# Find the torsion subgroup of $\mathbb{Z}\times (\mathbb{Z}/n\mathbb{Z})$

Let $G$ be an abelian group. $\left \{ g\in G||g|< \infty \right \}$ is a subgroup of $G$, called the torsion subgroup of $G$. Fix some $n\in \mathbb{Z}$ with $n>1$.

The question is to find the torsion subgroup of $\mathbb{Z}\times (\mathbb{Z}/n\mathbb{Z})$ and why?

Thank you.

Is that $0\times (\mathbb{Z}/n\mathbb{Z})$? But how to prove it?

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Take $(x,y)$ in your product. If $x\neq 0$, can $(x,y)$ be of finite order? And if $x=0$? – 1015 Feb 17 '13 at 22:55

Hint: the torsion group of $G$ is the group whose elements are those elements of $G$ which have finite order. They form a subgroup of $G$. The identity element is by definition, in this torsion group.

The torsion group may very well contain only the identity: the trivial subgroup, in that event.

Yes, the torsion subgroup of $\mathbb Z \times (\mathbb Z/n\mathbb Z)$ is $0 \times (\mathbb Z/n\mathbb Z)$.

And as Pete L. Clark suggests in his comment below: "To get a proof, just take it systematically: can you first show that every element that you've written down has finite order? That's almost obvious. Now write down any element other than yours and show that it does not have finite order."

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Is that $0\times (\mathbb{Z}/n\mathbb{Z})$? But I can't construct a proof. – i_a_n Feb 17 '13 at 22:58
A big +1 for giving a hint and not an answer to this question. – Pete L. Clark Feb 17 '13 at 22:58
@i_a_n: Yes, it is. To get a proof, just take it systematically: can you first show that every element that you've written down has finite order? That's almost obvious. Now write down any element other than yours and show that it does not have finite order. – Pete L. Clark Feb 17 '13 at 22:59