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

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Statement

Suppose G is a group, and H is a Subgroup of prime index (?) (specifically, a Normal subgroup of prime index (?)) that is also a Quotient-subisomorph-containing subgroup (?). Then, H is an 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

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.