I know that if $G$ is a group, $N < G$, then the condition that $|G:N|=2$ implies that$ N$ is normal in $G$. But what about the converse if we know that $N$ is normal in $G$ does that then imply that the index of $N$ in $G$ is $2$ ?
For some context from the author of this MSE question, see this comment:
@DietrichBurde I worked through another example actually where H is normal in G if the index of H in G is the smallest prime in p because then letting G act on H by right multiplication will induce an embedding in Sp and as G acts transitively we know p divides the image which implies the core of H and H itself are the same so therefore H must be normal in G – can'tcauchy
Every group is a normal subgroup of itself.
If you insist that the subgroup is not the group itself, then the subgroup containing only the identity is normal in every group.
If you insist that the subgroup is proper, then consider constructing a group by means of a semidirect product. In particular, as @egreg points out in the comments below, a direct product is easiest.