Let G be a group and suppose $g \in G$.

$\varphi:G\rightarrow Aut\left ( G \right )$

$g \mapsto i_{g}$

is a Homomorphism with image $Inn\left ( G \right )$ where $Inn\left ( G \right )=\left \{ i_{g}\mid g \in G \right \}$

The Kernel $Z\left ( G \right )=\left \{ x \in G \mid xg=gx, \forall x \in G \right \}$


Let G be a group and suppose $g, h \in G.$

The conjugation by g, $i_{g}$ is defined as

$i_{g}:G\rightarrow G$

$h\mapsto g^{-1}hg$

Could someone explain why $ker(\varphi)$ is the centre of a group G?

Thanks in advance


2 Answers 2


Suppose $i_{g}$ is the identity map on $G$, i.e. $g \in \ker(\varphi)$. Then $i_{g}(h) = g^{-1}hg = h$ for each $h \in G$, so $hg = gh$ for all $h \in G$. Hence, $g \in Z(G)$. On the other hand, suppose $g \in Z(G)$. Then $i_{g}(h) = g^{-1}hg = g^{-1}gh = h$, so $i_{g} = \mathrm{Id}_{G}$.

  • $\begingroup$ I have the idea. What is really confusing is notation; the sudden use of h from g. $\endgroup$ Apr 22, 2016 at 7:23
  • $\begingroup$ Let $i_{g}$ be the identity map The $ker\left ( \varphi \right )=\left \{ g \in G \mid \left ( g \right )\varphi=e=i_{g} \right \}$ $i_{g}\left ( g \right )=g^{-1}gg=g$ This implies $gg=gg$ Evidently, this is problematic. However, the switch of the g variables to the h variables is equally confusing. $\endgroup$ Apr 22, 2016 at 7:26
  • $\begingroup$ Figured it. Thanks $\endgroup$ Apr 22, 2016 at 7:38

Note that for any $g$, saying $gx=xg$ is equivalent to saying $gxg^{-1}=x$. So if $g$ is in the center then $gxg^{-1}=x$ for all $x\in G$, i.e. $i_g$ is the identity map.


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