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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##

## Homework Equations

## 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##.