Subgroups of $\mathbb Z \times(\mathbb Z/n\mathbb Z)$

So I am dealing with a problem from Dummit (specifically 2.1.7) and am having some issues. The part of the problem in question is:

Prove the set of elements of the direct product $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$ of infinite order together with the identity is not a subgroup of $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$.

I am assuming this question is referring to addition as the binary operation for both sets. I know the proof basically involves showing that the product is not closed; however, do not all the elements of the integers have infinite order (except the identity)? In that case, the elements of infinite order together with the identity would form all of $\mathbb{Z} \times (\mathbb{Z} / n \mathbb{Z})$ would it not?

Thanks

Consider the elements $(1,[ 1])$ and $(-1, [ 0])$; they each have infinite order but their sum has finite order and is not the identity.

HINT: $\langle 1,1\rangle$ and $\langle -1,0\rangle$ both have infinite order.