Notation 1. For a group $G$ we will denote by $Z(G), \text{Aut}(G)$ and $\text{Inn}(G),$ the center of $G$,  the group of automorphisms of $G$ and the group of inner automorphisms of $G$ respectively.

Notation 2. By $\tau_g$ we will mean the map $\tau_g : G \longrightarrow G$ defined by $\tau_g(x)=gxg^{-1},$ for all $x \in G.$ Recall that, by definition, $\text{Inn}(G)=\{\tau_g : \ g \in G \}.$

Remark. For every $g \in G$ and $\gamma \in \text{Aut}(G)$ we have $\gamma \tau_g \gamma^{-1} = \tau_{\gamma} \tau_g = \tau_{\gamma(g)}.$

Definition. A group $G$ is called complete if $Z(G)=\{1\}$ and $\text{Aut}(G)=\text{Inn}(G),$ i.e. every automorphism of $G$ is inner.

Problem 1. Let $G$ be a group with $Z(G)=\{1\}.$ Prove that the centralizer of $\text{Inn}(G)$ in $\text{Aut}(G)$ is trivial. In particular, $Z(\text{Aut}(G))=\{1\}.$

Solution. Let $A = \text{Aut}(G)$ and $B= \text{Inn}(G).$ Let $\sigma$ be in the centralizer of $B$ in $A.$ So for every $g \in G$ we have $\sigma \tau_g = \tau_g \sigma.$ Thus $\sigma \tau_g(\sigma^{-1}(x)) = \tau_g \sigma (\sigma^{-1}(x)),$ for all $x \in G.$ Hence $g^{-1} \sigma(g) x = xg^{-1} \sigma(g),$ for all $x \in G.$ Therefore $g^{-1} \sigma(g) \in Z(G)=\{1\}$ and so $\sigma = 1. \ \Box$

Problem 2. Let $G$ be a group with $Z(G)=\{1\}.$ Prove that if $\text{Inn}(G)$ is a characteristic subgroup of $\text{Aut}(G),$ then $\text{Aut}(G)$ is complete.

Solution. Let $A = \text{Aut}(G)$ and $B= \text{Inn}(G).$ So, by Problem 1 and the fact that $\text{Inn}(A)$ is a subgroup of $\text{Aut}(A),$ we only need to prove that $\text{Aut}(A) \subseteq \text{Inn}(A).$ Let $\varphi \in \text{Aut}(A).$ Since $B$ is characteristic in $A,$ for every $g \in G$ there exists some $g' \in G$ such that $\varphi(\tau_g) = \tau_{g'}.$ Define $\sigma : G \longrightarrow G$ by $\sigma(g)=g'.$ It is easy to see that $\sigma \in A.$ So

$\varphi(\tau_g) = \tau_{\sigma(g)}, \ \ \ \ \ \ \ \ \ \ \ (1)$

for all $g \in G.$ Let

$\psi = \varphi \tau_{\sigma^{-1}}$

and let $g \in G.$ Then by the above remark and $(1)$

$\psi(\tau_g)=\varphi \tau_{\sigma^{-1}} \tau_g = \varphi \tau_{\sigma^{-1}(g)} = \tau_g. \ \ \ \ \ \ \ \ \ \ \ \ \ (2)$

Thus, by the above remark and $(2)$

$\gamma \tau_g \gamma^{-1} = \tau_{\gamma(g)} = \psi(\tau_{\gamma(g)})=\psi(\gamma \tau_g \gamma^{-1})=\psi(\gamma) \tau_g \psi(\gamma^{-1}). \ \ \ \ \ \ \ \ \ \ \ \ (3)$

We can re-write $(3)$ as $\gamma^{-1} \psi(\gamma) \tau_g = \tau_g \gamma^{-1} \psi(\gamma),$ which means that $\gamma^{-1} \psi(\gamma)$ is in the centralizer of $B.$ Thus, by Problem 1, $\gamma^{-1} \psi(\gamma)=1$ and so $\psi(\gamma)=\gamma,$ for all $\gamma \in A.$ Thus $\psi = 1$ and hence $\varphi = \tau_{\sigma} \in \text{Inn}(A). \ \Box$

Problem 3. Prove that if $G$ is a non-abelian simple group, then $\text{Aut}(G)$ is complete.

Proof. Let $A = \text{Aut}(G)$ and $B= \text{Inn}(G).$ Since $Z(G)$ is a normal subgroup of $G$ and $G$ is non-abelian and simple, $Z(G)=\{1\}.$ By Problem 2, we only need to prove that $B$ is characteristic in $A.$ So let $\varphi \in \text{Aut}(A).$ We need to prove that $\varphi(B) \subseteq B.$ Let

$C = B \cap \varphi(B).$

Clearly $C$ is normal in $\varphi(B)$ because $B$ is normal in $A.$ We also have

$\varphi(B) \cong B \cong G/Z(G) \cong G,$

which implies that $\varphi(B)$ is simple, because $G$ is simple. So either $C=\{1\}$ or $C = \varphi(B).$

Case 1. $C = \varphi(B).$ Then $\varphi(B) \subseteq B$ and we are done.

Case 2. $C = \{1\}.$ Then $\varphi(B)$ will be in the centralizer of $B,$ because both $B$ and $\varphi(B)$ are normal in $A.$ Hence, by Problem 1, $\varphi(B) = \{1\}$ and so $B = \{1\},$ which is absurd because $G$ is non-abelian. $\Box$