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If $G$ is an abelian group of order $p_1p_2...p_k$ , where $p_1,p_2,...,p_k$ are distinct primes , then is it true that $G$ is cyclic ?

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Hint: Why does the group have an element of order $p_i$ for each $i$? Once you've figured this out, think about how to "combine" these elements to get a generator for the group.

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  • $\begingroup$ Short and simple! +1 $\endgroup$ – Swapnil Tripathi Jul 26 '14 at 13:12
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Yes, this is a direct consequence of the classification theorem for finitely generated abelian groups:

For every finite abelian group, there holds

$$ G \cong \mathbb{Z}_{n_1} \oplus \mathbb{Z}_{n_2} \oplus \cdots \oplus \mathbb{Z}_{n_k}$$ where $n_i$ divides $n_{i+1}$ Therefore, in your case, as all your multipliers are distinct primes, the only possibility is $n_1=p_1p_2\dots p_k$ and the rest zero.

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Yes, this is a particular case of the Chinese remainder theorem (where you consider the ring structure of $G$ as a $\mathbb{Z}$-algebra).

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