Question

Prove that for any group $\mathbf{G}$, the center$\mathbf{Z}\left(G\right)$ is a characteristic subgroup.

Short answer

It is proved that for any group $\mathrm{G}$, the center$\mathrm{Z}\left(\mathrm{G}\right)$ is a characteristic subgroup.

Step-by-step solution

Assume

Assume that $\mathrm{G}$be a group, we need to show that for every automorphism $\mathrm{f}$of $\mathrm{G}$, that is, $\mathrm{f}\left(\mathrm{Z}\left(\mathrm{G}\right)\right)\subset \mathrm{Z}\left(\mathrm{G}\right)$.

Prove

Consider that $\mathrm{g}$belongs to$\mathrm{Z}\left(\mathrm{G}\right)$ be an arbitrary element. We have to show that $\mathrm{f}\left(\mathrm{g}\right)\in \mathrm{Z}\left(\mathrm{G}\right)$.

Consider arbitrary element $\mathrm{h}\in \mathrm{G}$, as$\mathrm{f}$ is an automorphism of $\mathrm{G}$, there exists${\mathrm{h}}^{\text{'}}\in \mathrm{G}$ such that $\mathrm{f}\left({\mathrm{h}}^{\text{'}}\right)=\mathrm{h}$.

This implies that:

$\begin{array}{rcl}\mathrm{hf}\left(\mathrm{g}\right)& =& \mathrm{f}\left({\mathrm{h}}^{\text{'}}\right)\mathrm{f}\left(\mathrm{g}\right)\\ & =& \mathrm{f}\left({\mathrm{h}}^{\text{'}}\mathrm{g}\right)\\ & =& \mathrm{f}\left({\mathrm{gh}}^{\text{'}}\right)\\ & =& \mathrm{f}\left(\mathrm{g}\right)\mathrm{f}\left({\mathrm{h}}^{\text{'}}\right)\\ & =& \mathrm{f}\left(\mathrm{g}\right)\mathrm{h}\end{array}$

For every$\mathrm{h}\in \mathrm{G}$ it is verified that $\mathrm{f}\left(\mathrm{h}\right)=\mathrm{Z}\left(\mathrm{G}\right)$. This implies that for any group $\mathrm{G}$, the center$\mathrm{Z}\left(\mathrm{G}\right)$ is a characteristic subgroup.

Hence, it is proved that for any group $\mathrm{G}$, the center$\mathrm{Z}\left(\mathrm{G}\right)$ is a characteristic subgroup.