Normalizer of Center is Group
From ProofWiki
Theorem
Let $G$ be a group.
Let $Z \left({G}\right)$ be the center of $G$.
Let $x \in G$.
Let $N_G \left({x}\right)$ be the normalizer of $x$ in $G$.
Let $\left[{G : N_G \left({x}\right)}\right]$ be the index of $N_G \left({x}\right)$ in $G$.
Then:
- $Z \left({G}\right) = \left\{{x \in G: N_G \left({x}\right) = G}\right\}$
That is, the center of a group $G$ is the set of elements $x$ of $G$ such that the normalizer of $x$ is the whole of $G$.
Thus $x \in Z \left({G}\right) \iff N_G \left({x}\right) = G$, and so $\left[{G : N_G \left({x}\right)}\right] = 1$.
Proof
$N_G \left({x}\right)$ is the normalizer of the set $\left\{{x}\right\}$. Thus:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle N_G \left({x}\right)\) | \(=\) | \(\displaystyle G\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \forall a \in G: \left\{ {x}\right\}^a\) | \(=\) | \(\displaystyle \left\{ {x}\right\}\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Normalizer | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \forall a \in G: a x a^{-1}\) | \(=\) | \(\displaystyle x\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Conjugate | ||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle \forall a \in G: a x\) | \(=\) | \(\displaystyle x a\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \iff\) | \(\displaystyle \) | \(\displaystyle x\) | \(\in\) | \(\displaystyle Z \left({G}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | Definition of Center of a Group |
$\blacksquare$
Sources
- Allan Clark: Elements of Abstract Algebra (1971)... (previous)... (next): $\S 50$