Question

If $\mathbf{G}$ is a group, prove that every subgroup of$\mathbf{Z}\left(G\right)$ 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 $\text{Z}\left(\mathrm{G}\right)$by the definition as:

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

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 $\mathrm{G}$. 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 $\mathrm{G}$.