Question

If G is a finite abelian p-group such that $\mathbf{pG}\mathbf{=}<0>$, prove that $\mathbf{G}\mathbf{\cong }{\mathbf{Z}}_{\mathbf{p}}\mathbf{\oplus }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\oplus }{\mathbf{Z}}_{\mathbf{p}}$ for some finite number of copies of ${\mathbf{Z}}_{\mathbf{p}}$ .

Short answer

$\mathrm{G}\cong {\mathrm{Z}}_{\mathrm{p}}\oplus .....\oplus {\mathrm{Z}}_{\mathrm{p}}$for some finite number of copies of ${\mathrm{Z}}_{\mathrm{p}}$.

Step-by-step solution

Fundamental theorem of finite abelian group

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

Step 2: G≅Zp⊕...⊕Zp

Let G be a finite abelian p-group. Then by the fundamental theorem of finite abelian group $\mathrm{G}={\mathrm{Z}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\oplus {\mathrm{Z}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\oplus ...\oplus {\mathrm{Z}}_{{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}}$.

Given thatthen${\mathrm{pZ}}_{{{\mathrm{p}}_1}^{{\mathrm{a}}_1}}\oplus {\mathrm{pZ}}_{{{\mathrm{p}}_2}^{{\mathrm{a}}_2}}\oplus ...\oplus {\mathrm{pZ}}_{{{\mathrm{p}}_{\mathrm{n}}}^{{\mathrm{a}}_{\mathrm{n}}}}=\left\{0\right\}$.

If ${\mathrm{p}}_{\mathrm{i}}\ne \mathrm{p}$or ${\mathrm{a}}_{\mathrm{i}}>1$then order of generator localid="1655706951014" ${\mathrm{g}}_{\mathrm{i}}$divides p that is ${\mathrm{pg}}_{\mathrm{i}}\ne 0$which is a contradiction.

Hence, each ${\mathrm{p}}_{\mathrm{i}}=\mathrm{p}$ implies $\mathrm{G}={\mathrm{Z}}_{\mathrm{p}}\oplus ...\oplus {\mathrm{Z}}_{\mathrm{p}}$.

Therefore, $\mathrm{G}={\mathrm{Z}}_{\mathrm{p}}\oplus ...\oplus {\mathrm{Z}}_{\mathrm{p}}$ for some finite number of copies of ${\mathrm{Z}}_{\mathrm{p}}$.