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 


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