Exercise 4, § 5

(a) Let G be a non-commutative simple group, $\Gamma$ the automorphism group of G and $\Delta$ the group of inner automorphisms of G (isomorphic to G). If s is an automorphism of the group $\Gamma$, show that $s(\Delta) = \Delta$ (using Exercise 3 above and § 4, no. 9, Proposition 15, note that the intersection $\Delta \cap s(\Delta)$ cannot consist only of the identity element of $\Gamma$). (b) Show that the only automorphism of $\Gamma$ leaving invariant each of the elements of $\Delta$ is the identity element (use the fact that this automorphism leaves invariant $\alpha$ and $\sigma \alpha \sigma^{-1}$ for all $\alpha \in \Delta$ and $\sigma \in \Gamma$ and use Exercise 3). (c) Let s be an automorphism of $\Gamma$, $\phi$ the isomorphism $x \mapsto \operatorname{Int}(x)$ of G onto $\Delta$, $\psi$ the inverse isomorphism and $\sigma$ the automorphism $\psi \circ s \circ \phi$ of G; show that the automorphism $\xi \mapsto \sigma^{-1}s(\xi)\sigma$ of $\Gamma$ is the identity automorphism (use (b), noting that, for all $x \in G$, $s(\operatorname{Int}(x)) = \operatorname{Int}(\sigma(x))$). Deduce that every automorphism of $\Gamma$ is an inner automorphism.