Question

Question: If $\mathit{G}$and ${\mathit{G}}_{\mathbf{i}}\left(1\le i\le n\right)$are abelian groups such that localid="1656570463331" $\mathit{G}\mathbf{=}{\mathit{G}}_{\mathbf{1}}\mathbf{\oplus }{\mathit{G}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\oplus }{\mathit{G}}_{\mathbf{n}}$, show that $\mathit{p}\mathit{G}\mathbf{=}\mathit{p}{\mathit{G}}_{\mathbf{1}}\mathbf{\oplus }\mathit{p}{\mathit{G}}_{\mathbf{2}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\oplus }\mathit{p}{\mathit{G}}_{\mathbf{n}}$ .

Short answer

It is proved that, $pG=p{G}_1\oplus ......\oplus p{G}_n$.

Step-by-step solution

Finite Abelian Group

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

Given that G and $G_i\left(1\le i\le n\right)$ are abelian group such that$G=G_1\oplus ......\oplus G_n$ .

Prove pG=pG1⊕......⊕pGn

The group $G=G_1\oplus ......\oplus G_n$ implieslocalid="1656571048958" $G_i\cap G_j=\left\{0\right\}$ for all $i\ne j$.

Since $p{G}_i$ is a subgroup of $G_i$, $p{G}_i\cap p{G}_j=\left\{0\right\}$ for all$i\ne j$.

For any$pg\in pG$, $g\in G$, which can be written as $g=g_1+g_2+......+g_n$.

Then, find $pg$as:

$$\begin{gathered} pg=p{g}_1+p{g}_2+.....+p{g}_n \\ \in p{G}_1+p{G}_2+.....+p{G}_n \end{gathered}$$$$\begin{gathered} pg=p{g}_1+p{g}_2+.....+p{g}_n \\ \in p{G}_1+p{G}_2+.....+p{G}_n \end{gathered}$$

Thus, the subgroup $pG$ can be written as$pG=p{G}_1\oplus ......\oplus p{G}_n$.

Therefore, if and $G_i\left(1\le i\le n\right)$ are abelian groups such that $G=G_1\oplus ......\oplus G_n$ then,$pG=p{G}_1\oplus ......\oplus p{G}_n$.