 Solution

For $|G|=12$, the third Sylow theorem gives $n_3\equiv 1(\mathrm{mod}3)$ and $n_3\mid 4$, so $n_3=1$ or $4$. In the first case the Sylow $3$-subgroup is unique.

If $n_3=4$, conjugation acts transitively on the four Sylow $3$-subgroups. Each normalizer has order $|G|/4=3$ and is the subgroup it normalizes. The action kernel lies in every such subgroup, whose intersection is trivial, so $G$ embeds in $S_4$. Its image has order $12$ and hence index two. The unique index-two subgroup of $S_4$ is $A_4$, so