Given a free group $F$ on $d$ generators and a normal subgroup $H$ of $F$ whose index is finite of prime power order, is there a systematic way to find the numbers of generators of $H/[H,F]$ and of $H/[H,F]H^p$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ The answer is yes, in the sense that there are algorithms to solve these problems, and it would not be particularly difficult to write programs in a language like GAP or Magma to do so. Is this what you are looking for, or is this more of a theoretical question? $\endgroup$– Derek HoltCommented May 9, 2013 at 12:15
-
$\begingroup$ Thank you Professor Derek Holt, I would like to know if there is an explicit formula (perhaps depending on d and the index of H) to calculate this number of generators. $\endgroup$– Yassine GuerboussaCommented May 9, 2013 at 12:44
-
2$\begingroup$ These numbers can certainly depend on the isomorphism type of $F/H$ (and not just on its order). $\endgroup$– Derek HoltCommented May 9, 2013 at 15:11
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
This is really a cohomological question and has a simple cohomological answer. Recall that if a group $G$ acts on an abelian group $M$, then $M_G$ denotes the coinvariants of the action, that is, the quotient of $M$ by the subgroup generated by $\{\text{$m-g(m)$ $|$ $m \in $M, $g \in G$}\}$. The group $F$ acts on $H$ by conjugation, and thus there is an induced action of $F$ on $H_1(H)$.
Key Observation : $H/[H,F] \cong (H_1(H;\mathbb{Z}))_F$ and $H/[H,F]H^p \cong (H_1(H;\mathbb{Z}/p))_F$.
Indeed, we have $H/[H,H] \cong H_1(H;\mathbb{Z})$ and $H/[H,H]H^p \cong H_1(H;\mathbb{Z}/p)$ by definition, and quotienting by $[H,F]$ just kills off the $F$-action.
The other needed ingredient is the 5-term exact sequence in group homology. Given a short exact sequence
$$1 \longrightarrow A \longrightarrow B \longrightarrow C \longrightarrow 1$$
of groups and a ring of coefficients $R$, this 5-term exact sequence takes the form
$$H_2(B;R) \longrightarrow H_2(C;R) \longrightarrow (H_1(A;R))_B \longrightarrow H_1(B;R) \longrightarrow H_1(C;R) \longrightarrow 0.$$
Letting $Q = F/H$, we will apply this to the short exact sequence
$$1 \longrightarrow H \longrightarrow F \longrightarrow Q \longrightarrow 1.$$
The key simplification that occurs is that $H_2(F;R) = 0$ since $F$ is free. We thus get exact sequences
$$0 \longrightarrow H_2(Q;\mathbb{Z}) \longrightarrow H/[F,H] \longrightarrow H_1(F;\mathbb{Z}) \longrightarrow H_1(Q;\mathbb{Z}) \longrightarrow 0$$
and
$$0 \longrightarrow H_2(Q;\mathbb{Z}/p) \longrightarrow H/[F,H]H^p \longrightarrow H_1(F;\mathbb{Z}/p) \longrightarrow H_1(Q;\mathbb{Z}/p) \longrightarrow 0.$$
If you understand $Q$ enough to calculate its first and second homologies, these short exact sequences allow you to determine $H/[F,H]$ and $H/[F,H]H^p$.
answered May 9, 2013 at 16:54
-
$\begingroup$ Thank you dear professor andy Putman. clearly we can start by a minimally d-generated group Q (I'm interested to the case when Q is a finite p-group), in that case $H_1(Q,Z/p) is isomorphic to the frattini quotient of Q, so it is elementary abelian of rank d. $\endgroup$ Commented May 10, 2013 at 10:53
-
$\begingroup$ also it is not difficult to prove that $F/[F,F]F^p$ is elementary abelian of rank d, and so is $H_1(F,Z/p)$. It follows from your last exact sequence that $H/[F,H]H^p$ and $H_2(Q,Z/p)$ are isomorphic. So we have only to compute $H_2(Q,Z/p)$. I may ask how much harder to do this? $\endgroup$ Commented May 10, 2013 at 11:00
-
$\begingroup$ It's nontrivial to compute it, but there is a huge literature on group cohomology, so there are many tools available. To help your search, you should be aware that $H_2(G;\mathbb{Z})$ is also known as the Schur multiplier of $G$. For a particular finite group of reasonable size, by the way, you should be able to compute $H_2$ using GAP. The relevant packages are cohomolo (see gap-system.org/Packages/cohomolo.html) and hap (see gap-system.org/Packages/hap.html). $\endgroup$ Commented May 10, 2013 at 16:43