Semidirect product with self-normalizing subgroup of automorphism group of coprime order implies every automorphism is inner

Statement
Suppose $$N$$ is a finite group, and $$H$$ is a nontrivial fact about::self-normalizing subgroup of the automorphism group $$\operatorname{Aut}(N)$$, i.e., the normalizer of $$H$$ in $$\operatorname{Aut}(N)$$ is $$H$$. Further, suppose the orders of $$N$$ and $$H$$ are relatively prime. Consider the external semidirect product:

$$G = N \rtimes H$$.

Then, $$G$$ is a fact about::group in which every automorphism is inner. Note that $$G$$ need not be a fact about::complete group, though it is a complete group in many practical situations. One example where $$G$$ is not complete is when $$N$$ is a cyclic group of order two.

Related facts

 * Centerless and characteristic in automorphism group implies automorphism group is complete
 * Characteristically simple and non-abelian implies automorphism group is complete
 * Holomorph of cyclic group of odd prime order is complete

Facts used

 * 1) uses::Equivalence of definitions of normal Hall subgroup
 * 2) uses::Schur-Zassenhaus theorem: We use here the conjugacy part of the theorem: any two permutable complements to a normal Hall subgroup are conjugate.
 * 3) uses::Automorphism group action lemma: Suppose $$G$$ is a group, and $$N,H \le G$$ are subgroups such that $$H \le N_G(N)$$. Suppose $$\sigma$$ is an automorphism of $$G$$ such that the restriction of $$\sigma$$ to $$N$$ gives an automorphism $$\alpha$$ of $$N$$, and such that $$\sigma$$ also restricts to an automorphism of $$H$$, say $$\sigma'$$. Consider the map:

$$\rho: H \to \operatorname{Aut}(N)$$

that sends an element $$h \in H$$ to the automorphism of $$N$$ induced by conjugation by $$h$$ (note that this is an automorphism since $$H \le N_G(N)$$). Then, we have:

$$\rho \circ \sigma' = c_\alpha \circ \rho$$

where $$c_\alpha$$ denotes conjugation by $$\alpha$$ in the group $$\operatorname{Aut}(N)$$.

First part
Given: A finite group $$N$$, a self-normalizing subgroup $$H$$ of $$\operatorname{Aut}(N)$$, such that the order of $$N$$ and $$H$$ are relatively prime. $$G$$ is the semidirect product of $$N$$ and $$H$$.

To prove: Every automorphism of $$G$$ is inner.

Proof:


 * 1) $$N$$ is a normal Hall subgroup of $$G$$: Since the order of $$N$$ and $$H$$ are relatively prime, we see that the index $$[G:N]$$, which equals the order of $$H$$, is relatively prime to the order of $$N$$. Thus, $$N$$ is a Hall subgroup of $$G$$. It is also clearly normal, by the definition of semidirect product.
 * 2) $$N$$ is characteristic in $$G$$: This follows from step (1) and fact (1).
 * 3) If $$\sigma$$ is an automorphism of $$G$$, then $$\sigma(N) = N$$: This follows from step (2).
 * 4) If $$\sigma$$ is an automorphism of $$G$$, then there exists a $$g \in G$$ such that if $$c_g$$ denotes conjugation by $$G$$, $$c_g(H) = \sigma(H)$$. In particular, if $$\tau = c_{g^{-1}} \circ \sigma$$, then $$\tau(H) = H$$: By the previous step, $$\sigma(N) = N$$, so $$\sigma$$ must send $$H$$ to a permutable complement of $$N$$. By fact (2) (Schur-Zassenhaus theorem) any two such permutable complements are conjugate by some $$g \in G$$, yielding the required result.
 * 5) If $$\tau$$ is an automorphism of $$G$$ such that $$\tau(H) = H$$, then there exists $$h \in H$$ such that $$\tau = c_h$$: Note that since $$N$$ is characteristic in $$G$$, $$\tau(N) = N$$ as well. We see that the conditions are satisfied for the automorphism group action lemma (fact (3)), with $$\rho$$ being the inclusion of $$H$$ in $$\operatorname{Aut}(N)$$. So, if $$\alpha$$ is the restriction of $$\tau$$ to $$N$$, we obtain that the action of $$\tau$$ on $$H$$ is given by conjugation by $$\alpha$$ in $$\operatorname{Aut}(N)$$. Thus, we have $$\alpha \in \operatorname{Aut}(N)$$ such that $$\alpha$$ conjugates $$H$$ to itself. By the assumption that $$H$$ is self-normalizing in $$\operatorname{Aut}(N)$$, $$\alpha$$ must be equal to an element inside $$H$$. Call this element $$h$$. Then, $$\tau$$ agrees with conjugation by $$h$$ on $$H$$ as well as on $$N$$. Since $$G = NH$$, we get $$\tau = c_h$$ on $$G$$.
 * 6) If $$\sigma$$ is an automorphism of $$G$$, $$\sigma$$ is inner: By step (4), $$\sigma = c_g \circ\tau$$ for $$g \in G$$ and $$\tau$$ an automorphism that preserves $$H$$, and by step (5), $$\tau = c_h$$ for some $$h \in H$$. Thus, $$\sigma = c_g \circ c_h = c_{gh}$$, so $$\sigma$$ is an inner automorphism of $$G$$.