Can ${\rm Aut}(H)$ be replaced with ${\rm Inn}(H)$ in the N/C theorem?

Question

Could we assert that if $H$ is a subgroup of $G$, then the factor group $N_G(H)/C_G(H)$ is isomorphic to a subgroup of ${\rm Inn}(H)$ instead of ${\rm Aut}(H)$?

Answer 1

There is another approach that is less known than the "N/C" theorem.
The outer automorphisms of a group $H$, ${\rm Out}(H)$, are defined as the quotient group ${\rm Aut}(H)/{\rm Inn}(H)$. Note that the elements of ${\rm Out}(H)$ are cosets of automorphisms of $H$, and not themselves automorphisms. Since $H$ and $C_G(H)$ are both normal in $N_G(H)$, the subgroup $HC_G(H) \unlhd N_G(H)$. Observe that $N_G(H)/HC_G(H) \cong (N_G(H)/C_G(H))/(HC_G(H)/C_G(H))$ and $HC_G(H)/C_G(H) \cong H/(H \cap C_G(H)) = H/C_H(H)= H/Z(H) \cong {\rm Inn}(H).$ From this, one can easily show that

$N_G(H)/HC_G(H) \hookrightarrow {\rm Out}(H)$.

And this is interesting for instance when both ${\rm Out}(H)$ and $Z(H)$ are trivial (such groups are called complete), then $N_G(H)$ becomes a direct product of $H$ and $C_G(H)$. By the way, the symmetric groups $S_n$ are all complete, except for $n=2$ or $6$.

Answer 2

No, it is not necessary for (the image of) $N_G(H)/C_G(H)$ to be inner in ${\rm Aut}(H)$.

This means that "external" conjugation can defy imitation by "internal" conjugation.

In fact, every element of ${\rm Aut}(H)$ can be realized as conjugation by an element in some overgroup containing $H$. In particular, set $G$ to be the holomorph ${\rm Hol}(H):=H\rtimes{\rm Aut}(H)$ (for necessary background information see semidirect product). If $\varphi\in{\rm Aut}(H)$ then "$\varphi\in N_G(H)$" and the image of $\varphi$ in ${\rm Aut}(H)$ (under $N_G(H)\to N_G(H)/C_G(H)\to{\rm Aut}(H)$) is just itself, $\varphi$.