# Quotients of Hurwitz group

Since my question at Simplicity of infinite groups was not answered (well, at least, my second question), instead of trying to find the isomorhism type of those groups, I will instead try to find the quotients of $G := \langle a, b \ | \ a^2, b^3, (ab)^7, [a,b]^{10} \rangle$. I know of three quotients: PSL(2,41), the Janko group J(1), and the Janko group HJ. What other finite quotients are there?

Edit: I have done some research, and this paper http://www.heldermann-verlag.de/gcc/gcc02/gcc028.pdf seems to suggest that the chevalley group G(2,5) of order 5859000000 is a quotient of G. So I update my question to: What other simple quotients are there?

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These are the only finite quotients of order up to $5 \times 10^8$ (I am trying to extend that, but it is getting harder.) I know also that $q=41$ gives the only finite quotient of form ${\rm PSL}(2,q)$. – Derek Holt Sep 9 '13 at 11:20
Ah, that would explain why those were the only ones I found. All I did was go through my (short) list of interesting elements that I could not relate the order to some earlier element. For each one I tried to set the order of it. If you want, I'll give you my presentation for the second Janko group. I was quite surprised that HJ was a quotient, since the relations given on the other sites have the order of [a,b] equal to 12. – Thomas Sep 9 '13 at 11:31
Sorry, I meant those are the only known finite simple quotients. The kernel of the homomorphism onto ${\rm PSL}(2,41)$ has an infinite free abelian quotient of rank 42, so there are arbitrarily large finite quotients. – Derek Holt Sep 9 '13 at 13:41
Wow, rank 42?! So, what other non-simple quotients are there? – Thomas Sep 9 '13 at 13:58
All direct products of distinct simple quotients of a group are themselves quotients of that group. – Derek Holt Sep 10 '13 at 15:17

Just to summarize, here is all I know about the group $G = \langle x,y \mid x^2=y^3=(xy)^7=[x,y]^{10}=1>$ and its quotients.

It has (essentially unique) homomorphisms onto the simple groups ${\rm PSL}(2,41)$, $J_1$ and ${\rm HJ} = J_2$. There are no other finite simple quotients of order up to $2 \times 10^9$ and no others of type ${\rm PSL}(2,q)$. All direct products of these three groups, like $J_1 \times J_2$ or ${\rm PSL}(2,41) \times J_1 \times J_2$ are also quotients.

Computationally, we can also examine the kernels of the homomorphisms onto the three simple groups. (In principle we could study the kernels onto the direct products, but that would be much harder, because the images are too large.)

The abelianization of the kernel $K$ of the map onto ${\rm PSL}(2,41)$ is free abelian of rank $42$ (this proves that $G$ is infinite) and the associated rational module for the action of ${\rm PSL}(2,41)$ is irreducible. However, on reduction mod 2, this module has submodules of dimensions $1$ and $21$. We can also study the $p$-quotients of $K$ for primes $p$, and it appears to have class $2$ $p$-quotients of order $p^{495}$ for all $p$, so it looks as though there are many very large $p$-quotients!

The abelianization of the kernel of the map onto $J_1$ is elementary abelian of order $11^{14}$ and the action of $J_1$ on this quotient is irreducible. There is also a class $2$ quotient of order $11^{28}$ and a class $3$ quotient of order $11^{42}$, which suggests a pattern.

I have not managed to compute the abelianization of the kernel of $G$ onto $J_2$ yet, but it certainly involves a large elementary abelian 2-group. Update: it is elementary abelian of order $2^{41}$.

These computations work by using the Reidemeister-Schreier algorithm to compute a presentation of a subgroup of finite index in a given finitely presented group. This presentation is initially on the Schreier generators and, for a 2-generator group, the number of these is roughly equal to the index of the subgroup. The presentation can be simplified by eliminating redundant generators, but this tends to make the group relations longer, and so the larger the index of the subgroup, the more complex is the computed presentation. This limits our ability to perform computations with the subgroup, like computing its abelianization. (There are alternative approaches to computing subgroup presentations, but in my experience they all ultimately have similar limitations.)

Added later: It turns out that the group $G$ also has quotients isomorphic to (at least) one of the simple groups $G_2(p)$, $G_2(p^2)$, $G_2(p^4)$ for almost all primes $p$. These are finite quotients of the images of 7-dimensional representations over number fields of degree 4 over ${\mathbb Q}$, constructed by Plesken and Souvignier. See the discussion in Other Quotient of Hurwitz group

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I didn't know that a semidirect product of non-abelian simple groups was isomorphic to the direct product before. It explains a lot. I conjecture that those are the only simple groups. Now all we have to do is find the kernel of the homomorphism for each product (lets start with J1) and we're good. Is there a way to find the presentation of the direct product of two quotients of a group with trivial intersection? That would speed things up a lot. – Thomas Sep 11 '13 at 13:23
It's true that a semidirect product of nonabelian simple groups is isomorphic to the direct product (that's by the Schreier Conjecture, which was proved true by the classification of finite simple groups), but I don't see how that is relevant, and we do not need that for the result about quotients that are direct products. I will have to think about the problem of computing presentations of the larger quotients. They would not be very nice presentations, so there may not be much point in doing that. – Derek Holt Sep 11 '13 at 15:54
I was just confused about why the quotients were direct products instead of semidirect products, but that explains it. I guess we don't need the presentations, but it would have been nicer, and it might have made it easier to prove that there wasn't any more simple quotients. – Thomas Sep 12 '13 at 5:02
It might not be so hard to prove that there aren't any more simple groups. You have already proved that there are no more PSL(2,q) quotients, so that is a start. We could try to prove it for more general classes of simple groups. Also, it is surprising that the action of J1 on 11^14 is irreducible, because J1 has a representation in terms of 7x7 matrices over the field of order 11. – Thomas Sep 12 '13 at 5:19
Is there a way for me to investigate this, or has it gotten so complicated that we can only do it with a computer program? – Thomas Sep 12 '13 at 8:25