Let $G$ be a group and $x \in G$ such that $x$ has infinite order. Prove carefully that the elements $x^n $ where $n \in \mathbb{Z}$ are all ${\bf distinct}$
Proof. (attempt)
We are given $|x| = \infty$. Now, say $x^n = x^m$ for some $n \neq m $ integers. So that $x^{n-m} = e$ but this means that $|x| \leq n - m $ which is a contradiction.
Is this correct?
Now, what if $|x| = N$? instead of $\infty$ as in problem above. I think we can use similar argument to argue that $e, x, x^2,..., x^{N-1}$ are distinct. For instance, say we have $m,n \in Z$ less than $N$ so obviously $n-m < N$. Therefore, if $x^n = x^m $, then $x^{n-m} = e$ but this is contradiction since $N$ is the least with $x^N = e$. QED
Is this correct?