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?
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$\begingroup$ At first sight, it looks like $H=G$. $\endgroup$– Alex DegtyarevCommented Nov 2, 2014 at 0:50
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$\begingroup$ That can't be possible, because the order of $([a,b]^2[a,b^2])$ is not 8 (otherwise when n=6 the group would be trivial). Also, the Janko group J1 is a quotient of G, but not of H (when n=8), since the Janko group contains no elements of order 8. $\endgroup$– ThomasCommented Nov 2, 2014 at 1:39
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$\begingroup$ Woops, sorry, I made an error in the question. Corrected now. $\endgroup$– ThomasCommented Nov 2, 2014 at 1:40
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$\begingroup$ I suppose you know already that $G$ maps onto ${\rm PSL}_2({\bf F}_{41})$. (The exponent of $\ [a,b]^2[a,b^2]$ in the image is $21$.) $\endgroup$– Noam D. ElkiesCommented Nov 4, 2014 at 3:18
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2$\begingroup$ Well, if that's true then it certainly implies that your group is infinite (since no finite group is isomorphic to a proper subgroup of itself). $\endgroup$– HJRWCommented Nov 4, 2014 at 11:27
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1 Answer
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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.
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$\begingroup$ Are those the highest powers of each of those simple groups that are quotients of G? It was found (in another question) that G2(3^4), G2(7^2), G2(11^2), G2(13^4) are also quotients of G (and there should be more quotients corresponding to higher primes). Is it possible to figure out the highest power of each of those that are quotients of G, and the exponent of $[a,b]^2[a,b^2]$ for each? $\endgroup$– ThomasCommented Nov 5, 2014 at 23:17
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$\begingroup$ I had noticed that the simple quotients were there for different exponents. Does that always imply that there are multiple powers of the group appearing as a quotient? $\endgroup$– ThomasCommented Nov 6, 2014 at 0:06
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1$\begingroup$ Yes, there are two (independent) copies each of J1, J2 and G2(5). $\endgroup$– ahulpkeCommented Nov 6, 2014 at 3:31
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$\begingroup$ and I should have said only two -- assuming my program is correct it found all normal subgroups such that $G/N$ is of the given isomorphism type. $\endgroup$– ahulpkeCommented Nov 6, 2014 at 4:11
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2$\begingroup$ Yes it is possible to compute the order of that element in the image of $G$ in $G_2(p^e)$ (where $e=1,2$ or $4$) for any prime $p$. The $7$-dimensional representation over a number field of degree $7$ over the rationals was computed explicitly by Plesken and Robertz, so you just have to reduce the matrices modulo $p$. I tried this for primes up to about $10000$, but the smallest orders I found were $30$, $31$, $108$,... $\endgroup$ Commented Nov 6, 2014 at 8:58