Simple non-abelian group is isomorphic to subgroup of symmetric group on left coset space of proper subgroup

Statement
Suppose $$G$$ is a simple non-abelian group and $$H$$ is a proper subgroup of $$G$$. Then, there is an injective homomorphism:

$$G \to \operatorname{Sym}(G/H)$$.

Thus, $$G$$ is isomorphic the symmetric group on $$G/H$$.

Related facts

 * Simple non-abelian group is isomorphic to subgroup of alternating group on left coset space of proper subgroup of finite index
 * order of simple non-abelian group divides factorial of index of proper subgroup
 * order of simple non-abelian group divides factorial of every Sylow number

Facts used

 * 1) uses::Group acts on left coset space of subgroup by left multiplication

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

To prove: There is an injective homomorphism $$G \to \operatorname{Sym}(G/H)$$.

Proof: By fact (1), $$G$$ acts on the left coset space $$G/H$$. This action is transitive, so we get a homomorphism:

$$\varphi: G \to \operatorname{Sym}(G/H)$$.

Since $$H$$ is proper, $$G/H$$ has size more than one, and the action being transitive, the homomorphism is nontrivial. Thus, the kernel of $$\varphi$$ is not the whole group. Since $$G$$ is simple, the kernel must be trivial, so the homomorphism $$\varphi$$ is injective.