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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 $k\geq3$?

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What do you mean with basis? Just a generating set? –  sebigu Aug 19 '12 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. –  Mustafa Gokhan Benli Aug 19 '12 at 20:44
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1 Answer 1

up vote 5 down vote accepted

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 $G'$.

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The commutator subgroup has infinite index. You began by assuming finite index. Is that relevant? –  Grumpy Parsnip Aug 20 '12 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. –  user1729 Aug 20 '12 at 12:48
    
@user1729: Thanks for the clarification! –  Grumpy Parsnip Aug 20 '12 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. –  Derek Holt Aug 20 '12 at 14:44
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