This is a classical question in the basic group theory and I try to write down the proof :
Without loss of generality that we assume the order of $G$ is $n\in\mathbb{N}$.Then Cayley theorem says that the group $G$ can be embedded in the symmetry group $S_n$. In other words, $G\cong\iota(G)\leq S_n\,,$ where $\iota$ is the induced embedding map from $G$ to $S_n$.
Then $S_n$ also can be embedded in $A_{n+2}$.
To this end,we define the following map $\varphi :S_n\longrightarrow A_{n+2}$ by $$\varphi : \alpha \longmapsto \begin{cases}\alpha& ,whenever\;\;\alpha\in A_n\\{}\\ \alpha(n+1\;n+2)&,whenever\;\alpha\in S_n- A_n\end{cases}$$ Then the map $\varphi$ is well-defined monomorphism.
Henceforth,combine these two results to obtain an embedding mapping $\varphi\,\circ\iota :G\longrightarrow A_{n+2}$
we then have $G\cong\varphi(\iota(G))\leq A_{n+2}$ , so our conclusion follows.
Is there any problem for the above proof ? Any advice will appreciate it.Thanks for considering my request and patiently reading.
I take note under the line and everybody can ignore it since there is some tedious,or, any comment will appreciate it.
$\rule{18cm}{2pt}$
$\varphi$ is well-defined :
For each $\alpha,\beta\in S_n$ with $\alpha=\beta$.
(i): If $\alpha,\beta\in A_n$,then $\varphi(\alpha)=\alpha=\beta=\varphi(\beta)$
(ii): If $\alpha,\beta\in S_n -A_n$,then $\varphi(\alpha)=\alpha(n+1\;n+2)=\beta(n+1\;n+2)=\varphi(\beta)$
$\varphi$ is a group homomorphism :
(i): If $\alpha,\beta\in A_n$,then $\alpha\beta\in A_n$,moreover, $\varphi(\alpha\beta)=\alpha\beta=\varphi(\alpha)\varphi(\beta)$
(ii): If $\alpha,\beta\in S_n - A_n$,then $\alpha\beta\in A_n$,moreover, $$\varphi(\alpha\beta)=\alpha\beta(n+1\;n+2)^2=\alpha (n+1\;n+2)\beta(n+1\;n+2)=\varphi(\alpha)\varphi(\beta)$$
(iii): If $\alpha\in S_n-A_n\, ,\beta\in A_n$,then $\alpha\beta\in S_n-A_n$,moreover, $$\varphi(\alpha\beta)=\alpha\beta(n+1\;n+2)=\alpha(n+1\;n+2)\beta=\varphi(\alpha)\varphi(\beta)$$
(iv): it is just like the case (iii)
$\varphi$ is monomorphism:
Clearly, $(\,e_{S_n}\,)\subseteq \ker\varphi$
Now, let $\pi\in S_n$ is not the identity one with $\varphi(\pi)=e_{A_{n+2}}\,\,\,.$ Then either $\pi\in A_n$ or $\pi\in S_n-A_n.$
If $\pi\in A_n$ then $\pi=\varphi(\pi)=e_{A_{n+2}}=e_{S_{n}}$ ,a contradiction.
If $\pi\in S_n-A_n$ then $\pi(n+1\;n+2)=\varphi(\pi)=e_{A_{n+2}}\,\,$ this implies that $\pi=(n+1\;n+2)\not\in S_n,$ a contradiction.
Then $\ker\varphi=(e_{S_n})$
I wrote this proof explicitly, in other words, I checked everything in this problem we needed. So I think there is a quite some difference with the three answers.
