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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?

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5  
This seems a reasonable question to me. –  Derek Holt Aug 17 '13 at 18:18
5  
@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

1 Answer 1

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

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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
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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

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