Subgroup of Subgroup with Prime Index

Theorem

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

Let $H$ be a subgroup of $G$.

Let $K$ be a subgroup of $H$.

Let:

$\index G K = p$

where:

$p$ denotes a prime number

$\index G K$ denotes the index of $K$ in $G$.

Then either:

$H = K$

or:

$H = G$

Corollary

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

From the Tower Law for Subgroups:

$\index G K = \index G H \index H K$

As $\index G K = p$ is prime, either $\index G H = p$ or $\index H K = p$.

Thus either $\index G H = 1$ or $\index H K = 1$.

The result follows from Index is One iff Subgroup equals Group.

$\blacksquare$

Sources

• 1978: John S. Rose: A Course on Group Theory ... (previous) ... (next): $0$: Some Conventions and some Basic Facts: Exercise $11 \ \text{(ii)}$