# 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 $0 \times (\mathbb Z/n\mathbb Z)$.
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