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Let $G$ be a group, and let $g \in G$ . Prove that the function $\gamma_g: G \to G$ defined by $(\forall a \epsilon g): \gamma_g(a)=g a^{-1} g $ is an automorphism of G.

The automorphisms $\gamma_g$ are called 'inner' automorphisms.

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    $\begingroup$ Check your notes: it should be $\gamma(a)=gag^{-1}$ rather than what you wrote. Other than that, this is very straightforward to verify. What did you try so far? $\endgroup$ – rschwieb Apr 17 '13 at 20:37
  • $\begingroup$ Do you know what should be shown about $\gamma$ to prove that it is an automorphism? $\endgroup$ – Hagen von Eitzen Apr 17 '13 at 20:38
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Hints:

If $b\in G$, notice that $b=gg^{-1}bgg^{-1}$.

If $gag^{-1}=1$, solve for $a$.

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