GAP semidirect product algorithm

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) ])

• 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
• 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
• @JackSchmidt Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Jun 23 '13 at 8:20

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 it does the subdirect product smooshy thing with $H$'s original rep: that is $H \ltimes K$ acts on $X \dot\cup K$ where $H$ acts on $X$ as usual, $K$ centralizes $X$, and $H,K$ have the previously described action on $K$.
In particular, this method is impractical with $S_{20} as the normal subgroup. If$K$acts on$Y$and$|H|$is small, then$H \ltimes K$can act on$H \times Y$as well, but this is not implemented. There is a special method implemented on line 961 to catch a nice case: if$K$acts on$Y$, and if the action of$H$on$K$lifts to a homomorphism from$H$to$\operatorname{Sym}(Y)$, then one can view$H$more compactly as a subgroup of$\operatorname{Sym}(X \dot\cup Y)$. This method is not guaranteed to succeed even if applicable if$C_{\operatorname{Sym}(Y)}(K)\$ is large, as the method to check if the homomorphism lifts is just to choose a random section and see if it a homomorphism.