# 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 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 Group in which every automorphism is inner (?). Note that $G$ need not be a 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.

## Facts used

1. Equivalence of definitions of normal Hall subgroup
2. Schur-Zassenhaus theorem: We use here the conjugacy part of the theorem: any two permutable complements to a normal Hall subgroup are conjugate.
3. 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 $\operatorname{Aut}(N)$.

## Proof

### 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 $G$.