Simple non-abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index

Statement
Suppose $$G$$ is a simple non-abelian group and $$H$$ is a proper subgroup of $$G$$ that is a subgroup of finite index in $$G$$. Then, $$G$$ is isomorphic to a subgroup of the alternating group on the left coset space $$G/H$$.

Facts used

 * 1) uses::Simple non-abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup
 * 2) uses::Normality satisfies transfer condition: The intersection of a normal subgroup and any subgroup is normal in the latter subgroup.
 * 3) uses::Second isomorphism theorem

Proof
Given: A simple non-Abelian group $$G$$, a proper subgroup $$H$$ of finite index in $$G$$.

To prove: $$G$$ is isomorphic to a subgroup of the alternating group on $$G/H$$.

Proof:


 * 1) By fact (1), $$G$$ is isomorphic to a subgroup, say $$K$$, of $$\operatorname{Sym}(G/H)$$.
 * 2) By definition, the alternating group, $$\operatorname{Alt}(G/H)$$, is normal in $$\operatorname{Sym}(G/H)$$. Thus, by fact (2), $$\operatorname{Alt}(G/H) \cap K$$ is normal in $$K$$.
 * 3) Since $$K$$ is simple, $$\operatorname{Alt}(G/H) \cap K = K$$ or $$\operatorname{Alt}(G/H)$$ intersects $$K$$ trivially. We consider both cases:
 * 4) $$\operatorname{Alt}(G/H) \cap K = K$$: In this case, $$K \le \operatorname{Alt}(G/H)$$, so $$G$$ is isomorphic to a subgroup, $$K$$, of $$\operatorname{Alt}(G/H)$$.
 * 5) $$\operatorname{Alt}(G/H) \cap K$$ is trivial: In this case, the second isomorphism theorem (fact (3)) yields that $$K \operatorname{Alt}(G/H)/\operatorname{Alt}(G/H) \cong K$$. The left side is a subgroup of $$\operatorname{Sym}(G/H)/\operatorname{Alt}(G/H)$$, which is a group of order two. But a group of order two has no simple non-Abelian subgroups, so this case is not possible.