Subgroup of Subgroup with Prime Index/Corollary

Theorem
Let $\struct {G, \circ}$ be a group.

Let $H$ and $K$ be subgroups of $G$.

Let $K \subsetneq H$.

Let:
 * $\index G K = p$

where:
 * $p$ denotes a prime number
 * $\index G K$ denotes the index of $K$ in $G$.

Then:
 * $H = G$

Proof
As $K \subsetneq H$ and $K$ is a subgroups of $G$, it follows that $K$ is a proper subgroup of $H$.

That is, $K \ne H$

Hence from Subgroup of Subgroup with Prime Index:
 * $H = G$