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**jackie** Can someone give me a hand on this problem? I want to show that $\displaystyle G_1 \cong G_2 \Leftrightarrow G_1(p_i) \cong G_2(P_i)$ for all $\displaystyle i \in \{1,2,...,n\}$. Given that $\displaystyle G_1, G_2$ are abelian groups such that $\displaystyle ord(G_1)=ord(G_2)={p_1}^{k_1}{p_2}^{k_2}...{p_n}^{ k_n}$

Suppose $\displaystyle G_1 \cong G_2$. We have $\displaystyle G_1 \cong G_1(p_1) \times ...\times G_1(p_n)$ and $\displaystyle G_2 \cong G_2(p_1) \times ...\times G_1(p_n)$.

But we know each of $\displaystyle G_1(p_i)$ and $\displaystyle G_2(p_i)$ can be decomposed into a product of cyclic groups. So, $\displaystyle G_1(p_i) \cong Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}}$ where $\displaystyle Z_{x}$ is a cyclic group of order x.

Also, $\displaystyle G_2(p_i) \cong Z_{{p_i}^{j_1}} \times Z_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$

So, I try to show that $\displaystyle Z_{{p_i}^{l_1}} \times Z_{{p_i}^{l_2}} \times...\times Z_{{p_i}^{l_s}} \cong Z_{{p_i}^{j_1}} \times C_{{p_i}^{j_2}} \times...\times Z_{{p_i}^{j_s}}$

But I don't really know how to apply the hypothesis to prove this here.