Normalizer is Subgroup
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Theorem
Let $G$ be a group.
The normalizer of a subset $S \subseteq G$ is a subgroup of $G$.
- $S \subseteq G \implies \map {N_G} S \le G$
Proof
Let $a, b \in \map {N_G} S$.
Then:
\(\ds S^{a b}\) | \(=\) | \(\ds \paren {S^b}^a\) | Conjugate of Set by Group Product | |||||||||||
\(\ds \) | \(=\) | \(\ds S^a\) | Definition of Normal Subgroup | |||||||||||
\(\ds \) | \(=\) | \(\ds S\) | Definition of Normal Subgroup |
Therefore $a b \in \map {N_G} S$.
Now let $a \in \map {N_G} S$:
- $a \in \map {N_G} S \implies S^{a^{-1} } = \paren {S^a}^{a^{-1} } = S^{a^{-1} a} = S$
Therefore $a^{-1} \in \map {N_G} S$.
Thus, by the Two-Step Subgroup Test, $\map {N_G} S \le G$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.20$
- 1966: Richard A. Dean: Elements of Abstract Algebra ... (previous) ... (next): $\S 1.9$: Exercise $5.15$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
- 1971: Allan Clark: Elements of Abstract Algebra ... (previous) ... (next): Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 48$
- 1978: Thomas A. Whitelaw: An Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $12 \ \text{(i)}$