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Let $G$ be a finite groupe and $n = \text{Card}(G)$.
Caley's theorem states that there exists $\phi \in \text{Hom}(G, S_n)$ which is injective, which means that $G$ is isomorphic to a subgroup of $S_n$.
We will make $S_n$ "act" on the vector space $\mathbb{R}^n$. We will define for every $\sigma \in S_n$ the matrix :
$$M_{\sigma} = (\delta_{\sigma(i), j})$$
where $\delta$ is the Kronecker delta defined by : $\delta_{a, b} = \begin{cases}
1 \text{ if } a = b \\
0 \text{ otherwise}
\end{cases}$
These $M_{\sigma}$ are elements of $\text{GL}_n(\mathbb{R})$ because $\forall \sigma \in S_n, \ M_{\sigma} \times M_{\sigma^{-1}} = M_{\sigma^{-1}} \times M_{\sigma} = I_n$.
Let $\phi' : \sigma \mapsto M_{\sigma}$. $\phi' \in \text{Hom}(S_n, \text{GL}_n(\mathbb{R}))$ and is injective, so $S_n$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$.
then, let $\theta = \phi' \circ \phi$. $\phi$ and $\phi'$ are injective morphisms, so $\theta$ is an injective morphism from $G$ to $\text{GL}_n(\mathbb{R})$. That means $G$ is isomorphic to a subgroup of $\text{GL}_n(\mathbb{R})$, which is, in our case, $\text{Im}(\theta)$.
Adding a non-GAP argument.
A key ingredient is the fact that $S_4=V_4\rtimes S_3$, where $V_4\unlhd S_4$ is the copy of Klein-4 whose non-trivial elements are the three products of two disjoint 2-cycles. Furthermore, conjugation by elements of $S_3$ gives all the six permutations of those three type $(2,2)$ permutations.
Let use view $N=C_2\times C_2\times C_2$ additively as a 3-dimensional vector space over $\Bbb{F}_2$:
$$
N=\{(x_1,x_2,x_3)\mid x_1,x_2,x_3\in\Bbb{F}_2\}.
$$
We can write $N$ as a direct sum $N=Z\oplus V$ where $$Z=\langle(1,1,1)\rangle\qquad \text{and}\qquad V=\{(x_1,x_2,x_3)\in N\mid x_1+x_2+x_3=0\}.$$
Both $Z$ and $V$ are stable under the action of $S_3$. Also, $S_3$ acts trivially on $Z$, and permutes the elements of $V$ the same way it permutes that copy of $V_4\le S_4$ - all the six ways of permuting the vectors $110,101,011$ are gotten by permuting the coordinates.
Putting all this together gives
$$
N\rtimes S_3\simeq Z\times (V\rtimes S_3)\simeq C_2\times S_4.
$$
The subgroup $Z$ is the center of this group.
Best Answer
Let be $a\in S_n$, $a^5=e$, $a\ne e$. By the theorem of Lagrange, $5\,\big|\,n!$ and we conclude that $n\ge 5$.