Question

If $\mathbf{G}$is a abelian group, prove that $\mathbf{G}\left(p\right)$ is a subgroup.

Short answer

It is proved that, $\mathrm{G}\left(\mathrm{p}\right)$ is a subgroup.

Step-by-step solution

Finite Abelian Group

Every finite abelian group G is the direct sum of cyclic groups, each of prime power orders.

Group G and a subgroup  H

Let G be an additive group and H be a subset of G.

Then, H is called a subgroup if $\mathrm{a},\mathrm{b}\in \mathrm{H}$ implies $\mathrm{a}+{\mathrm{b}}^{-1}\in \mathrm{H}$.

Given that G is an abelian group and $\mathrm{G}\left(\mathrm{p}\right)=\left\{\mathrm{g}\in \mathrm{G}:\left|\mathrm{g}\right|={\mathrm{p}}^{\mathrm{n}}\mathrm{for}\mathrm{some}\mathrm{n}\in \ge 0\right\}$

G(p) is a subgroup

$$\begin{gathered} \mathrm{Let},\mathrm{a},\mathrm{b}\in \mathrm{H}. \\ \mathrm{Then},\left|\mathrm{a}\right|={\mathrm{p}}^{\mathrm{n}},\left|\mathrm{b}\right|={\mathrm{p}}^{\mathrm{m}}\mathrm{and}\left|{\mathrm{b}}^{-1}\right|={\mathrm{p}}^{\mathrm{m}}\mathrm{for}\mathrm{some}\mathrm{n},\mathrm{m}\ge 0. \\ \mathrm{Take},{\mathrm{p}}^{\mathrm{m}+\mathrm{n}}\left(\mathrm{a}+{\mathrm{b}}^{-1}\right). \\ \mathrm{Then},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}{\mathrm{p}}^{\mathrm{m}+\mathrm{n}}\left(\mathrm{a}+{\mathrm{b}}^{-1}\right)\mathrm{as}: \\ {\mathrm{p}}^{\mathrm{m}+\mathrm{n}}\left(\mathrm{a}+{\mathrm{b}}^{-1}\right)={\mathrm{p}}^{\mathrm{m}+\mathrm{n}}\mathrm{a}+{\mathrm{p}}^{\mathrm{m}+\mathrm{n}}{\mathrm{b}}^{-1} \\ ={\mathrm{p}}^{\mathrm{m}}\left({\mathrm{p}}^{\mathrm{n}}\mathrm{a}\right)+{\mathrm{p}}^{\mathrm{n}}\left({\mathrm{p}}^{\mathrm{m}}{\mathrm{b}}^{-1}\right) \\ =0+0 \\ =0 \end{gathered}$$

Thus, the order of ${\left(\mathrm{a}+\mathrm{b}\right)}^{-1}$ divides ${\mathrm{p}}^{\mathrm{m}+\mathrm{n}}$ implies n must be the power of p.

Hence, $\mathrm{a}+{\mathrm{b}}^{-1}\in \mathrm{G}\left(\mathrm{p}\right)$ implies $\mathrm{G}\left(\mathrm{p}\right)$ is a subgroup of G.

Thus, $G\left(p\right)$ is a subgroup of G.