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Aug 19, 2013 at 1:40 comment added Thomas Oh, right, but other than those, are there any extra quotients?
Aug 18, 2013 at 16:14 comment added Derek Holt You will always get the quotients from divisors of $p$. So, for example, $G(15)$ will have $G(5)$ and $G(3)$ as quotients.
Aug 18, 2013 at 10:27 comment added Thomas 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?
Aug 18, 2013 at 10:22 comment added Derek Holt 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.
Aug 18, 2013 at 9:07 comment added Thomas What about when p is not prime, for example $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^9, (([a,b]^4)b)^{8} \rangle$?
Aug 18, 2013 at 7:34 comment added Derek Holt 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.
Aug 18, 2013 at 0:51 comment added Thomas 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.
Aug 18, 2013 at 0:40 vote accept Thomas
Aug 17, 2013 at 18:40 history edited Derek Holt CC BY-SA 3.0
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Aug 17, 2013 at 18:18 history answered Derek Holt CC BY-SA 3.0