## The inner automorphism group is a normal subgroup of the full automorphism group

If $\sigma \in \mathsf{Aut}(G)$ and $\varphi_g$ is conjugation by $g$, prove that $\sigma \varphi_g \sigma^{-1} = \varphi_{\sigma(g)}$ Deduce that $\mathsf{Inn}(G) \vartriangleleft \mathsf{Aut}(G)$. (The quotient $\mathsf{Aut}(G)/\mathsf{Inn}(G)$ is called the outer automorphism group of $G$.)

Note that $(\sigma \varphi_g \sigma^{-1})(x) = \sigma(\varphi_g(\sigma^{-1}(x)))$ $= \sigma(g \sigma^{-1}(x) g^{-1})$ $= \sigma(g) x \sigma(g)^{-1}$ $= \varphi_{\sigma(g)}(x)$. Hence $\sigma \varphi_g \sigma^{-1} = \varphi_{\sigma(g)} \in \mathsf{Inn}(G)$.

More generally, this implies that $\sigma \mathsf{Inn}(G) \sigma^{-1} \subseteq \mathsf{Inn}(G)$ for all $\sigma \in \mathsf{Aut}(G)$; hence $\mathsf{Inn}(G) \vartriangleleft \mathsf{Aut}(G)$.