[GAP Forum] Modified ConjugacyClassesSubgroups?

Joe Bohanon jbohanon2 at gmail.com
Thu Oct 9 21:14:48 BST 2008

This seems like it will get the job done, however it appears to be 
dependent on being able to at least find zuppos of G.  For instance, say 
G is the Suzuki sporadic group and H is its smallest maximal subgroup 
(an A7).  The filter doesn't apply to the zuppos, and for that group, I 
can imagine it would take a very long time to get all of them.  Is there 
anything that would get lost by computing zuppos of A7 then collapsing 
under conjugacy in Suz?

I'm specifically asking because sometimes the maximal subgroups of some 
simple groups have enormous rank elementary abelian subgroups with tons 
of subgroups that aren't conjugate in H but end up being conjugate in 
G.  I'm just trying to save some time by testing for conjugacy in G on 
the front end.  I had one group that was extending class 5,000 to get 
class 10,000, and would probably have gobbled up all memory if I'd let it.


Alexander Hulpke wrote:
> Dear Forum,
> Joe Bohanon asked:
>> I have a group G and a subgroup H.  I want to compute the conjugacy 
>> classes of subgroups H where conjugacy is taken in G, instead of H.  
>> The way I've been doing it is the following:
>> List(ConjugacyClassesSubgroups(H),Representative);
>> Then I manually check to see how which classes are fused in G and 
>> remove redundacies.
>> Is there a way to get the same results without doing that?  In other 
>> words, could I use the cyclic extension method, except compute the 
>> zuppos of H as conjugacy classes of G and do all the required 
>> calculations in G instead of H?
> You can do so by explicitly calling the cyclic extension method, this 
> lets you apply a further filter function. For example:
> gap> g:=SymmetricGroup(7);
> Sym( [ 1 .. 7 ] )
> gap> h:=DerivedSubgroup(g);
> Group([ (1,3,2), (2,4,3), (3,4,5), (1,5,6,4,3), (1,3,6,7,5) ])
> selector function:
> gap> func:=i->IsSubset(h,i);
> function( i ) ... end
> gap> l:=LatticeByCyclicExtension(g,func);
> <subgroup lattice of Sym( [ 1 .. 7 ] ), 37 classes,
> 3786 subgroups, restricted under further condition l!.func>
> gap> c:=ConjugacyClassesSubgroups(l);;
> gap> Length(c);
> 37
> # check in a naive way
> gap> Length(ConjugacyClassesSubgroups(h));
> 40
> gap> 
> Number(ConjugacyClassesSubgroups(g),i->IsSubset(h,Representative(i)));
> 37
> Best,
>    Alexander Hulpke
> -- Colorado State University, Department of Mathematics,
> Weber Building, 1874 Campus Delivery, Fort Collins, CO 80523-1874, USA
> email: hulpke at math.colostate.edu, Phone: ++1-970-4914288
> http://www.math.colostate.edu/~hulpke

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