Inn(G) is a Normal Subgroup of Aut(G)

# Inn(G) is a Normal Subgroup of Aut(G)

Proposition 1: Let $G$ be a group. Then $\mathrm{Inn}(G)$ is a normal subgroup of $\mathrm{Aut}(G)$. |

**Proof:**We have already proved that $\mathrm{Inn}(G)$ is a subgroup of $\mathrm{Aut}(G)$ on The Subgroup of Inner Automorphisms of a Group, Inn(G) page. All that remains to show is that $\mathrm{Inn}(G)$ is normal in $\mathrm{Aut}(G)$.

- Let $f \in \mathrm{Aut}(G)$ and let $i_a \in \mathrm{Inn}(G)$. Then for all $g \in G$ we have that:

\begin{align} \quad (f \circ i_a \circ f^{-1})(g) &= f(i_a(f^{-1}(g)) \\ &= f(af^{-1}(g)a^{-1}) \\ &= f(a)[f(f^{-1}(g))][f(a^{-1})] \\ &= f(a)gf(a^{-1}) \\ &= f(a)g[f(a)]^{-1} \\ &= i_{f(a)}(g) \\ \end{align}

- Thus $f \circ i_a \circ f^{-1} = i_{f(a)} \in \mathrm{Inn}(G)$ for each $f \in \mathrm{Aut}(G)$ and for each $i_a \in \mathrm{Inn}(G)$. So $f\mathrm{Inn}(G)f^{-1} \subseteq \mathrm{Inn}(G)$, i.e., $\mathrm{Inn}(G)$ is a normal subgroup of $\mathrm{Aut}(G)$.