# 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

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The answer to your last question: The element $(0,a)\in \mathbb{Z}\times\mathbb{Z}/n\mathbb{Z}$ has finite order, but is not the identity unless $a$ is. – Jason DeVito Sep 17 '12 at 19:44
Well this is embarrassing, I cannot believe I overlooked such a simple thing. Thanks guys. – KF Gauss Sep 18 '12 at 2:11
@KFGauss Just FYI, if one of the answers settles your question satisfactorily you are encouraged to "accept" the answer. To do this, click on the grey check-mark to the left of it, which will then turn green. This rewards the user who answered your question and shows other users that the question has been answered. – Alex Becker Sep 19 '12 at 21:49

## 2 Answers

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

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You need $n>2$ for this to work. – Brian M. Scott Sep 17 '12 at 19:46
I see that now, thank you. I have changed it. – Tarnation Sep 17 '12 at 19:51

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

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