**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 2.2 Exercise 2.2.9**

Let $G$ be a group, $H \leq G$, and $A \subseteq G$. Define $$N_H(A) = \{ h \in H \ |\ hAh^{-1} = A \}.$$ Show that $N_H(A) = N_G(A) \cap H$ and deduce that $N_H(A) \leq H$.

Solution:

($\subseteq$) That $N_H(A) \subseteq N_G(A)$ is clear, as is $N_H(A) \subseteq H$. Thus $N_H(A) \subseteq N_G(A) \cap H$.

($\supseteq$) Suppose $h \in N_G(A) \cap H$. Then we have $h \in H and hAh^{-1} = A$, so that $h \in N_H(A)$.

Now $N_H(A) \subseteq H$. By a lemma to Exercise 2.2.6, then, $N_H(A) \leq H$.