1. ## cyclic group problem??

How to prove that the direct product of group Z is not cyclic?

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

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

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

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

5. 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 $\displaystyle \mathbb{Z}\times\mathbb{Z}$ is cyclic or not?

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

6. 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 $\displaystyle \mathbb{Z}\times\mathbb{Z}$ is cyclic or not?

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