Exercise 2.2.9 (Normalizer of $A$ in a subgroup $H$)

Question

For any subgroup $H$ of $G$ and any nonempty subset $A$ of $G$ define $N_H(A)$ to be the set ${\{h\in H \mid hAh^{-1} = A\}}$. Show that $N_H(A) = N_G(A) \cap H$ and deduce that $N_H(A)$ is a subgroup of $H$ (note that $A$ does not need to be a subset of $H$).

Richard Ganaye · Accepted Answer

\begin{proof} 
If $h \in N_H(A)$, then $h \in H \subseteq G$, and $hAh^{-1} = A$, so $h \in N_G(H)$. Therefore $h \in H \cap N_G(H)$, so
$$N_H(A) \subseteq H \cap N_G(A).$$
Conversely, if $h \in H \cap N_G(A)$, then $h \in H$ satisfies $h A h^{-1} = A$, thus $h \in N_H(A)$, so
$$H \cap N_G(A)\subseteq N_H(A).$$
In conclusion
$$N_H(A) = H \cap N_G(A).$$
Since $H$ and $N_G(A)$ are subgroups of $G$, so is $N_H(A) = H \cap N_G(A)$. Moreover $N_H(A) \subseteq H$, therefore
$$N_H(A) \leq H.$$
\end{proof}

Exercise 2.2.9 (Normalizer of $A$ in a subgroup $H$)

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Exercise 2.2.9 (Normalizer of $A$ in a subgroup $H$)

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