Subgroup of Subgroup with Prime Index
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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 $11 \ \text{(ii)}$