Group of order 24 with no element of order 6 is isomorphic to $S_4$

Proposition: Given a group $G$ with $|G|=24$ such that $\nexists g\in G$ with $|g|=6$, then $G\cong S_4$.

I understand methods you can employ to deduce the number of Sylow $p$-groups in $G$ by counting elements or reasoning about permutation representations. But how can we construct an isomorphism to $S_4$ given that $n_{2-\text{Sylow}}=3$ and $n_{3-\text{Sylow}}=4$? Or otherwise, rule out all other possible cases (perhaps reasoning with Cayley's Theorem)? I am also interested in finding a proof that is perhaps less direct, but more elegant, in particular one that may involve the irreducible representations of $S_4$. Any insight appreciated.

Let's assume that the number of 3-Sylow subgroups is 4*

Let $G$ act on the set of 3-Sylow subgroups, and this will give a homomorphism $$ f : G\to S_4 $$ We claim that this homomorphism is injective, and so an isomorphism.

a) Let $K = \ker(f)$, then $K < N_G(P)$ where $P$ is some fixed 3-Sylow subgroup. Now, $$ [G:N_G(P)] = 4 \Rightarrow |N_G(P)| = 6 \Rightarrow N_G(P) \cong \mathbb{Z}_6 \text{ or } S_3 $$ If $N_G(P) \cong \mathbb{Z}_6$, then $G$ would have an element of order 6, and so $N_G(P) \cong S_3$, and so $$ |K| \in \{1,3,6\} $$ (Note $|K| \neq 2$ since $S_3$ has no normal subgroups of order 2)

b) If $|K| = 1$ we're done and if $|K| = 6$, then $K = N_G(P) \triangleleft G$, but $N_G(N_G(P)) = N_G(P)$ which is a contradiction. Hence, assume $|K| = 3$

c) If $|K| =3$, then by the N/C theorem, $$ G/C_G(K) = N_G(K)/C_G(K) \cong \text{ a subgroup of } Aut(\mathbb{Z}_3) \cong \mathbb{Z}_2 $$ In particular, $2 \mid C_G(K)$, so $C_G(K)$ contains an element of order 2, which, when multiplied with a generator of $K$ will give an element of order 6 - a contradiction.

Thus, $|K| = 1$ and we're done.

$\ast$ All that remains to show is that $n_3 = 4$, or equivalently, $n_3 \neq 1$ : I believe that a unique (and hence normal) 3-Sylow subgroup would end up producing an element of order 6 (perhaps by looking at the product $HK$ for some suitable $K$), but I am not sure about that.