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If [itex]\sigma\in Aut(G)[/itex] and [itex]\varphi_g[/itex] is conjugation by g prove [itex]\sigma\varphi_g\sigma^{-1}=\varphi_{\sigma(g)}[/itex]. Deduce [itex]Inn(G)\trianglelefteq Aut(G)[/itex]

Let [itex]x\in G[/itex].

[tex]\sigma\varphi_g\sigma^{-1}(x)=\sigma(g\sigma^{-1}(x)g^{-1})=\sigma(g)x\sigma(g)^{-1}[/tex]

Why is this:

[tex]\sigma(g\sigma^{-1}(x)g)=\sigma(g)x\sigma(g)^{-1}[/tex]

Let [itex]x\in G[/itex].

[tex]\sigma\varphi_g\sigma^{-1}(x)=\sigma(g\sigma^{-1}(x)g^{-1})=\sigma(g)x\sigma(g)^{-1}[/tex]

Why is this:

[tex]\sigma(g\sigma^{-1}(x)g)=\sigma(g)x\sigma(g)^{-1}[/tex]

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