= Solution
Let $G$ act by left multiplication on the set $G/H$ of $n$ left <coset>[cosets]. This gives a <group action> homomorphism
$$
\varphi:G\longrightarrow S_n.
$$
Its kernel
$$
K=\bigcap_{g\in G}gHg^{-1}
$$
is a <normal subgroup> of $G$, and the <first isomorphism theorem> gives $G/K\cong\operatorname{im}\varphi$. By <Lagrange's theorem>, $|G/K|$ divides $n!$. The image acts transitively on the $n$ cosets, so the <orbit-stabilizer theorem> shows that $n$ divides $|G/K|$; in particular,
$$
|G/K|\geq n.
$$
Now suppose $G$ is nonabelian and <simple group>[simple]. Since $H$ is proper, the coset action is nontrivial, so $K\ne G$. Simplicity gives $K=\{e\}$, and $\varphi$ embeds $G$ into $S_n$. Composing with the <sign homomorphism> gives
$$
G\longrightarrow S_n\longrightarrow\{\pm1\}.
$$
Its kernel is normal. A nontrivial map would embed the simple group $G$ into the abelian group of order two, which is impossible because $G$ is nonabelian. The sign is therefore always $+1$, so
$$
\boxed{G\cong\varphi(G)\leq A_n}.
$$
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