# Cartesian product and cyclic groups

• Nov 15th 2009, 05:02 PM
flabbergastedman
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?
• Nov 15th 2009, 05:32 PM
Jose27
$\displaystyle \mathbb{Z}_2 \times \mathbb{Z}_2$ is a group of order 4 where all it's elements have order 2
• Nov 15th 2009, 07:06 PM
flabbergastedman
Excellent! Thank you very much
• Nov 16th 2009, 08:03 AM
Drexel28
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 PM
flabbergastedman
Quote:

Originally Posted by Drexel28
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.

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 PM
Drexel28
Quote:

Originally Posted by flabbergastedman
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!

(Yes)