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:

$H = G$

$\blacksquare$

Sources

• 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Cosets and Lagrange's Theorem: Exercise $13$