I am sorry to say that my answer to message from Bruce Colletti
SemidirectProduct was partially wrong, which I only discovered after Bruce
contacted me again. Please take my apologies for any confusion I may have
The homomorphism a is, of course, not one from G to H, but from G to a
group A operating on H.
However, the point is that "operation", "operates" means that every element
x of A (viz. of the image of a in A) is an automorphism on H, which is
given by h -> h^x. Here, it is assumed (without verification) that the map
h -> h^x is, indeed, an automorphism of H.
In Bruce's case, this is definitely not the case, since (1,2)^(1,2,3) =3D (2,3) is not contained in the subgroup H =3D <(1,2)>.
Sorry for any inconvenience.