Exercise 3.1.31 ($N_G(N) = \max \{H \leq G \mid N \unlhd H\}$)

Question

Prove that if $H \leq G$ and $N$ is a normal subgroup of $H$ then $H \leq N_G(N)$. Deduce that $N_G(N)$ is the largest subgroup of $G$ in which $N$ is normal (i.e., is the join of all subgroups $H$ for which $N \unlhd H$).

Richard Ganaye · Accepted Answer

\begin{proof} 
Suppose that $N \unlhd H \leq G$. Then for every $h \in H$ and for every $a \in N$, $hah^{-1} \in N$, where $h \in G$, thus $h \in N_G(N)$. This shows that
$$H \subseteq N_G(N).$$
$N_G(N)$ contains every subgroup $H$ of $G$ such that $N\unlhd H$. Moreover $N \unlhd N_G(N)$. So $N_G(N)$ is the maximum (for the relation of inclusion) of the set of subgroups
$$\{H \leq G \mid N \unlhd H\}.$$
\end{proof}

Exercise 3.1.31 ($N_G(N) = \max \{H \leq G \mid N \unlhd H\}$)

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Exercise 3.1.31 ($N_G(N) = \max \{H \leq G \mid N \unlhd H\}$)

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