# Thread: Cartesian product and cyclic groups

1. ## Cartesian product and cyclic groups

A pressing question:

If we are given two cyclic groups, and we take a cartesian product of the two, is the result also a cyclic group?

2. $\mathbb{Z}_2 \times \mathbb{Z}_2$ is a group of order 4 where all it's elements have order 2

3. Excellent! Thank you very much

4. In fact, $\mathbb{Z}_n\oplus\mathbb{Z}_m$ is cyclic iff $(m,n)=1$. To think of why this is we need some element of $\mathbb{Z}\oplus\mathbb{Z}_m$ to be of exactly order $mn$ but it is clear the order of $\mathbb{Z}_n\oplus\mathbb{Z}_m$ is $\text{lcm}(m,n)$. Therefore lastly noting that $\text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $(m,n)=1$. And since all cyclic groups of order $\ell$ are isomorphic to $\mathbb{Z}_{\ell}$ this is the case for any cyclic groups.

5. Originally Posted by Drexel28
In fact, $\mathbb{Z}_n\oplus\mathbb{Z}_m$ is cyclic iff $(m,n)=1$. To think of why this is we need some element of $\mathbb{Z}\oplus\mathbb{Z}_m$ to be of exactly order $mn$ but it is clear the order of $\mathbb{Z}_n\oplus\mathbb{Z}_m$ is $\text{lcm}(m,n)$. Therefore lastly noting that $\text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $(m,n)=1$. And since all cyclic groups of order $\ell$ are isomorphic to $\mathbb{Z}_{\ell}$ this is the case for any cyclic groups.
thank you for further elucidating the problem. We had recently learned about this part: " $\text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $(m,n)=1$" So it was pretty clear from the previous post with Z2 x Z2, but your explanation was definitely something that went into my notes. Thank you!

6. Originally Posted by flabbergastedman
thank you for further elucidating the problem. We had recently learned about this part: " $\text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $(m,n)=1$" So it was pretty clear from the previous post with Z2 x Z2, but your explanation was definitely something that went into my notes. Thank you!