If $H \trianglelefteq G$, then there is a homomorphism $\theta : G / H \rightarrow \text{Aut}(H)$ such that $(G/H) \rtimes_\theta H \cong G$?
I know that in general $(G/H) \times H \not{\cong} G$ so I was wondering if you can always "recover" the structure lost by quotienting by taking a semidirect product.
This definitely works in some cases (like $D_{2n} / \langle r \rangle \cong C_2$ but if you take the nontrivial homomorphism from $C_n$ to Aut$(C_2)$ you get $C_n \rtimes C_2 \cong D_{2n}$, but can you do this in general? I feel like yes, but I don't know how to approach it.