Question

If $\mathbf{G}$is a group, prove that every subgroup of$\mathbf{Z}\mathbf{\left(}\mathbf{G}\mathbf{\right)}$is normal in $\mathbf{G}$. [Compare with Exercise 14.]

Short answer

It is proved that every subgroup of$\mathrm{Z}\left(\mathrm{G}\right)$is normal subgroup of$\mathrm{G}.$

Step-by-step solution

Assume

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

Define $\mathrm{Z}\left(\mathrm{G}\right)$by the definition as:

$\mathrm{Z}\left(\mathrm{G}\right)=\{\mathrm{a}\in \mathrm{G}:\mathrm{ag}=\mathrm{ga},\forall \mathrm{g}\in \mathrm{G}\}$

Prove

Consider that$\mathrm{H}$be subgroup of $\mathrm{Z}\left(\mathrm{G}\right)$. For every element $\mathrm{h}\in \mathrm{H}$from the definition of $\mathrm{Z}\left(\mathrm{G}\right)$, we have:

$\mathrm{hg}=\mathrm{gh},\forall \mathrm{g}\in \mathrm{G}$

For all $\mathrm{h}\in \mathrm{H}$and $\mathrm{g}\in \mathrm{G}$, we have ${\mathrm{ghg}}^{-1}=\mathrm{h}$this implies that ${\mathrm{gHg}}^{-1}=\mathrm{H},\forall \mathrm{g}\in \mathrm{G}$.

This implies that $\mathrm{H}$is a normal subgroup of $G$. Therefore, every subgroup of $\mathrm{Z}\left(\mathrm{G}\right)$is normal subgroup of $\mathrm{G}$.

Hence, it is proved that every subgroup of $\mathrm{Z}\left(\mathrm{G}\right)$is normal subgroup of $\mathrm{G}$.