4

The output of WreathProduct of two permutation groups in GAP is a permutation group, I want to compute the wreath product of two general linear groups.

l:=group of 3x3 matrices of size 6 over GF(2)
g:=GL(2,2)
gap> WreathProduct(PSL(2,2),l);
<group of size 279936 with 4 generators>
gap> GeneratorsOfGroup(last);
[ WreathProductElement((2,3),(),(),(),(),(),()), 
  WreathProductElement((1,2),(),(),(),(),    (),()), 
  WreathProductElement((),(),(),(),(),(),(1,2)(3,5)(4,6)), 
  WreathProductElement((),(),(),(),(),(),(1,3)(2,4)(5,6)) ]

how will the "WreathProductElements" of generator be considered as matrix group?

Olexandr Konovalov
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Nil
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1 Answers1

3

I'm assuming you want the Kronecker product.

Is this the group you want?

gap> l := Group( [
>   [[1,0,0],[0,1,1],[0,0,1]],
>   [[1,0,0],[0,1,0],[0,1,1]]
> ] * One(GF(2)) );;
gap> g := GL(2,2);;
gap> gw := List( GeneratorsOfGroup( g ), x -> KroneckerProduct( x, One(l) ) );;
gap> lw := List( GeneratorsOfGroup( l ), x -> KroneckerProduct( One(g), x ) );;
gap> w := Group( Concatenation( gw, lw ) );;
gap> Size(w);
36
gap> Perform( GeneratorsOfGroup(w), function(x) Display(x); Print("\n"); end );
 1 . . 1 . .    . . . 1 . .    1 . . . . .    1 . . . . .
 . 1 . . 1 .    . . . . 1 .    . 1 1 . . .    . 1 . . . .
 . . 1 . . 1    . . . . . 1    . . 1 . . .    . 1 1 . . .
 . . . 1 . .    1 . . . . .    . . . 1 . .    . . . 1 . .
 . . . . 1 .    . 1 . . . .    . . . . 1 1    . . . . 1 .
 . . . . . 1    . . 1 . . .    . . . . . 1    . . . . 1 1

This is all matrices of the form: $$w = \left\{ \begin{bmatrix} A & B \\\ C & D \end{bmatrix} : A,B,C,D \in \ell \right\}$$

If you swap the order of the arguments to KroneckerProduct then you get: $$v = \left\{ \begin{bmatrix} I & 0 & 0 \\ 0 & A & B \\ 0 & C & D \end{bmatrix} : A,B,C,D \in g, I = 1_g \right\}$$

Here g is all $2\times 2$ matrices, $$g = \operatorname{GL}(2,2) = \left\{ \begin{bmatrix} A & B \\ C & D \end{bmatrix} : A,B,C,D \in \mathbb{Z}/2\mathbb{Z}\right\} $$ and l is the unique (up to iso) 3-dimensional rep of $\operatorname{GL}(2,2)$, $l=\operatorname{SL}(1,2) \times \operatorname{GL}(2,2)= \left\{ \begin{bmatrix} I & 0 & 0 \\ 0 & A & B \\ 0 & C & D \end{bmatrix} : A,B,C,D \in \mathbb{Z}/2\mathbb{Z}, I=1 \right\}$.

Jack Schmidt
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