Exercise 4.4.1 ($\mathrm{Inn}(G) \unlhd \mathrm{Aut}(G)$)

Question

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

Richard Ganaye · Accepted Answer

\begin{proof} Suppose that $\sigma \in \mathrm{Aut}(G)$. For any $y \in G$, put $x = \sigma^{-1}(y)$. Then
\begin{align*}
(\sigma \varphi_g \sigma^{-1})(y)&=\sigma (\varphi_g(x))\
&=\sigma(g x g^{-1})\
&= \sigma(g) \sigma(x) \sigma(g)^{-1}\
&=\sigma(g) y \sigma(g)^{-1}\
&= \varphi_{\sigma(g)}(y).
\end{align*}
This shows that
$$\sigma \varphi_g \sigma^{-1} = \varphi_{\sigma(g)}.$$
If $	au \in \mathrm{Inn}(g)$, then by definition there is some $g \in G$ such that $	au = \varphi_g$. For every $\sigma \in \mathrm{Aut}(G)$
\begin{align*}
\sigma 	au \sigma^{-1} &= \sigma \varphi_g \sigma^{-1}\
&= \varphi_{\sigma(g)},
\end{align*}
thus $\sigma 	au \sigma^{-1} \in \mathrm{Inn}(G)$. So 
$$\mathrm{Inn}(G) \unlhd \mathrm{Aut}(G).$$

\end{proof}

Exercise 4.4.1 ($\mathrm{Inn}(G) \unlhd \mathrm{Aut}(G)$)

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Exercise 4.4.1 ($\mathrm{Inn}(G) \unlhd \mathrm{Aut}(G)$)

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