[GAP Forum] Comparing character tables in GAP

[Andreas Bächle]

Dear Alexander and John, dear Forum,

thanks a lot for the helpful answers.  I misinterpreted the function 
"TransformingPermutationsCharacterTables", as I thought it just performs 
the permutation on the rows and columns of the character table, but it 
also takes care of the power maps.  Hence, this is exactly the function 
I was looking for with my first question.

For the second question I had the following situation in mind:

gap> c1 := CharacterTable(AlternatingGroup(6));
CharacterTable( Alt( [ 1 .. 6 ] ) )
gap> c2 := CharacterTable("A6") mod 2;
BrauerTable( "A6", 2 )

So c1 and c2 are two character tables of the alternating group of 
degree 6.  Is there a GAP-function checking if the two tables (ordinary 
character tables or Brauer tables) are tables of isomorphic groups?  Or 
do I have to be lucky that for both tables the attribute 
UnderlyingGroup(-) is known (respectively 
UnderlyingGroup(OrdinaryTable(-)), if it is a Brauer table)?
(I am aware of the possibility to call "CharacterTableWithStoredGroup" 
to attach a group to a character table, but I was wondering if this is 
not somehow done automatically for tables coming from the ATLAS?)

Thanks again for your answers!

Kind regards,
   Andreas


On 2014-07-01 17:36, Alexander Hulpke wrote:
> Dear Forum,
> 
> On Jul 1, 2014, at 7/1/14 1:44, Andreas Bächle <ABachle at 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