## § Semidirect product is equivalent to splitting of exact sequence

Consider the exact sequence
$0 \rightarrow N \xrightarrow{\alpha} G \xrightarrow{\pi} H \rightarrow 0$
• We want to show that if there exists a map $s: H \rightarrow G$ such that $\forall h, \pi(s(h)) = h$ (ie, $\pi \circ s = id$), then G $\simeq N \ltimes H$. So the splitting of the exact sequence decomposes $G$ into a semidirect product.
• The idea is that elements of $G$ have an $N$ part and a $K$ part. We can get the $K$part by first pushing into $K$ using $\pi$ and then pulling back using $s$. So define $k: G \rightarrow G; k(g) \equiv s(\pi(g))$. This gives us the "K" part. To get the $N$part, invert the "k part" to annihiliate it from $G$. So define a map $n: G \rightarrow G; n(g) \equiv g k(g)^{-1} = g k(g^{-1})$.
• See that the image of $n$ lies entirely in the kernel of $\pi$, or the image of $n$indeed lies in $N$. This is a check:
\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}
• Hence, the image of $n$ is entirely in the kernel of $\pi$. But the kernel of $\pi$ is isomorphic to $N$, and hence the image of $n$ is isomorphic to $N$. So we've managed to decompose an element of $G$ into a $K$part and an $N$ part.
• Write $G$ as $N \ltimes K$, by the map $\phi: G \rightarrow N \ltimes K; \phi(g) = (n(g), k(g))$. Let's discover the composition law.
\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 $k(gh) = k(g) k(h)$, so that composes in an entirely straightforward fashion. For the other component, we need:
\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 $n$ of $h$ to be twisted by the $k$ component of $g$ by a conjugation. So we define the semidirect structure as:
\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 $\phi: G \rightarrow N \ltimes K$. 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 $(n, k) \in N \ltimes K$. Create the element $g = \alpha(n) s(k) \in G$. We get $\pi(g) = \pi(\alpha(n)s(k)) = \pi(\alpha(n)) \pi(s(k)) = e k = k$. We get $n(g) = g k$