> < ^ Date: Fri, 17 Jun 1994 11:59:00 +0200
> < ^ From: Joachim Neubueser <joachim.neubueser@math.rwth-aachen.de >
< ^ Subject: Re: order of conjugacy classes

Dear Forum members,

Jacob Hirbawi asked:

Does Dixon's algorithm for calculating character tables always order
classes the same way that "ConjugacyClasses" does?

The answer is "yes".

Suppose I have a permutation representation of a group and I use "CharTable"
to calculate the character table which GAP does using Dixon's algorithm.
If I then use "ConjugacyClasses" to get (what else!) the conjugacy classes.
Is it *guaranteed* that the classes are sorted in the same order in both
cases. My guess, and my hope, is that they are. In fact I have been relying
on this fact for some calculations, is this dangerous practice?

The answer is "no".

My other question is a long shot but here it goes :

I have a matrix group generated by M1,M2,M3,and M4. <M1,M2,M3> generate
a subgroup for which I found a permutation representation <p1,p2,p3>.
<M1,M2,M4> generate another subgroup for which I can also get a
permutation reprersentation <q1,q2,q4>. The p's may or may not be of
the same degree as the q's. I can't get a permutation rep for the entire
group directly because it's a huge group, so I'm hoping to approach it
using the reps of the two subgroups. Are there any facilities in GAP that
can help? Thanks.

There can't be such facilities: Take the symmetric group on 5 letters
(in some matrixgroup representation if you like) and take the elements
M1 = (1), M2 = (1,3)(2,4), M3 = (1,2)(3,4), M4 = (1,5). Then M1, M2,
and M3 generate a Klein four group, while M1, M2, and M4 generate a
dihedral group of order 12. But if you have just independent
permutation representations of the Klein four group and a dihedral
group of order 12 then you have no information into which group they
embed; for all you know this constellation may come from the direct
product of a dihedral group of order 12 with a cyclic group of order
two, that is, you may have elements N1 = (1), N2 = (1,3)(2,4), N3 =
(6,7), and N4 = (1,5) Then also N1, N2, and N3 generate a Klein four
group and N1, N2, and N4 still generate a dihedral group of order 12,
but all N's together generate the above claimed direct product.

Kind regards Joachim Neubueser

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