Dear Mrs. and Mr. Forum,

Peter Blanchard wrote

The following seems to be a problem in which
the function TableAutomorphisms misses a symmetry:

He gives the following example.

gap> DisplayCharTable( CharTablePGroup( TwoGroup( 16, 3 ) ) );

2  4  4  4  4  3  3  3  3  3  3

   1a 2a 2b 2c 2d 2e 4a 4b 4c 4d
2P 1a 1a 1a 1a 1a 1a 2a 2a 2c 2c

X.1      1  1  1  1  1  1  1  1  1  1
X.2      1  1  1  1 -1 -1  1  1 -1 -1
X.3      1 -1  1 -1  1 -1  A -A  A -A
X.4      1 -1  1 -1 -1  1  A -A -A  A
X.5      1  1  1  1  1  1 -1 -1 -1 -1
X.6      1  1  1  1 -1 -1 -1 -1  1  1
X.7      1 -1  1 -1  1 -1 -A  A -A  A
X.8      1 -1  1 -1 -1  1 -A  A  A -A
X.9      2  2 -2 -2  .  .  .  .  .  .
X.10     2 -2 -2  2  .  .  .  .  .  .

A = E(4)
  = ER(-1) = i
gap> TableAutomorphisms(tbl,tbl.irreducibles,"closed");
Group( ( 7, 8)( 9,10), ( 5, 6)( 9,10), ( 2, 4)( 7, 9)( 8,10) )
gap> (2,4) in last;
false

Permuting the classes with (2,4) leaves the set of irreducible characters
invariant, namely it induces the permutation (X.9,X.10).
But we have computed the table automorphisms, that is, those automorphisms
of the matrix of irreducible characters that respect the power maps.
In our example this means that if we want to swap the classes '2a' and '2c'
then also their preimages under the 2nd power map must be swapped.
So the desired permutation must map { '4a', '4b' } to { '4c', '4d' }.
This leads to one of the permutations (2,4)(7,9)(8,10), (2,4)(7,10)(8,9)
of classes, which induce the permutations (X.2,X.6)(X.4,X.8)(X.9,X.10)
and (X.2,X.6)(X.3,X.7)(X.9,X.10) of characters.

Kind regards
Thomas Breuer