[GAP Forum] Modified ConjugacyClassesSubgroups?
Alexander Hulpke
ahulpke at gmail.com
Sun Oct 12 17:17:18 BST 2008
Dear Forum, Dear Joe,
I first have to amend the explanation of what the routines I mentioned
do:
If H is not normal in G, it is not guaranteed that this approach (or
any other approach that computes subgroups inductively) will find all
classes of H, when working up to G-conjugacy.
The reason is that two subgroups U,V of H might be conjugate in G, but
not in H, and one has subgroups or supergroups in H and the other not
(it has these subgroups in G, but they don't lie in H).
For example take G=S6 and H=<(1,2,3,4),(1,3)(56)> (isomorphic to D8)
and U=<(1,3)(2,4)>, V=<(1,3)(5,6)>.
Then U is conjugate to V in G, but there is a subgroup of H, namely
<(1,2,3,4)> which contains U, but V is not contained in any such
subgroup of H. One can build similar examples for subgroups.
I cannot see how to avoid this problem for any kind of algorithm --
this means one cannot discard G-conjugate subgroups early, but might
need to extend/process them later, even if they themselves are G-
conjugate.
Best,
Alexander
On Oct 9, 2008, at 2:14 PM, Joe Bohanon wrote:
> 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.
>
> Thanks
> Joe
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