11
$\begingroup$

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?

$\endgroup$
2
  • 5
    $\begingroup$ This seems a reasonable question to me. $\endgroup$
    – Derek Holt
    Aug 17, 2013 at 18:18
  • 5
    $\begingroup$ @DerekHolt: Indeed. -- Maybe the reason for the votes to close was the way the question was formulated. $\endgroup$
    – Stefan Kohl
    Aug 17, 2013 at 19:25

1 Answer 1

17
$\begingroup$

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 ${\rm PSL}(2,8)$.

$\endgroup$
7
  • $\begingroup$ 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. $\endgroup$
    – Thomas
    Aug 18, 2013 at 0:51
  • 2
    $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Aug 18, 2013 at 7:34
  • $\begingroup$ What about when p is not prime, for example $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9, (([a,b]^4)b)^{8} \rangle$? $\endgroup$
    – Thomas
    Aug 18, 2013 at 9:07
  • $\begingroup$ 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. $\endgroup$
    – Derek Holt
    Aug 18, 2013 at 10:22
  • $\begingroup$ 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? $\endgroup$
    – Thomas
    Aug 18, 2013 at 10:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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