-1

Fix some $n\in{\mathbb{Z}}$ with $n>1$. Find the torsion subgroup of $\mathbb{Z}\times\mathbb{Z}_n$. Show that the set of elements of infinite order together with the identity is not a subgroup.

I've seen a solution of this where $0\times\mathbb{Z}_n$ is the torsion subgroup, but there was no explanation. I don't understand where this comes from or what it means, and I'm not even sure what the group operation is.

mrose
  • 533
  • $0\times \mathbb Z_n$ is the set of elements of the form $(0,k)$ where $k\in\mathbb Z_n$. It might be clearer if you write it ${0}\times \mathbb Z_n$. – Cheerful Parsnip Sep 15 '17 at 04:53

1 Answers1

2

The group operation for a product of groups is just "do the operations for each group separately". So if $(a, x)$ and $(b, y)$ are in $\Bbb Z \times \Bbb Z_n$, then their product* is $(a + b, x + y)$.

An element $g \in G$ is "torsion" if $g \cdot g \cdots g = 1$ for some non-zero number of $g$s. If $G$ is abelian, then the set of all torsion elements forms a subgroup. This is called, naturally, the torsion subgroup.

So, knowing this, what is the torsion subgroup of $\Bbb Z \times \Bbb Z_n$? What elements, when repeatedly added to themselves, eventually make their way back to $(0, 0)$?

*Of course, since this is abelian, we'll usually call it a sum.

Henry Swanson
  • 12,972
  • Thanks. Any insight on why the elements of infinite order don't form a subgroup? – mrose Sep 15 '17 at 12:32
  • Can you find a way to combine two elements of infinite order and end up in the torsion subgroup? Hint: you only need to care about one component of the pair. – Henry Swanson Sep 15 '17 at 15:21