[GAP Forum] subgroups of GL(3,7)

[Joe Bohanon]

Actually, I think this can be done at least for L3(11), but you'll have 
to get down-and-dirty with the source code.

If you call ConjugacyClassesSubgroups on either of these groups, the 
function that ends up being called (assuming you have the latest version 
of GAP) is LatticeViaRadical.  In turn, this determines that the radical 
group of GL(3,11) is a Z10, and it will try to construct the lattice of 
L(3,11) first.  That's what's causing the exceeded memory errors.  If 
GAP did eventually finish constructing these, it would then try 
determining G:which of the lifted subgroups contain non-trivial 
subgroups that do not contain the radical and do "lifting".  If you look 
in the "lib" directory, these functions are in "grplatt.gi".

As Dmitiri suggested, you can look at the maximal subgroups, but even 
better, you can look at the table of marks in GAP's library.  The bigger 
ones in there were actually constructed using the maximal subgroup idea 
(presented in a paper by Pfeiffer on the lattice of M24).  Here's a 
basic list of commands that you can run.  Here, I'm assuming G is the 
permutation representation of GL(3,7):

R:=RadicalGroup(G);
hom:=NaturalHomomorphismByNormalSubgroup(G,R);
H:=Image(hom);
tom:=TableOfMarks("L3(11)");
K:=UnderlyingGroup(tom);
iso:=IsomorphismGroups(H,K);
ccls:=List([1..Size(SubsTom(tom))],i->ConjugacyClassSubgroups(K,i));

At this point, you would copy the source code for "LatticeViaRadical" 
into a new function, find the point at which it is trying to construct 
the lattice of L(3,11), and skip that and instead of computing it, let 
it be one of the arguments you feed in.

Again, I'm not 100% sure this will work, but I know that it is taking 
what is probably the hardest part of the computation out of it.

This works on L3(11) and not L3(7) because in the case of GL(3,7), 
modding out by the radical group gives you PGL which, for q=7 is not the 
same as PSL.  GAP's TOM library does have L3(7) in there.  L3(13) is not 
in there, but I constructed it using Pfeiffer's methods a few years 
ago.  I can probably dig it up if you want it.

I have another way that I can construct lattices for groups like this 
that I will try out and possible be able to send you, for GL(3,7).  (It 
involves code that was given to me by someone else that I probably 
shouldn't release).

Joe



Asst. Prof. Dmitrii (Dima) Pasechnik wrote:
> Dear Alex,
>
> As stated, your problem is pretty hopeless --- the group is too big to
> do this by brute force.
> You have to "help" GAP -- the maximal subgroups of GL(3,q) were determined by
> Dickson, so you can generate the maximal subgroups "by hand".
> And then consider each one in turn...
>
> HTH,
> Dima
>
> 2010/3/26 Alex Trofimuk <trof1984 at rambler.ru>:
>   
>>
>> -- Dear Gap forum,
>>  Alex Trofimuk asked:
>>
>> How to compute all subgroups of GL(3,7) and GL(3,11)?
>>
>> I write the following commands:
>> G:=GL(3,7);
>> GL(3,7)
>> gap> imG:=Image(IsomorphismPermGroup(G));;
>> gap> c:=ConjugacyClassesSubgroups(imG);;
>>
>> and GAP says:
>>
>> exceeded the permitted memory (`-o' command line option) at
>> sorted := ShallowCopy( list );
>>  called from
>> Collected( StructuralCopy( types ) ) called from
>> rbase.nextLevel( rbase.partition, rbase ); called from
>> PBEnumerate( 1, not repr ) called from
>> PartitionBacktrack( G, Pr, false, rbase, data, L, L ) called from
>> oper( Parent( D ), D ) called from …..
>>
>> Alex Trofimuk.
>>
>> --
>> Скачай новую ICQ 7.0 http://icq.rambler.ru/
>>
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>>     
>
>
>
> --
> Dmitrii Pasechnik
> -----
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