During reading a book, I have faced to this problem telling:
$G$ is a group of order $12$ such that $Z(G)$ has no element of order $2$ . Then $G≅A_4$.
Obviously, this group is not abelian and I think some information about $S_4$ is involved here because of the desired deduction. Can we say $|\frac{G}{Z(G)}|≠3$? And if so, is it useful for the problem?
If $H$ is a subgroup of order $3$, then $[G:H] = 4$ and there is a homomorphism from $G$ to $S_4$ with the kernel contained in $H$. Show that the kernel is trivial, implying that $G$ is isomorphic to a subgroup of $S_4$ of order $12$, which implies that $G \cong A_4$.