Split short exact sequence equivalent to internal semidirect product

Question

Given a short exact sequence $1 \to N \to G \to H \to 1$, $\alpha: N \to G, \beta: G \to H$ which is split: there exists $s: H \to G$, s.t. $\beta \circ s = \text{id}_H$, I want to see that $G$ is the internal semidirect product of $H$ and $N$: first, how to see that $G = NH$? Of course, there is a subgroup $\tilde N = \alpha(N)$ inside $G$ isomorphic to $N$. Now why is $s$ injective and $G = \tilde N \cdot \tilde H$, where $\tilde H = s(H)$?

Answer 1

Trivially $s$ is a injective since if $s(h)=s(h')$, then $h = \beta(s(h))=\beta(s(h'))=h'$. If $g$ is in $G$, then $g = (g(s(\beta(g)))^{-1}) s(\beta(g))$. Now, clearly $s(\beta(g))$ is in $s(H)$. Can you see why $g(s(\beta(g)))^{-1}$ is in $\alpha(N)$? Note that you also need to check that $s(H)\cap\alpha(N)=\{e\}$. To do so suppose that $g \in s(H)\cap\alpha(N)$, what can you say about $\beta(g)$? Since $g=s(\beta(g))$ what does this tell you about $g$?