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Can anybody guide me towards, or possibly even explain here, the algorithm that GAP uses to compute the semidirectproduct of two permutation groups which outputs another permutation group?

EXAMPLE:

gap> C3:=CyclicGroup(IsPermGroup,3);
Group([ (1,2,3) ])
gap> C7:=CyclicGroup(IsPermGroup,7);
Group([ (1,2,3,4,5,6,7) ])
gap> A:=AutomorphismGroup(C7);
< group with 1 generators >
gap> elts := Elements(A);
[ IdentityMapping( Group([ (1,2,3,4,5,6,7) ]) ),
[ (1,2,3,4,5,6,7) ] -> [ (1,3,5,7,2,4,6) ],
[ (1,2,3,4,5,6,7) ] -> [ (1,4,7,3,6,2,5) ],
[ (1,2,3,4,5,6,7) ] -> [ (1,5,2,6,3,7,4) ],
[ (1,2,3,4,5,6,7) ] -> [ (1,6,4,2,7,5,3) ],
[ (1,2,3,4,5,6,7) ] -> [ (1,7,6,5,4,3,2) ] ]
gap> sigma := elts[2];
[ (1,2,3,4,5,6,7) ] -> [ (1,3,5,7,2,4,6) ]
gap> sigma^3;
[ (1,2,3,4,5,6,7) ] -> [ (1,2,3,4,5,6,7) ]
gap> map := GroupHomomorphismByImages(C3, A, GeneratorsOfGroup(C3), [sigma]);
[ (1,2,3) ] -> [ [ (1,2,3,4,5,6,7) ] -> [ (1,3,5,7,2,4,6) ] ]
gap> SDP := SemidirectProduct(C3, map, C7);
Group([ (2,3,5)(4,7,6), (1,2,3,4,5,6,7) ])

share|improve this question
The code is on line 919 of lib/gprdperm.gi I'll look over it to see if it is easy to explain. – Jack Schmidt Aug 3 '12 at 20:11
2  
Oh my, yes it is very simple. A little scary. It rewrites the normal subgroup in its regular action (so $K$ acting on $|K|$ points), and then of course the complement subgroup acts on the normal subgroup giving the semidirect product of $H/C_H(K)$ with $K$. If $C_H(K) \neq 1$, then do the subdirect product smooshy thing with $H$'s original rep. – Jack Schmidt Aug 3 '12 at 20:15
In particular, please don't try this with $S_{20}$ as the normal subgroup. :-) – Jack Schmidt Aug 3 '12 at 20:15

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