Prove that if a subgroup $H$ generated by $x$ is of infinity order, then $H=\langle x^a\rangle$ iff $a = ±1$

Question

I am trying to prove a proposition in Dummit and Foote book which says:

Let $H = \langle x \rangle$. Assume $|x| = \infty$.Then $H = \langle x^a \rangle$ if and only if $=±1$.

My attempt was: If $=±1$, then obviously $H = \langle x^{-1} \rangle = \langle x \rangle$, since for $H$ to be a group it must also contain inverses of $x$.

Conversely, if $H = \langle x^a \rangle$, we have $|x| = \infty$ hence $|H| = \infty$, which implies $|x^a| = \infty$, from here I don't know how to prove that $=±1$. I know that $x^a \neq 1$ for all non zero $a$

Answer 1

Here's a hint: If $H = \langle x^a \rangle$, then we must have $x \in H$. Therefore, there is some $k \in \mathbb{Z}$ such that $x^{ak} = x$.

Can you finish it from here?