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.
 A: 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.
A: Here is an alternative proof: 
This proof is a little wordy, but I think it's an alternative way of tackling the problem. THe premise is similar to @PrahlanVaidyanathan 's. 
Assume that $\exists$ 4 3-Sylow subgroups in $G.$ Let $H$ be a fixed 3-Sylow subgroup. Now, $N_G(H)$ is isomorphic to $S_3$ by the same reasoning as above provided by @PrahladVaidyanathan. Now, $G$ acts by conjugation on both the Sylow 3-subgroups and their normalizers. Considering the group action on the normalizers there is a homomorphism of $G \rightarrow S_4$. If $g \in G$ acts trivially, then $g$ normalizes each $S_3$. Thus, $g$ is in each $N_G(Syl_3(G))$ and must have order $1, 2$ or $3$. The intersection of all normalizers is trivial and then the kernel is trivial which implies $G$ is isomorphic to $S_4$. Otherwise the kernel is a normal subgroup with two elements and $G4$ has an element of order 6 by considering the product of the kernel of $f$ and any Sylow 3-subgroup. Thus, if $G$ has 4 Sylow 3-subgroups and an element of order 6, then the Sylow 2-subgroup is normal which implies $G$ is not isomorphic to $S_4$. Then, use the reasoning provided by @TobiasKildetoft to illustrate the number of 3-Sylow subgroups is 4 and not 1. 
