Dear Forum,
On Jul 1, 2014, at 7/1/14 1:44, Andreas Bächle <abac...@vub.ac.be> wrote:
>
> I want to check whether two character tables are "equal" (in the sense that
> their irreducible characters coincide and also the corresponding power maps
> match, of course, everything up to permutation). According to Thomas
> Breuer's answer from 2000
> (http://www.gap-system.org/ForumArchive/Breuer.1/Thomas.1/Re__Comp.8/1.html)
> there is no built in function to compare character tables in GAP,
I think this is a misunderstanding. The archived email says that the only way
to test such equivalence is by an explicit search for permutations, there is no
cheaper test that would only need to check some properties without possibly
testing for permutations.
> but one has to do the checks manually. Is this still the case? What would
> be the best way to do it? In particular, I want to recognize, for example,
> the two character tables CharacterTable("A4") and
> CharacterTable(AlternatingGroup(4)) as "equal". As I want to use this in a
> function, I want to automatize it as far as possible.
gap> c1:=CharacterTable("A4");
CharacterTable( "a4" )
gap> c2:=CharacterTable(AlternatingGroup(4));
CharacterTable( Alt( [ 1 .. 4 ] ) )
gap> TransformingPermutationsCharacterTables(c1,c2);
rec( columns := (), group := Group([ (3,4) ]), rows := (2,3) )
The result tells you that the classes are already in correspondence, but
characters 2 and 3 were swapped.
>
> In case only the "heads" of the character tables are known and I want to
> avoid to calculate the irreducibles (as the groups might be very large and
> the calculations needed can be done with induced characters), there is
> probably no hope to see if two character tables are tables of the same group,
> except if the attribute "UnderlyingGroup" is stored for both tables?
Since there are Brauer Pairs you will never be able to determine group
isomorphism based on equivalence of the character tables (or do I misunderstand
the question)?
Best,
Alexander Hulpke
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