Let $G$ be a finite non-Abelian group of order $12$. Prove that if $| Z (G) | = 1$, then $G$ is isomorphic to $A_4$ Let $G$ be a finite non-Abelian group of order $12$. Prove that:
a) $| Z (G) | \in \{{1,2}\}$
b) If $| Z (G) | = 1$, then $G$ is isomorphic to $A_4$
I responded to in paragraph a) in this way: $ | Z (G) |$ is not 4 or 6 or 12 otherwise $G / Z (G)$ has a prime order and therefore $G$ is cyclic and therefore abelian.
As absurd as the order of $Z (G)$ was $3$ then $n_3 (G) = 1 $ ($Z (G)$ is normal) and $n_2 (G) = 3$ (otherwise G would be abelian). Let $H = \{id, a, b​​, c\}$ a 2-Sylow (abelian then!). We know that $Z (G) <C (H) <N (H)$;
$| N (H) |$ is 4 because $[G: N (H)] $ is 3 (number of 2-Sylow). But $C (H) = (id, z1, z2, a, b​​, c, ..) $, then $ | C (H) |\ge 6$, absurd! (from the fact that $| C (H) | \le |N (H)|$).
Is it okay? I could do it in another way?
any suggestions for b)?
 A: You can finish a) like this: $Z(G)<N(H)$, so $3=|Z(G)|\mid 4$, a contradiction. It remains $|Z(G)|=1$ or $2$. 
A: First, note that part A is false (see the link on the classifications at the bottom to find several counterexamples, the comment example of $A_4$ works wonderfully).
As for part B, I see no 'simple' way to do this immediately. This group must have a nontrivial subgroup, for it did not then $G$ would be cyclic contradicting the fact that $|Z(G)|=1$. If $H$ is a subgroup of order $3$, then $[G:H]=4$ and there is a homomorphism from $G$ to $S_4$ with $H$ containing the kernel. Then it would be simple to show that $G$ would have to be isomorphic to a subgroup of $_4$ of order $12$, but then that is clearly $A_4$, so then $G \cong A_4$. We can do something similar if the exists a subgroup of order $4$. If there were a subgroup of order $6$, it would be normal and the center would be nontrivial. We can continue this. 
Notice also, we can use Sylow's Theorem to 'easily' (an overstatement) to classify the groups of order $12$. Once we know that $|Z(G)|=1$, that leaves one and only one possibility. To see the process for the classification, look here
