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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