# Show that $G\to\operatorname{Aut}(G)$, $g\mapsto (x\mapsto gxg^{-1})$ is a homomorphism [closed]

Let $G$ be a group, and let $g\in G$. Prove that the function $G \rightarrow \operatorname{Aut}(G)$ defined by $g \mapsto \gamma_g$ is a homomorphism.

We know from a previous proof: Prove that the function $\gamma_g\colon G \rightarrow G$ defined by $(\forall a\in G)(\gamma_g(a)=g ag^{-1})$ is an automorphism of $G$.

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

## closed as off-topic by Julian Kuelshammer, Lord_Farin, Dominic Michaelis, user1729, rschwiebSep 11 '13 at 20:19

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• "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." – Julian Kuelshammer, Lord_Farin, Dominic Michaelis, user1729, rschwieb
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• The previous proof being posted 5 minutes ago, showing a similar level of effort. This question suffers from the same mistken definition of inner automorphism (should be $gag^{-1}$) – rschwieb Apr 17 '13 at 20:40
• What is the precise spelled-out meaning of "this map is a homomorphism"? At which step is it possible to fail to continue? – Hagen von Eitzen Apr 17 '13 at 20:42

We denote the function by $$\Phi : G \to \operatorname{Aut}(G),\quad g\mapsto \gamma_g.$$ Now for all $g,h,x\in G$, $$\Phi(gh)(x) = \gamma_{gh}(x) = (gh)x(gh)^{-1} = ghxg^{-1}h^{-1} = g(hxh^{-1})g^{-1} = g\Phi(h)(x)g^{-1} = \Phi(g)(\Phi(h)(x)) = (\Phi(g)\circ \Phi(h))(x).$$ and hence $$\Phi(gh) = \Phi(g)\circ\Phi(h).$$ So $\Phi$ is an homomorphism.