MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Conjecture: Let $p$ be a prime. Then the group

$G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9, (([a,b]^4)b)^{2p} \rangle$

has a composition series of the form ${\rm PSL}(2,8) - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p - {\rm Z}_p$.

Is there any literature on this subject, and if not, how can this conjecture be proved?

share|cite|improve this question
This seems a reasonable question to me. – Derek Holt Aug 17 '13 at 18:18
@DerekHolt: Indeed. -- Maybe the reason for the votes to close was the way the question was formulated. – Stefan Kohl Aug 17 '13 at 19:25
up vote 16 down vote accepted

For some reason, people seem to be voting to close this, so I will give a quick reply. Your conjecture is true and it can be proved mainly by computer.

The group $G=\langle a,b \mid a^2, b^3, (ab)^7, [a,b]^9 \rangle$ has a homomorphism onto ${\rm PSL}(2,8)$. Let $K$ be the kernel. Then it can be shown that $K$ is nilpotent of class 2, with $|Z(K)|=2$ and $K/Z(K)$ free abelian of rank 7. The element $x:=([a,b]^4b)^2$ lies in $K$ and maps onto a generator of $K/Z(K)$. So factoring out the normal closure of $x^p$ in $G$ will result in an extension of an elementary abelian group of order $p^7$ by ${\rm PSL}(2,8)$.

share|cite|improve this answer
Thanks! I would also like to know whether there are any quotients of this group other than the trivial group, PSL(2,8), and the whole group. I know of one other quotient if p=2, but I think that is a special case. – Thomas Aug 18 '13 at 0:51
There are no other quotients when $p>2$. The action of ${\rm PSL}(2,8)$ on the 7-dimensional integral module is irreducible and reduces mod $p$ to an irreducible module for $p>2$. For $p=2$ the reduction is reducible and uniserial with a submodule of dimension 1. – Derek Holt Aug 18 '13 at 7:34
What about when p is not prime, for example $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9, (([a,b]^4)b)^{8} \rangle$? – Thomas Aug 18 '13 at 9:07
I think the kernel of $G \to {\rm PSL}(2,8)$ is a abelian of order $p^7$ when $4$ does not divide $p$ and, when $4|p$, it has order $2p^7$ and is nilpotent of class 2 with centre of order 2. – Derek Holt Aug 18 '13 at 10:22
So, when 4 does not divide p, are there any other quotients? Also, when 4 does divide p, are there any other than the trivial group, PSL(2,8), the quotient of index two, and the whole group? – Thomas Aug 18 '13 at 10:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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