$|G|=24$ and $Z(G)=1$ $\implies$ $G$ has $4$ Sylow $3$-subgroups Suppose $|G|=24$ and $Z(G)=1$.
How can we prove $G$ has $4$ Sylow $3$-subgroups?
The final goal is to prove such  $G$ is isomorphic to $S_4$.
If $N(3) = 4$, then conjugation action of $G$ on Sylow $3$-subgroups yields homomorphism $\rho: G \to S_4$, and by some analysis we know $\ker \rho = 1$, so $G \cong S_4$.
Sylow's theorem says $N(3) = 3k+1 \mid 8$, so $N(3) = 1$ or $N(3) = 4$.
How can we prove $N(3) \ne 1$ as concise as possible?
Update:
Suppose $N(3) = 1$ and the Sylow $3$-subgroup is $A$. $N(2) = 1$ or $3$.
If $N(2) = 1$, denote the Sylow $2$-subgroup by $B$.
$A \triangleleft G$, $B \triangleleft G$, $A \cap B = \{1\}$, $G \cong A \times B$, $1 \ne A \le Z(G)$. Contradiction.
So how can we proceed by showing $N(2) = 3$ is impossible?
 A: I would have just gone with the other, perfectly serviceable proof, but the question wanted as concise as possible, and I think I can do slightly better.
By Sylow's theorem, if $n_3\neq 4$ it is $1$. If $P$ is the Sylow $3$-subgroup and $Q$ is a Sylow $2$-subgroup, then $Q$ acts on $P$ by conjugation. We have
$$|Q/C_Q(P)|\leq |\mathrm{Aut}(P)|=2,$$
so $C_Q(P)$ has index at most $2$ in $Q$, hence is normal in $Q$. Thus $C_Q(P)\cap Z(Q)>1$, and this is contained in $Z(G)$ (as $G=PQ$).
A: If $n_3=1$, then conjugation induces a homomorphism $\phi\colon G\to C_2$ permuting the two order-3 elements of $G$.  Pick a Sylow-2 $P_2$ of $G$.  We have $Z(G)\supseteq Z(P_2)\cap\ker\phi$ since $P_2,P_3$ generates $G$. The centre of $P_2$ (note order $2^3$) is either the whole Sylow-2 or $\pm 1$.  If $Z(P_2)=P_2$ then we have $\#(Z(P_2)\cap\ker\phi)\geq 4$, so the Sylow-2 is either $D_{2\cdot 4}$ or $Q_8$.  But in both cases the $-1\in Z(P_2)$ is a square in $P_2$ so must be in $\ker\phi$, contradicting $Z(G)=1$.
