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.

  • 1
    $\begingroup$ 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


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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.