Centerless and characteristic in automorphism group implies automorphism group is complete

Statement
Suppose $$G$$ is a fact about::centerless group such that, with the natural embedding of $$G$$ in its fact about::automorphism group $$\operatorname{Aut}(G)$$, $$G$$ is a fact about::characteristic subgroup. Then, the automorphism group $$\operatorname{Aut}(G)$$ is a fact about::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.

For the hands-on proof

 * 1) uses::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) uses::Inner automorphism group is normal in automorphism group
 * 3) uses::Centralizer-free and normal implies automorphism-faithful: If $$A \le B$$ 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
uses::Automorphism group action lemma: Suppose $$K$$ is a group and $$N, H \le K$$ are subgroups such that $$H \le N_G(N)$$ 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 $$\operatorname{Aut}(N)$$.

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 $$H \le N_G(N)$$. 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 $$\sigma$$.