# cyclic group problem??

• Sep 8th 2009, 11:19 PM
Mathventure
cyclic group problem??
How to prove that the direct product of group Z is not cyclic?
• Sep 9th 2009, 12:15 AM
Taluivren
Can $\mathbb{Z}\times\mathbb{Z}$ be generated by a single element? Suppose it can be generated by an element $(a,b)$ and arrive at contradiction.
Spoiler:

if $a=0$, you'll never get the element $(1,0)$, if $a \neq 0$, you'll never get the element $(a,b+1)$
• Sep 9th 2009, 04:21 AM
Mathventure
• Sep 9th 2009, 04:25 AM
Mathventure
Quote:

Originally Posted by Taluivren
Can $\mathbb{Z}\times\mathbb{Z}$ be generated by a single element? Suppose it can be generated by an element $(a,b)$ and arrive at contradiction.
Spoiler:

if $a=0$, you'll never get the element $(1,0)$, if $a \neq 0$, you'll never get the element $(a,b+1)$

• Sep 9th 2009, 04:38 AM
Taluivren
Hi, what exactly is not clear?

by definition, a cyclic group is a group that can be generated by a single element. As I understand, you ask whether $\mathbb{Z}\times\mathbb{Z}$ is cyclic or not?

If $\mathbb{Z}\times\mathbb{Z}$ were cyclic, then there would be an element $(a,b) \in \mathbb{Z}\times\mathbb{Z}$ such that every element $(c,d) \in \mathbb{Z}\times\mathbb{Z}$ can be written as $(c,d) = n(a,b) = (na,nb)$ for some integer $n$. The proof in the previous post shows that if you choose any $(a,b) \in \mathbb{Z}\times\mathbb{Z}$ then there exists an element from $\mathbb{Z}\times\mathbb{Z}$ such that no such $n$ exists. This contradicts the statement that $\mathbb{Z}\times\mathbb{Z}$ is cyclic.
• Sep 10th 2009, 05:19 AM
Mathventure
Quote:

Originally Posted by Taluivren
Hi, what exactly is not clear?

by definition, a cyclic group is a group that can be generated by a single element. As I understand, you ask whether $\mathbb{Z}\times\mathbb{Z}$ is cyclic or not?

If $\mathbb{Z}\times\mathbb{Z}$ were cyclic, then there would be an element $(a,b) \in \mathbb{Z}\times\mathbb{Z}$ such that every element $(c,d) \in \mathbb{Z}\times\mathbb{Z}$ can be written as $(c,d) = n(a,b) = (na,nb)$ for some integer $n$. The proof in the previous post shows that if you choose any $(a,b) \in \mathbb{Z}\times\mathbb{Z}$ then there exists an element from $\mathbb{Z}\times\mathbb{Z}$ such that no such $n$ exists. This contradicts the statement that $\mathbb{Z}\times\mathbb{Z}$ is cyclic.

thanks.........(Rofl)