§ Semidirect product is equivalent to splitting of exact sequence

Consider the exact sequence
0NαGπH0 0 \rightarrow N \xrightarrow{\alpha} G \xrightarrow{\pi} H \rightarrow 0
π(n(g))=π(gk(g1))=π(g)π(k(g1))=π(g)π(s(π(g1)))=π(s(x))=x:=π(g)π(g1)=e \begin{aligned} &\pi(n(g)) \\ = \pi(g k(g^{-1})) \\ = \pi(g) \pi(k(g^{-1})) \\ = \pi(g) \pi(s(\pi(g^{-1}))) \\ = \text{$\pi(s(x)) = x$:}\\ = \pi(g) \pi(g^{-1}) = e \end{aligned}
ϕ(gh)=?ϕ(g)ϕ(h)(n(gh),k(gh))=?(n(g),k(g))(n(h),k(h))(ghk((gh)1),k(gh))=?(gk(g1),k(g))(hk(h1),k(h)) \begin{aligned} &\phi(gh) =^? \phi(g) \phi(h) \\ &(n(gh), k(gh)) =^? (n(g), k(g)) (n(h), k(h)) \\ &(ghk((gh)^{-1}), k(gh)) =^? (gk(g^{-1}), k(g)) (hk(h^{-1}), k(h)) \\ \end{aligned}
We need the second to be k(gh)=k(g)k(h)k(gh) = k(g) k(h), so that composes in an entirely straightforward fashion. For the other component, we need:
ghk((gh)1)=?gk(g1)hk(h1)ghk((gh)1)=?gk(g1)k(g)hk(h1)k(g1)ghk((gh)1)=?g[k(g1)k(g)]h[k(h1)k(g1)]ghk((gh)1)=?ghk((gh)1)ghk((gh)1)=ghk((gh)1) \begin{aligned} &ghk((gh)^{-1}) =^? gk(g^{-1}) \cdot hk(h^{-1}) \\ &ghk((gh)^{-1}) =^? gk(g^{-1}) k(g) \cdot hk(h^{-1}) k(g^{-1}) \\ &ghk((gh)^{-1}) =^? g [k(g^{-1}) k(g)] \cdot h [k(h^{-1}) k(g^{-1})] \\ &ghk((gh)^{-1}) =^? g \cdot h k((gh)^{-1}) \\ &ghk((gh)^{-1}) = gh k((gh)^{-1}) \\ \end{aligned}
So we need the nn of hh to be twisted by the kk component of gg by a conjugation. So we define the semidirect structure as:
(n(g),k(g))(n(h),k(h))(n(g)k(g)n(h)k(g)1,k(g)k(h))=(n(g)n(h)k(g),k(g)k(h)) \begin{aligned} (n(g), k(g)) \cdot (n(h), k(h)) \equiv (n(g) k(g) n(h) k(g)^{-1}, k(g) k(h)) \\ &= (n(g) n(h)^{k(g)}, k(g) k(h)) \end{aligned}
We've checked that this works with the group structure. So we now have a morphism ϕ:GNK\phi: G \rightarrow N \ltimes K. we need to check that it's an isomorphism, so we need to make sure that this has full image and trivial kernel.
  • Full image: Let (n,k)NK(n, k) \in N \ltimes K. Create the element g=α(n)s(k)Gg = \alpha(n) s(k) \in G. We get π(g)=π(α(n)s(k))=π(α(n))π(s(k))=ek=k\pi(g) = \pi(\alpha(n)s(k)) = \pi(\alpha(n)) \pi(s(k)) = e k = k. We get n(g)=gkn(g) = g k