Let $ G $ be a random group, $ H \leq G$ and $ N \triangleleft G$. Then $ HN $ is equal to $ (H^\pi)^{\pi^{-1}} $ where $\pi$ is the canonical projection from $ G $ to $G/N$. This proves also that $ HN \leq G $.
Why is $HN$ equal to $ (H^\pi)^{\pi^{-1}} $? I understand that $ H^\pi$ is $ Ng$, just don't see how the inverse of $ \pi $ would make it $HN$. And why does it prove $ HN \leq G $?
If we unwrap things, $$H^{\pi}=\{\pi(h)|h\in H\}=\{hN|h\in H\},$$ and thus $g\in (H^{\pi})^{\pi^{-1}}$ if and only if $\pi(g)\in H^{\pi}$, which is equivalent to saying that $gN=hN$ for some $h\in H$; and this is equivalent to saying that $h^{-1}gN=N$, and thus that $h^{-1}g\in N$. Finally, this is the same as saying that $g\in HN$, and this proves that $(H^{\pi})^{\pi^{-1}}=HN$.
This proves that $HN$ is a subgroup of $G$, because it is a special case of the fact that if $\varphi:G_1\to G_2$ is a group homomorphism and $K\leq G_2$, then $\varphi^{-1}(K)\leq G_1$.