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

## 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.

## 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 $H = K$.