I am trying to prove the following theorem. I need some guidance to prove that $G = \sigma(H)N$.
Let $G$ be a group, $N$ a normal subgroup of $G$. If there exists a split short exact sequence of groups $$ 1 \longrightarrow N \overset{i}{\longrightarrow}G \overset{\pi}{\longrightarrow}H \longrightarrow 1 $$
then $N$ has a complement $H$ in $G$, i.e., a subgroup $H$ of $G$ such that $HN = G$ and $H\cap N = 1$.
Proof:
Choose a split short exact sequence of groups $$ 1 \longrightarrow N \overset{i}{\longrightarrow}G \overset{\pi}{\longrightarrow}H \longrightarrow 1 $$ and a homomorphism $\sigma: H \to G$ with the property that $\pi \circ \sigma = \text{id}_H$. Clearly, $\sigma(H) \leq G$. We claim first that $\sigma(H) \cap N = 1$. Notice that $N = \text{im}(i) = \ker(\pi)$. Let $g \in \sigma(H) \cap \ker(\pi)$. Then $g = \sigma(h)$ for some $h \in H$ and $\pi(g) = 1$. Hence, we have $1 = \pi(g) = (\pi \circ \sigma)(h)= \text{id}_H(h) = h$. Hence, $g = \sigma(h) = \sigma(1) = 1$, as desired.
Next, we need to show that $G = \sigma(H)N = \sigma(H) \ker(\pi)$. This is where I am stuck.