Exercise 3.1.28 (If $N = \langle S \rangle$ is finite, $g \in N_G(N) \iff gSg^{-1} \subseteq N$)

Question

Let $N$ be a {\bf finite} subgroup of a group $G$ and assume $N = \langle S \rangle$ for some subset $S$ of $G$. Prove that an element $g \in G$ normalizes $N$ if and only if $g S g^{-1} \subseteq N$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $g \in G$. By Exercise 27, $g$ normalizes $N$ if and only if $gNg^{-1} \subseteq N$. Since $S \subseteq N$, $gSg^{-1} \subseteq gNg^{-1} \subseteq N$, so $gSg^{-1} \subseteq  N$.

\bigskip
Conversely, suppose that $gSg^{-1} \subseteq  N$. Then $S \subseteq g^{-1} N g$, and $g^{-1} N g = \gamma_{g^{-1}}(N)$ is a subgroup of $G$. Hence $\langle S \rangle \subseteq g^{-1} N g$, since by definition $\langle S \rangle$ is the smallest subgroup of $G$ which contains $S$. Therefore $N \subseteq g^{-1} N g$ or equivalently $gNg^{-1} \subseteq N$.

This shows that if $N$ is a finite subgroup of $G$, for all $g \in G$,
$$ g \in N_G(N) \iff gSg^{-1} \subseteq N.$$

\end{proof}

Exercise 3.1.28 (If $N = \langle S \rangle$ is finite, $g \in N_G(N) \iff gSg^{-1} \subseteq N$)

Answers

Comments

Exercise 3.1.28 (If $N = \langle S \rangle$ is finite, $g \in N_G(N) \iff gSg^{-1} \subseteq N$)

Answers

Comments

Add answer