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 $k\geq3$?

  • $\begingroup$ What do you mean with basis? Just a generating set? $\endgroup$ – sebigu Aug 19 '12 at 19:17
  • $\begingroup$ $F_k'$ is a free group since subgroups of free groups are free. I am asking for a set of free generators. $\endgroup$ – Mustafa Gokhan Benli Aug 19 '12 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 $G'$.

  • $\begingroup$ The commutator subgroup has infinite index. You began by assuming finite index. Is that relevant? $\endgroup$ – Cheerful Parsnip Aug 20 '12 at 11:01
  • $\begingroup$ @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. $\endgroup$ – user1729 Aug 20 '12 at 12:48
  • $\begingroup$ @user1729: Thanks for the clarification! $\endgroup$ – Cheerful Parsnip Aug 20 '12 at 13:07
  • $\begingroup$ 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. $\endgroup$ – Derek Holt Aug 20 '12 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.