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

Question

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?

Thanks in advance

Answer 1

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 $i_{g} = \mathrm{Id}_{G}$.

Answer 2

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.