> < ^ From:

> ^ Subject:

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 31a 2a 2b 2c 2d 2e 4a 4b 4c 4d 2P 1a 1a 1a 1a 1a 1a 2a 2a 2c 2cX.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

> < [top]