Centerless and characteristic in automorphism group implies automorphism group is complete

Statement

Suppose $G$ is a Centerless group (?) such that, with the natural embedding of $G$ in its Automorphism group (?) $\operatorname {Aut} (G)$, $G$ is a Characteristic subgroup (?). Then, the automorphism group $\operatorname {Aut} (G)$ is a Complete group (?): it is centerless and every automorphism of it is inner.

Related facts

• Centerless implies inner automorphism group is centralizer-free in automorphism group: This states that if $G$ is centerless, $C_{\operatorname {Aut} (G)}\operatorname {Inn} (G)$ is trivial. In particular, it shows that $\operatorname {Aut} (G)$ is centerless. This idea leads to the notion of an automorphism tower.
• Wielandt's automorphism tower theorem: This states that for any finite centerless group, the tower obtained by repeatedly taking automorphism groups eventually stabilizes. Note that our result gives a condition under which it stabilizes after just one step: the automorphism group is itself complete, so the tower stabilizes after that.

Facts used

For the hands-on proof

1. Centerless implies inner automorphism group is centralizer-free in automorphism group: If $G$ is centerless, then $\operatorname {Inn} (G)$ (identified naturally with $G$) has trivial centralizer in $\operatorname {Aut} (G)$. In other words, no non-identity automorphism commutes with every inner automorphism.
2. Inner automorphism group is normal in automorphism group
3. Centralizer-free and normal implies automorphism-faithful: If $A\leq B$ is a normal subgroup and $C_{B}(A)$ is trivial, then any automorphism of $B$that fixes every element of $A$ is the identity map.
4. The basic idea that, if $c_{g}$ denotes conjugation by $g$ and $\tau$ denotes an automorphism of $G$, we have:

$c_{\tau (g)}=\tau \circ c_{g}\circ \tau ^{-1}$

In other words, if $\tau$ is an automorphism of $G$, the action of $\tau$ on $\operatorname {Inn} (G)$ by conjugation in $\operatorname {Aut} (G)$ precisely mimics the action of $\tau$ as an automorphism on $G$.

For the proof using the automorphism group action lemma

Automorphism group action lemma: Suppose $K$ is a group and $N,H\leq K$ are subgroups such that $H\leq N_{G}(N)$ and $\sigma$ is an automorphism of $K$ that restricts to an automorphism of $N$ as well as of $H$. Let $\alpha$ be the induced automorphism of $N$ and $\sigma '$ be the induced automorphism of $H$. Then, if

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

denotes the homomorphism induced by the action by conjugation, we have:

$\rho \circ \sigma '=c_{\alpha }\circ \rho$.

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

Proof

Hands-on proof

Given: A centerless group $G$, embedded naturally as $G=\operatorname {Inn} (G)$ in the automorphism group $\operatorname {Aut} (G)$. Further, $G$ is a characteristic subgroup of $\operatorname {Aut} (G)$ under this embedding.

To prove: $\operatorname {Aut} (G)$ is complete: the center of $\operatorname {Aut} (G)$ is trivial, and any automorphism of $\operatorname {Aut} (G)$ is inner.

Proof:

1. (Facts used: fact (1)): $\operatorname {Aut} (G)$ is centerless: By fact (1), $G=\operatorname {Inn} (G)$ is centralizer-free in $\operatorname {Aut} (G)$. Thus, $\operatorname {Aut} (G)$ is centerless.
2. (Facts used: facts (1), (2), (3)): $G$ is automorphism-faithful in $\operatorname {Aut} (G)$: By fact (1), $G$ is centralizer-free in $\operatorname {Aut} (G)$, and by fact (2), $G$ is normal in $\operatorname {Aut} (G)$. Combining these with fact (3), we obtain that $G$ is automorphism-faithful in $\operatorname {Aut} (G)$.
3. (Given data used: $G$ is characteristic in $\operatorname {Aut} (G)$): For any automorphism $\sigma$ of $\operatorname {Aut} (G)$, there exists an element $\tau \in \operatorname {Aut} (G)$ such that conjugation by $\tau$ has the same effect as $\sigma$ on the subgroup $G$: First, observe that since $G$ is characteristic in $\operatorname {Aut} (G)$, the restriction of $\sigma$ to $G$ equals an automorphism of $G$, so there is an automorphism $\tau$ of $G$ having the same effect. Note that (by fact (4)) an automorphism $\tau$ of $G$ manifests itself as the inner automorphism by $\tau$ on $G=\operatorname {Inn} (G)$. Thus, $c_{\tau }$ and $\sigma$ have the same effect on $G$, viewed as a subgroup of $\operatorname {Aut} (G)$.
4. $\sigma =c_{\tau }$, and thus, $\sigma$ is inner: By step (3), $\sigma ^{-1}c_{\tau }$ is an automorphism of $\operatorname {Aut} (G)$ whose restriction to $G=\operatorname {Inn} (G)$ is the identity map. Step (2) now forces $\sigma ^{-1}c_{\tau }$ to be the identity map on the whole group $\operatorname {Aut} (G)$, yielding $\sigma =c_{\tau }$.
5. Step (1) shows that $\operatorname {Aut} (G)$ is centerless, and steps (3) and (4) show that every automorphism is inner. This completes the proof of completeness.

Proof using automorphism group action lemma

Given: A centerless group $G$, embedded naturally as $G=\operatorname {Inn} (G)$ in the automorphism group $\operatorname {Aut} (G)$. Further, $G$ is a characteristic subgroup of $\operatorname {Aut} (G)$ under this embedding.

To prove: $\operatorname {Aut} (G)$ is complete: the center of $\operatorname {Aut} (G)$ is trivial, and any automorphism of $\operatorname {Aut} (G)$ is inner.

Proof: Set $K=H=\operatorname {Aut} (G)$ and $N=\operatorname {Inn} (G)$. By assumption, $N$ is characteristic in $K=H$, so $H\leq N_{G}(N)$. Let $\sigma$ be any automorphism of $K$. Observe that:

• $\sigma$ restricts to an automorphism of $H$, because $H=K$.
• $\sigma$ restricts to an automorphism of $N$, because by assumption $N$ is characteristic in $K$.

Further, the natural map

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

is in this case the identity map. Thus, the automorphism group action lemma tells us that for any automorphism $\sigma$ of $K=\operatorname {Aut} (G)$, we have:

$\sigma =c_{\alpha }$

where $\alpha$ is the automorphism of $N=G$ induced by $\sigma$.