TITLE: New relator in hurwitz group
QUESTION [7 upvotes]: I have found that $([a,b]^2[a,b^2])^n$ is a good relator to use in my search for quotients of $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. For n<=5 $H := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10}, ([a,b]^2[a,b^2])^n \rangle$ is the trivial group, but when n=6, it is the Janko group J1, and when n=7, it is the Hall-Janko group J2. I have tried finding what the group is when n = 8, but without success (magma failed, and GAP used too much memory and crashed my computer). Is there a way of finding what this group is?

REPLY [8 votes]: A brute-force calculation in GAP (searching for homomorphisms) shows that the group G has a quotient $PSL_2(41)\times J1^2\times J2^2\times G_2(5)^2$.
There are no other quotients that are simple groups of order $\le 10^{10}$, and (as you surely know) the group is perfect.
The order of $[a,b]^2[a,b^2]$ under the simple quotients is 21 (PSL), 6 and 10 (J1), 7 and 15 (J2) and both times 31 (G2(5)). Thus this gives you information for $n=10$, but not $n=8$.
Its only $PSL_2$ quotient is $PSL_2(41)$. (Noam Elkies mentioned this already, but the statement might not have been clear about that there are no others) , this is by the Plesken/Fabianska algorithm in Magma.