Question

Prove that for any group $\mathbf{G}$, the center $\mathbf{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}$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,$\mathrm{f}\left(\mathrm{Z}\right(\mathrm{G}\left)\right)\subset \mathrm{Z}\left(\mathrm{G}\right)$.

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:

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

For every $\mathrm{h}\in \mathrm{G}$it is verified that $\mathrm{f}\left(\mathrm{g}\right)\in \mathrm{Z}\left(\mathrm{G}\right)$. 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.