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Does $X^m,m \in Z$ form a group under multiplication? What is it's order?

I found this exercise and it doesn't specify what X is.I checked all the conditions and i found that it is a group and actually an abelian group. And i said that it has an infinite order.

Are my conclusions correct?

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    Can you prove that this group is in fact isomorphic to the additive $\mathbb{Z}$? – 2017-02-18
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    well yes since m is any integer. – 2017-02-18
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    Yes $X^m \mapsto m$ is an isomorphism, so your group is an infinite cyclic one and your conclusions are OK. – 2017-02-18
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    This really isn't a meaningful question without saying what $X$ is or what "$X^m$" means. – 2017-02-19
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    That's what i was thinking but it is in the intro section of the book so i assume $X$ is either an integer or a square matrix and $X^m$ is the power of $X$. – 2017-02-19
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    @figarozo No need to make such assumptions about $X$, it can be an arbitrary element of any form. Since the question is about multiplication, $X^m$ naturally means $X$ multiplied by itself $m$ times. That's enough to be able to answer the question (which you did). – 2017-02-20
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    oh ok thanks @Leppala – 2017-02-20

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