Let $\phi: G \rightarrow H$ be a surjective homomorphism of groups. Let $\sigma: H \rightarrow G, \,\,\phi\, \sigma = id_H.$ I need to show that $G$ is isomorphic to the semidirect product $\ker(\phi)\rtimes H.$
We know that $\ker(\phi)$ is normal in $G.$ I do not see how $G$ is the product of the subgroups $\ker(\phi), H$ and that their intersection is trivial.
Do you have any suggestion how to prove the statement or a solution proposal?
Thanks.
Since $\phi \circ \sigma = \mbox{id}_H$, we can conclude that $\sigma$ is injective. (Recall that for arbitary functions $f: X \to Y$ and $g: Y \to X$, $g \circ f$ being bijective implies that $f$ is injective.) So, $\sigma$ injects $H$ into $G$, and most importantly, $H \cong \sigma(H)$. Now, to prove the theorem, we need to show two things: $\sigma(H) \cap \ker\phi = 1$ and $\sigma(H)\ker\phi = G$.
First assume for a contradiction that $\sigma(H) \cap \ker\phi \neq 1$. We choose an $g \neq 1$ lying in this intersection. We then choose the unique nonidentity $h \in H$ such that $\sigma(h) = g$. Now, since $g \in \ker\phi$, we have that $$\mbox{id}_H(h) = \phi\circ\sigma(h) = \phi(g) = 1$$ but this implies $h = 1$, a contradiction. These two subgroups intersect trivially.
Now, by the First Iso. Thm., we have that $|G:\ker\phi| = |H|$. We obtain that
$$ |\sigma(H)\ker\phi| = |\ker\phi||H|/|\ker\phi \cap \sigma(H)| = |\ker\phi||H| = |G| $$ implying $G = \sigma(H)\ker\phi$. In other words, $G = \ker\phi \rtimes H$.