clarify your arguments to get to an answer. The homomorphisms from a finitely generated group on $n$ generators to $\mathbb{Z}_2$ are defined by their values on the generators. Since there are two choices for each generator, there are (at most) $2^n$ homomorphisms. If the rank was greater than $n$ (possible infinite) then I would have more than $2^n$ options, a contradiction. That seems correct and elegant. – user128787 May 20 '17 at 10:18

If we would have a free group $F$ with rank $|X|$ where $|X|$ is infinite, then we can see $F$ as the group of all words over $X\cup X^{-1}$ with the usual concatenation (the 'standard' free group). Suppose we would have a finite generating set $\{g_1,\dots,g_k \}$. Then every $g_i$ is a word $x_{i_1}^{\pm}\dots x_{i_{n_i}}^{\pm}$ which uses only finitely many letters of $X$, so there will always be an element $x \in X$ not appearing in any of the generators. This implies that $x$ is not in the subgroup generated by $\{g_1,\dots,g_k\}$ so no such generating set can exist.

Alternatively, you could prove that the abelianization of the free group $F$ of rank $|X|$ is the abelian free group of rank $|X|$, i.e. $$\bigoplus_{x\in X} \mathbb{Z}$$ using the universal property. Tensoring with $\mathbb{Q}$ gives you a vector space of dimension $|X|$, which has clearly no finitely generating set (because such a set should contain a basis).

The commutator group is the kernel of the canonical surjection $F_2\to Z^2$ given by abelianization. It is therefore isomorphic to the Cayley graph of $Z^2$ endowed with a standard set of generators. From this, drawing a picture of this Cayley graph in the plane, one sees that it is freely generated by the elements $a^nb^m[a,b](a^nb^m)^{-1}$, corresponding to all squares in the plane. Note that these elements are commutators $=[a^nb^ma(a^nb^m)^{-1},a^nb^mb(a^nb^m)^{-1}]$. Geometrically, if an element belongs to the commutator group, it defines a closed curve in the graph, and this curve decomposes into squares of size 1. One can translate this into an algebraic proof that these elemnts is a free basis.

• Thanks for your answer but my question was not if the commutator subgroup is finitely generated. As the title says, I want to prove that a finitely generated free group is of finite rank... – user128787 May 20 '17 at 10:03

The commutators $[x,y^n]=xy^nx^{-1}y^{-n}$ for $n=1,2,3,\ldots$ generate freely an infinite rank subgroup of the free group with generators $x$ and $y$.

• So what ? A free group of finite rank contains infinite rank free subgroups. – Thomas May 20 '17 at 9:41
• @Thomas which is what the OP wants to prove. – Lord Shark the Unknown May 20 '17 at 9:44
• No he wants to prove that the commutator subgroup is not f.g. – Thomas May 20 '17 at 9:55
• Seems that my question is not clear... I only want to prove that a finitely generated free group is of finite rank. I know how to prove that the commutator subgroup is of infinite rank. – user128787 May 20 '17 at 10:06
• @user128787 What does finite rank mean if not "finitely generated"? – Lord Shark the Unknown May 20 '17 at 10:07