Unique Subgroup of a Given Order is Normal
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Theorem
Let a group $G$ have only one subgroup of a given order.
Proof
Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.
Let $H$ be the only subgroup of $G$ whose order is $\order H$.
Let $g \in G$.
From Conjugate of Subgroup is Subgroup:
- $g H g^{-1} \le G$
From Order of Conjugate of Subgroup:
- $\order {g H g^{-1} } = \order H$
But $H$ is the only subgroup of $G$ of order $\order H$.
Hence any subgroup whose order is $\order H$ must in fact be $H$.
That is, $g H g^{-1} = H$.
The result follows from Subgroup equals Conjugate iff Normal.
$\blacksquare$
Sources
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $8 \ \text{(i)}$