1. ## product of groups

Prove that the product of two infinite cyclic groups is not inifinite cyclic.

could anyone help me to prove this ?? please?

2. Originally Posted by jin_nzzang
Prove that the product of two infinite cyclic groups is not inifinite cyclic.

could anyone help me to prove this ?? please?
An infinite cyclic group is isomorphic to a group of integers $\mathbb{Z}$ and the product of two infinite cyclic group is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. First, Z and Z \timex Z are not isomorphic. The latter is not generated by a single element. If it were generated by a single element, it has the cardinality of a set of natural numbers, but it is not.

3. Originally Posted by algtop
An infinite cyclic group is isomorphic to a group of integers $\mathbb{Z}$ and the product of two infinite cyclic group is isomorphic to $\mathbb{Z} \times \mathbb{Z}$. First, Z and Z \timex Z are not isomorphic. The latter is not generated by a single element. If it were generated by a single element, it has the cardinality of a set of natural numbers, but it is not.
But $\mathbb{Z} \times \mathbb{Z}$ does have the same cardinality as the set of natural integers!

Nevertheless, the fact that $\mathbb{Z} \times \mathbb{Z}$ can't be generated by a single element is true.

4. Originally Posted by Taluivren
But $\mathbb{Z} \times \mathbb{Z}$ does have the same cardinality as the set of natural integers!

Nevertheless, the fact that $\mathbb{Z} \times \mathbb{Z}$ can't be generated by a single element is true.
Right, a strange diagonal argument popped up in my mind when I saw this problem. I mixed that up with $\{0,1\}^\omega$.

A finite product of countable sets is countable.

For instance, there is an injective map from $\mathbb{Z}_+ \times \mathbb{Z}_+$ to $\mathbb{Z}_+$ such that $f:\mathbb{Z}_+ \times \mathbb{Z}_+ \rightarrow \mathbb{Z}_+$ by the equation $f(n,m) = 2^n3^m$. If a set has an injective map to $\mathbb{Z}_+$, it is countable.