I want to prove Theorem 7(2) in section 2.3 from Abstract Algebra 3rd Edition by Dummit & Foote. This theorem states:
Let $H = \langle x \rangle$ be a cyclic group. If $|H| = \infty$, then for any distinct nonnegative integers $a$ and $b$, $\langle x^a \rangle \neq \langle x^b \rangle$. Furthermore, for every integer $m$, $\langle x^m \rangle = \langle x^{|m|} \rangle$, where $|m|$ denotes the absolute value of $|m|$, so that nontrivial subgroups of $H$ correspond bijectively with the integers $1, 2, 3, \ldots$.
This is my attempt at the proof, but I don't think it is rigorous enough. I feel like I am missing something.
Suppose $|H| = \infty$ and $\langle x^a \rangle = \langle x^b \rangle$ for distinct nonnegative integers $a$ and $b$. According to Prop. 2(2), this implies $x^a \neq x^b$. And $|H| = \infty$ implies that $x^a$ and $x^b$ are distinct, so $\langle x^a \rangle \neq \langle x^b \rangle$. We also have $\langle x^m \rangle = \langle x^{|m|} \rangle$, since for any $m \in \mathbb{Z}$, $\langle x^m \rangle = \langle x^{-m} \rangle$.
The Prop. 2(2) referenced in the proof is:
If $|H| = \infty$, then $x^n \neq 1$ for all $n \neq 0$ and $x^a \neq x^b$ for all $a \neq b$ in $\mathbb{Z}$.