$N/C$ Theorem: Kernel of $f:N(H)\rightarrow{\rm Aut}(H)$ by for all $a \in N(H), f(a)=\theta_a\vert _H$

Question

Let $G$ be a group and $H$ be a subgroup of $G$. I want to prove that $N(H)/C(H)$ is isomorphic to a subgroup of ${\rm Aut}(H)$, by defining a function $f: N(H) \rightarrow {\rm Aut}(H)$ for all $a \in N(H), f(a)=\theta_a\vert _H$ and using the first isomorphism theorem. I thought $\ker f = \{a \in N(H):f(a)=i_H \}$, however my book (Fundamentals of Abstract Algebra - Malik, Mordeson, Sen) says $\ker f = \{a \in G:f(a)=i_H \}$. But the domain of the function $f$ is $N(H)$, not $G$. I am wondering that if I am missing some points or there is a typo in my book.

Answer 1

I haven't read your book but I think you're right.

You define $f:N_G(H)\to{\rm Aut}(H)$ with $f(g)(h)=ghg^{-1}\in H,\ \forall h\in H$.

Then $\ker(f)=\{g\in N_{G}(H)|\ f(g)=Id_H\}=\{g\in N_{G}(H)|\ f(g)(h)=h, \ \forall h\in H\}= \\ \{g\in N_{G}(H)|\ ghg^{-1}=h \ \forall h\in H\}=C_G(H)$.

Hence $N_G(H)/C_G(H)\cong{\rm Aut}(H)$.

Maybe the book uses some different notation.