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Let $G$ be a group. Then $G$ acts on itself by conjugation, which corresponds to a homomorphism $K\colon G\to\operatorname{Aut}(G)$. Show that the kernel of $K$ is $Z(G)$.

$K: G\to \operatorname{Aut}(G)$

$G\times G\to G$, $(g,x)\mapsto xgx^{−1}$.

$\ker(K)=Z(G)$? $\ker(K)=\{x\in G\mid K(x)=I_G\}$. How can I continue?

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    I hope my edits of your question did not change nay of the intended meaning ...2012-10-29

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The map $K$ is identity if and only of $xgx^{-1}=e$, where $e$ is the identity element. This means that $K$ is the identity map if and only if $xgx^{-1}=g$ for all $g\in G.$ Hence $K$ is identity precisely when $x$ commutes with all $g\in G$, i.e., $g\in Z(G).$

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    I see. appreciated2012-10-30
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There is no need to restrict to finite groups.

An element $g$ is in $\ker K$ iff the conjugation $x\mapsto gxg^{-1}$ is the identity, that is iff $x=gxg^{-1}$ for all $x\in G$, that is iff $xg=gx$ for all $x\in G$, that is iff $g\in Z(G)$.