Existence of Subgroup whose Index is Prime Power
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Theorem
Let $G$ be a finite group.
Let $H$ be a normal subgroup of $G$ which has a finite index in $G$.
Let:
- $p^k \divides \index G H$
where:
- $p$ is a prime number
- $k \in \Z_{>0}$ is a (strictly) positive integer
- $\divides$ denotes divisibility.
Then $G$ contains a subgroup $K$ such that:
- $\index K H = p^k$
Proof
The order $\order {G / H}$ of the quotient group $G / H$ is $\index G H$.
Hence $p^k$ divides $\order {G / H}$.
By Group has Subgroups of All Prime Power Factors, $G / H$ has a subgroup of order $p^k$.
By Correspondence Theorem, this subgroup is in the form $K / H$ where $H \le K \le G$.
Hence the result.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: The Sylow Theorems: $\S 59 \delta$