# 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). 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. 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. 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... 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. May 20 '17 at 9:41
• @Thomas which is what the OP wants to prove. May 20 '17 at 9:44
• No he wants to prove that the commutator subgroup is not f.g. 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. May 20 '17 at 10:06
• @user128787 What does finite rank mean if not "finitely generated"? May 20 '17 at 10:07