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Let $m,n$ be positive integers bigger than $1$. Show that $\mathbb{Z}_m \times \mathbb Z_n$ is cyclic if and only if $\gcd(m,n)=1$.

I have no idea on how to start. Anyone hints are much helpful.

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closed as off-topic by user26857, C. Falcon, Daniel W. Farlow, Trevor Gunn, JMP Jun 23 '17 at 7:13

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Hint 1: Suppose $d = \gcd(m,n) > 1$. Then $k = \frac{mn}{d}$ is an integer (why?) and every element of $\mathbb{Z}_m \times \mathbb{Z}_n$ has order dividing $k$ (why?). Conclude that $\mathbb{Z}_m \times \mathbb{Z}_n$ cannot be cyclic in this case.

Hint 2: Suppose $d = \gcd(m,n) = 1$. Consider the element $(1,1) \in \mathbb{Z}_m \times \mathbb{Z}_n$ and compute its order. Conlcude that $\mathbb{Z}_m \times \mathbb{Z}_n$ is cyclic in this case.

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  • $\begingroup$ Can you tell me why every element of Zm x Zn has order dividing k? $\endgroup$ – user370967 Jun 21 '17 at 22:27
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    $\begingroup$ @Math_QED: It is an easy calculation $k (a,b) = (0,0)$ for $(a,b) \in \mathbb{Z}_m \times \mathbb{Z}_n$, since $k = m \cdot \frac{n}{d} = n \cdot \frac{m}{d}$, with each of $\frac{n}{d}$ and $\frac{m}{d}$ being integers. Then it follows that $\text{ord}(a,b) \mid k$. $\endgroup$ – Michael Joyce Jun 22 '17 at 1:25
  • $\begingroup$ Ah yes, I didn't look at it that way! Thanks a lot haha :) $\endgroup$ – user370967 Jun 22 '17 at 7:53
  • $\begingroup$ @Math_QED: Happy to help. $\endgroup$ – Michael Joyce Jun 22 '17 at 14:10

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