Prime index and quotient-subisomorph-containing implies index-unique

Statement
Suppose $$G$$ is a group, and $$H$$ is a fact about::subgroup of prime index (specifically, a fact about::normal subgroup of prime index) that is also a fact about::quotient-subisomorph-containing subgroup. Then, $$H$$ is an fact about::index-unique subgroup: there is no other subgroup of $$G$$ with the same index.

When $$G$$ is a finite group, this is equivalent to saying that $$H$$ is an order-unique subgroup: there is no other subgroup of $$G$$ of the same order.

Related facts

 * Finite cyclic and homomorph-containing implies order-unique

Proof
Given: A group $$G$$, a prime number $$p$$, a subgroup $$H$$ of prime index in $$G$$ such that $$H$$ is contained in the kernel of any homomorphism from $$G$$ to $$G/H$$.

To prove: $$H$$ is the only subgroup of index $$p$$ in $$G$$.

Proof: Suppose $$K$$ is a subgroup of index $$p$$. Then, $$G/K \cong G/H$$. Thus, the quotient map $$G \to G/K$$ can be composed with this isomorphism, giving a map $$\alpha:G \to G/H$$ with kernel $$K$$. Since $$H$$ is quotient-homomorph-containing in $$G$$, this implies $$H \le K$$. But since both have index $$p$$, this forces $$H = K$$.