= Solution
\b[No such <finite group> exists.] <Cayley theorem> embeds every <finite group> $G$ of order $n$ into $S_n$: the left-regular <group action> sends $g$ to the <permutation> $x\mapsto gx$, and this is faithful because the image of the identity element is $g$.
To make every image permutation even, let $t=(n+1\ n+2)$, with support disjoint from the first $n$ letters. For $\sigma\in S_n$, let $e(\sigma)=0$ for an <even permutation> and $e(\sigma)=1$ for an <odd permutation>, and set
$$
\iota(\sigma)=\sigma t^{e(\sigma)}.
$$
The <sign homomorphism> gives $e(\sigma\tau)\equiv e(\sigma)+e(\tau)\pmod2$, and $t$ commutes with all permutations of the first $n$ letters. Thus $\iota$ is a <group homomorphism>. It is injective by restriction to those letters, and its sign is $(-1)^{e(\sigma)}(-1)^{e(\sigma)}=1$. Consequently
$$
\boxed{G\hookrightarrow S_n\hookrightarrow A_{n+2}.}
$$
This is the <alternating-group embedding of a finite group>.
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