Exercise 3.1.27 (If $G$ is a finite subgroup, then $N_G(N) = \{g \in G \mid gNg^{-1}\subseteq N \}$)

Question

Let $N$ be a {\bf finite} subgroup of a group $G$. Show that $gNg^{-1} \subseteq N$ if and only if $gNg^{-1} = N$. Deduce that $N_G(N) = \{g \in G \mid gNg^{-1}\subseteq N \}$.

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $g \in G$, and suppose that $gNg^{-1} \subseteq N$. Consider the map
$$
\gamma_g\ 
\left\{
\begin{array}{ccl}
G & 	o &G\
x & \mapsto & gx g^{-1}.
\end{array}
ight.
$$
Then $\gamma_g$ is bijective (since $\gamma_g \circ \gamma_{g^{-1}} = \gamma_{g^{-1}} \circ\gamma_g = \mathrm{id}_G$), and $\gamma_g$ is a homomorphism by Exercise 26 part (a), so 
$\gamma_g$ is an automorphism of $G$, called inner automorphism. Therefore $|g N g^{-1}| = |\gamma_g(N)| = |N| < \infty$, where $gNg^{-1} \subseteq N$, hence $gNg^{-1} = N$. Since the converse is always true, for all $g \in G$,
$$gNg^{-1} \subseteq N \iff gNg^{-1} = N\qquad (	ext{if $N$ is a finite subgroup of $G$}).$$
By definition of $N_G(N)$, for all $g\in G$,
$$ g \in N_G(N) \iff gN g^{-1} = N \iff gNg^{-1} \subseteq N,$$
so
$$N_G(N) = \{g \in G \mid gNg^{-1} \subseteq N\}.$$
\end{proof}

Exercise 3.1.27 (If $G$ is a finite subgroup, then $N_G(N) = \{g \in G \mid gNg^{-1}\subseteq N \}$)

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Exercise 3.1.27 (If $G$ is a finite subgroup, then $N_G(N) = \{g \in G \mid gNg^{-1}\subseteq N \}$)

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