Infinite cyclic group only has two generators

Homework Statement

Let ##H = \langle x \rangle##. Assume ##|x| = \infty##. Show that if ##H = \langle x^a \rangle## then ##a = \pm 1##

The Attempt at a Solution

Here is my attempt: Suppose that ##H = \langle x^a \rangle##. Then, for arbitrary ##b \in \mathbb{Z}##, ##x^b = (x^a)^m## for some ##m \in \mathbb{Z}##. Then ##x^{b-ma} = 1##, which is only true when ##b = ma##. But for the last equation ##b## is arbitrary, so there only exists an ##m## for arbitrary ##b## such that ##b = am## when ##a = \pm 1##. To the contrary, suppose that ##a \ne \pm 1## and ##b \ne a##; then we can choose ##b## to be relatively prime with ##a## so that there exists no ##m## s.t. ##b = am##.

Is this clear and rigorous? I feel that there could be a better way of showing that ##b = am \implies a = \pm 1##.

Answers and Replies

haruspex
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which is only true when b = ma.
That is not so in general. Why is it so in this case, though?

fresh_42
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Homework Statement

Let ##H = \langle x \rangle##. Assume ##|x| = \infty##. Show that if ##H = \langle x^a \rangle## then ##a = \pm 1##

The Attempt at a Solution

Here is my attempt: Suppose that ##H = \langle x^a \rangle##. Then, for arbitrary ##b \in \mathbb{Z}##, ##x^b = (x^a)^m## for some ##m \in \mathbb{Z}##. Then ##x^{b-ma} = 1##, which is only true when ##b = ma##. But for the last equation ##b## is arbitrary, so there only exists an ##m## for arbitrary ##b## such that ##b = am## when ##a = \pm 1##. To the contrary, suppose that ##a \ne \pm 1## and ##b \ne a##; then we can choose ##b## to be relatively prime with ##a## so that there exists no ##m## s.t. ##b = am##.

Is this clear and rigorous? I feel that there could be a better way of showing that ##b = am \implies a = \pm 1##.
No. That's a good example why I favor to write all dependencies explicitly. E.g. people write "##\forall \, \varepsilon \,\exists \,N ## such that ..." Wrong! It has to be "##\forall \, \varepsilon \,\exists \,N(\varepsilon) ## such that ..." because the choice of ##N## depends on the choice of ##\varepsilon \,.##

It is the same in what you wrote: ##\forall \,b\in \mathbb{Z} \,\exists \, m(b)\in \mathbb{Z} \ldots ## We thus have ##b=m(b)\cdot a##. So what? This is no surprise as it is exactly what we started with.

There is an easy way to check your proof, because ##(H,\cdot)=(\mathbb{Z},+)##. The exponents of the multiplicative ##x## are precisely the integers, so that is the isomorphism. Just assume ##a \neq \pm 1\,.## How could we ever reach points between ##1## and ##a\,?## Only if we had a relation, but ##x## doesn't have a relation, it is a free group.

Edit: Correction. Now I got your argument. ##a \,\mid \,b## for all ##b## and thus ##a## is a unit, hence ##a= \pm 1\,.## Sorry. It's a bit late here.

That is not so in general. Why is it so in this case, though?
It is true in this case because ##|x| = \infty##. If I add this in would the proof be fine? When I read it's a bit hard to follow (as evidenced by fresh_42's edit), so O just want to make sure there's not some clearer way that uses the same underlying logic.

haruspex
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It is true in this case because |x|=∞.
Yes, but that step bears a little more detail.

Mr Davis 97