# Commutator subgroup of a free group

Let $F_k$ be the free group of rank $k$.

If $k=2$ it is not hard to see that the set $\{[s_1^{n_1},s_2^{n_2}] \mid n_i\neq 0\}$ is a basis for $F_2'$. (Prime denotes the commutator subgroup).

What is a basis for $F_k'$ if $k\geq3$?

• What do you mean with basis? Just a generating set? Commented Aug 19, 2012 at 19:17
• $F_k'$ is a free group since subgroups of free groups are free. I am asking for a set of free generators. Commented Aug 19, 2012 at 20:44

It is straightforward to write down the Schreier generators of a subgroups of finite index of a group given by a finite presentation. When the group is free, this will give you a free basis, by the proof of Schreier that subgroups of free groups are free. The free basis depends on a choice of well-ordering of words in the generators of $G$ and on a transversal of the subgroup in $G$. It will not necessarily give the "nicest" free basis, and it gives a slightly more complicated basis than yours in the 2-generator case.

Let $G$ be free on $x_1,\ldots,x_k$. Then the obvious right transversal for $G'$ in $G$ is $\{x_1^{n_1}\cdots x_k^{n_k} \mid n_i \in \mathbb{Z} \}$ and (if I have got this right), this gives rise to the free basis

$\{ x_1^{n_1}\cdots x_m^{n_m} x_l (x_1^{n_1}\cdots x_l^{n_l+1} \cdots x_m^{n_m})^{-1} \mid n_i \in \mathbb{Z}, 1 \le l < m \le k, n_m \ne 0\}$

of $G'$.

• The commutator subgroup has infinite index. You began by assuming finite index. Is that relevant? Commented Aug 20, 2012 at 11:01
• @JimConant: I think in his first paragraph he was just saying that in the case of finite index this is easy. The algorithm still works for subgroups of infinite index. Commented Aug 20, 2012 at 12:48
• @user1729: Thanks for the clarification! Commented Aug 20, 2012 at 13:07
• That's right - the theory does not depend on the index being finite, and in some examples, like this one, the calculation can be carried out when the index is infinite. Commented Aug 20, 2012 at 14:44

The same works in the case where you have a free group $F$ on infinitely many generators, as long as those generators are ordered. In general, in the setting of Schreier transversals you would require a well-ordering but in the case of the commutator subgroup the Schreier transversal is so easy to write that just a total ordering is enough.