15
$\begingroup$

I have been working with Hurwitz groups, and I came across the group $G := \langle a, b \ | \ a^2, b^3, (ab)^7, ([a,b]^2ab)^6 \rangle$. I'm trying to figure out exactly what this group is. I know it contains two copies of ${\rm PSL}(2,13)$, and there seems to be a very large 2-group in there as well, of order $2^{28}$ at least. Is this group even finite, and if so, what is the order? Does it contain any other simple groups?

$\endgroup$
24
$\begingroup$

It is infinite.

As you pointed out yourself, the kernels of the maps onto ${\rm PSL}(2,13)$ have $2$-quotients of apparently unbounded order, which can be computed using the $p$-quotient algorithm.

I also found homomorphisms onto the Janko sporadic groups J1 and J2. Then I found a subgroup of index $525$ in J2 of which the inverse image in $G$ has infinite abelianization. Here are some Magma commands to do some of these computations.

> G<a,b> := Group<a,b|a^2,b^3,(a*b)^7,((a,b)^2*a*b)^6 >;
> SQ := SimpleQuotients(G,1000000: Limit:=10 );
> ChiefFactors(Image(SQ[1][1]));
    G
    |  A(1, 13)                   = L(2, 13)
    1
> P := pQuotient(Kernel(SQ[1][1]), 2, 3 : Print:=1);

Lower exponent-2 central series for $

Group: $ to lower exponent-2 central class 1 has order 2^14

Group: $ to lower exponent-2 central class 2 has order 2^42

Group: $ to lower exponent-2 central class 3 has order 2^70 

> ChiefFactors(Image(SQ[2][1]));
    G
    |  J1
    1
> h := SQ[3][1];
> I := Image(h);
> ChiefFactors(I);
    G
    |  J2
    1
> L := LowIndexSubgroups(I,1000);
> [Index(I,l): l in L];
[ 840, 560, 280, 525, 315, 100, 1 ]
> p := CosetAction(I,L[4]);
> S := sub< G | h*p >;
> Index(G,S);
525
> AbelianQuotientInvariants(S);
[ 6, 0 ]
| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ You said you found a subgroup of index 550 in J2. How is that possible, as 11 is not in the factorisation of |J2|=604800? $\endgroup$ – Thomas Feb 19 '18 at 21:52
  • $\begingroup$ Sorry, as you can see from the Magma code, the index is 525 not 550. I have corrected it now. $\endgroup$ – Derek Holt Feb 19 '18 at 22:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.