Question

If $G$ is a finite abelian p-group such that $p\mathrm{G}=⟨0⟩$ prove that $\mathbf{G}\cong {\mathrm{\mathbb{Z}}}_p\oplus \cdots \oplus {\mathrm{\mathbb{Z}}}_p$for some finite number of copies of ${\mathrm{\mathbb{Z}}}_p$

Short answer

$G\cong {\mathrm{\mathbb{Z}}}_p\oplus \cdots \oplus {\mathrm{\mathbb{Z}}}_p$for some finite number of copies of${\mathrm{\mathbb{Z}}}_p$

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 $\mathrm{G}$be a finite abelian p-group. Then by the fundamental theorem of finite abelian

$\text{group}G={\mathrm{\mathbb{Z}}}_{p_1{1}_1}\oplus {\mathrm{\mathbb{Z}}}_{{p_2}^{s_2}}\oplus \cdots \oplus {\mathrm{\mathbb{Z}}}_{{p_n}^{s_n}}$

Given that $pG=⟨0⟩$then $p{\mathrm{\mathbb{Z}}}_{p_1^{a_1}}\oplus p{\mathrm{\mathbb{Z}}}_{{p_2}^{s_2}}\oplus \cdots \oplus p{\mathrm{\mathbb{Z}}}_{{p_n}^{s_n}}=\left\{0\right\}$

If $p_i\ne p$or $a_i>1$then order of generator $g_i$divides $p$that is $p{g}_i\ne 0$which is a contradiction.

Hence, each $p_i=p$implies $G={\mathrm{\mathbb{Z}}}_p\oplus \cdots \oplus {\mathrm{\mathbb{Z}}}_p$

Therefore, $G\cong {\mathrm{\mathbb{Z}}}_p\oplus \cdots \oplus {\mathrm{\mathbb{Z}}}_p$for some finite number of copies of${\mathrm{\mathbb{Z}}}_p$