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?

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- Nov 15th 2009, 05:02 PMflabbergastedmanCartesian 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? - Nov 15th 2009, 05:32 PMJose27
$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$ is a group of order 4 where all it's elements have order 2

- Nov 15th 2009, 07:06 PMflabbergastedman
Excellent! Thank you very much

- Nov 16th 2009, 08:03 AMDrexel28
In fact, $\displaystyle \mathbb{Z}_n\oplus\mathbb{Z}_m$ is cyclic iff $\displaystyle (m,n)=1$. To think of why this is we need some element of $\displaystyle \mathbb{Z}\oplus\mathbb{Z}_m$ to be of exactly order $\displaystyle mn$ but it is clear the order of $\displaystyle \mathbb{Z}_n\oplus\mathbb{Z}_m$ is $\displaystyle \text{lcm}(m,n)$. Therefore lastly noting that $\displaystyle \text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $\displaystyle (m,n)=1$. And since all cyclic groups of order $\displaystyle \ell$ are isomorphic to $\displaystyle \mathbb{Z}_{\ell}$ this is the case for any cyclic groups.

- Nov 16th 2009, 02:34 PMflabbergastedman
thank you for further elucidating the problem. We had recently learned about this part: "$\displaystyle \text{lcm}(m,n)=\frac{|mn|}{(m,n)}$ we conclude that $\displaystyle (m,n)=1$" 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!

- Nov 16th 2009, 02:47 PMDrexel28