# Another point of view that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is cyclic.

I was thinking that the product of groups $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is not cyclic, but $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is cyclic if p is an odd prime. Then I found logic that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is certainly cyclic if $m$ and $n$ are coprimes.

I searched on google a proof of that, to compare with mine and the answer is that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/mn\mathbb{Z}$ (which is clearly cyclic) $\iff$ $m$ and $n$ are coprimes.

But I got another proof and I want to know what you think about it :

I claim that $([1]_m,[1]_n)$ generates $\mathbb{Z}/m\mathbb{Z}$ $\times\mathbb{Z}/n\mathbb{Z}$, where $[a]_d$ is the canonical projection of $a\in \mathbb{Z}$ into $\mathbb{Z}/d\mathbb{Z}$.

Let's take multiples of $([1]_m,[1]_n)$ until we get $m \cdot$ $([1]_m,[1]_n) = ([0]_m,[m]_n)$ (where I suppose $m=\min(m,n)$... otherwise, consider $\mathbb{Z}/n\mathbb{Z}$ $\times\mathbb{Z}/m\mathbb{Z}$ and $\min(m,n)=n$).

The next step is $([1]_m,[m+1]_n)$. This pair is not equal to $([1]_m,[1]_n)$, since $m+1 \not\equiv1$ mod $n$ ($m$ and $n$ are supposed to be coprimes). We can now continue adding $([1]_m,[1]_n)$ to the results.

The last step is precisely reached when you got $([mn]_m,[mn]_n)=([0]_m,[0]_n)$. At this point, the fact that $m$ and $n$ are coprimes is absolutely necessary because if they are not coprimes, $lcm(m,n)\ne m\cdot n$ and $([lcm(m,n)]_m,[lcm(m,n)]_n)=([0]_m,[0]_n)$ but all the $m \cdot n$ possible pairs are not reached, only the $lcm(m,n)$ are !

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Edited after the comment of Andreas Blass – Alan Simonin Jan 7 '13 at 20:17
I think now my proof is complete – Alan Simonin Jan 9 '13 at 14:14

## 1 Answer

This argument begins reasonably but jumps to a conclusion at "The last step is reached precisely ..." Notice that your only use of the "coprime" hypothesis is where you said $m+1\not\equiv 1\pmod n$, and this uses only that $n$ doesn't divide $m$, which is less than the full strength of coprimeness. Try a pair of numbers like 4 and 6 that are not coprime but neither divides the other, to see where your proof needs more justification.

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Yes you are right, I'm trying to change what you say. Thanks for your opinion ! – Alan Simonin Jan 7 '13 at 20:02