Subgroup is Subgroup of Normalizer
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Theorem
Let $G$ be a group.
A subgroup $H \le G$ is a subgroup of its normalizer:
- $H \le G \implies H \le \map {N_G} H$
Proof
Subset of Normalizer
First we show that $H$ is a subset of $\map {N_G} H$.
This follows directly from Left Coset Equals Subgroup iff Element in Subgroup:
- $x \in H \implies x H = H$
As $x \in H \implies x^{-1} \in H$ it also follows that $x \in H \implies H x^{-1} = H$.
Thus:
- $x \in H \implies x H x^{-1} = H^x = H$
and so:
- $x \in \map {N_G} H$
So:
- $H \subseteq \map {N_G} H$
as we wanted to show.
$\Box$
Subgroup of Normalizer
By hypothesis, $H$ is a subgroup of $G$.
Thus $H$ is itself a group.
So by definition of subgroup:
- $(1): \quad H \subseteq \map {N_G} H$
- $(2): \quad $ is a group
it follows that $H$ is a subgroup of $\map {N_G} H$.
$\blacksquare$
Sources
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.10$