I was reading Dummit and Foote and they left this assertion as an exercise to the reader, but I'm not sure how I would go about proving it. I imagine it would be something of the form of, for distinct positive integers a,b: x^a != x^b, meaning that repeating the operation on x^a yields something different for every positive a. Is this a fair way to go about this? How would I rigorously express this?
(sorry for the poor notation, I don't know how to format it properly here)
If $x^a$ generates $H$ then in particular $x$ itself can be expressed as $(x^a)^b$ for some $b\in\mathbb{Z}$. So
$$x^1=(x^a)^b=x^{ab}$$
Now if you already know that $x^c\neq x^d$ if $c\neq d$ (which is a straight forward consequence of the fact that $|x|=\infty$) then the conclusion is:
$$1=ab$$
And that has only two solutions over integers, namely $a=1, b=1$ or $a=-1, b=-1$.