# Map of an element in a group to the conjugation by g

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

Recall:

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?

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 $i_{g} = \mathrm{Id}_{G}$.
• 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. Apr 22, 2016 at 7:26
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.