TITLE: GAP semidirect product algorithm
QUESTION [8 upvotes]: 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) ])

REPLY [4 votes]: The code is on line 919 of lib/gprdperm.gi and is very easy to understand.
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.