N/C theorem : If $\Psi$ is a homomorphism, then $N(H)/ C(H) ~\approx~ \Psi[N(H)]$. Is $\Psi[N(H)]$ a subgroup of $\operatorname{Aut}(H)$ If $H$ is a subgroup of $G$, then if $N(H) = \{g \in G~|~gHg^{-1}=H\} ; ~~C(H) = \{g \in G ~|~ ghg^{-1}=h ~\forall~h \in H\}$, then I have proved that :
$\Psi: N(H) \rightarrow \operatorname{Aut}(H)$ given by $\Psi(n) = nhn^{-1} ~\forall~h \in H,~ n \in N(H)$  is a homomorphism and its kernel is $C(H)$
Hence, by the first isomorphism theorem, 
$N(H)/ C(H) ~\approx~ \Psi[N(H)]$
I am confused how is $\Psi[N(H)]$ a subgroup of $\operatorname{Aut}(H)$ as per the statement of the $N/C$ theorem.
I know that $Inn(H)$ is a subgroup of $\operatorname{Aut}(H)$
Is $\Psi[N(H)]$ a subgroup of $\operatorname{Inn}(H)$ by any chance?
Help will be appreciated. Thank you.
 A: Basically, $\Psi$ is telling you to think of elements of $N(H)$ as things that you can conjugate stuff in $H$ by to get back stuff in $H$. Conjugating stuff in $H$ by things outside $N(H)$ will generally not result in something in $H$, so only conjugation by elements of $N(H)$ can give you automorphisms of $H$.
Since you're defining these automorphisms by conjugation, of course anything in $C(H)$ will act trivially. Hence, $N(H)/C(H)$ gives you a subgroup of $\text{Aut}(H)$, in the sense that you have a homomorphism $N(H)\rightarrow\text{Aut}(H)$ which has kernel $C(H)$, so you have an injection $N(H)/C(H)\hookrightarrow\text{Aut}(H)$. (ie, any element $n\in N(H)$ gives you an automorphism of $H$, but $nc$ for any $c\in C(H)$ will give you the same automorphism. Ie, the automorphism determined by $n$ does not depend on the part that comes from $C(H)$).
$\Psi(N(H))$ is not a subgroup of $\text{Inn}(H)$, because $\text{Inn}(H)$ only allows conjugation by elements of $H$. Unless $N(H) = H\cdot C(H)$, $\Psi(N(H))$ will in general be bigger than $\text{Inn}(H)$.
For example, if you embed a group $G$ of size $n$ via the left regular representation inside $S_n$, then $\text{Aut}(G) = N_{S_n}(G)/C_{S_n}(G)$.
