# If an element of a group $G$ has infinite order, prove that powers of that element are distinct

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?

• Looks good. Minor point: in both cases you might want to say you assume $m<n$ so that $n-m>0$. Commented Jun 6, 2020 at 0:35
• Yes. That's correct. For the second part, where you consider finite order, $e, x, x^2,...,x^{N-1}$ have to be distinct by definition of order of an element as you have correctly shown.
– Koro
Commented Jun 6, 2020 at 0:36
• @rogerl, that assumption is not required, I think. Because even if $n-m\lt 0$, we can multiply both sides by $a^{N}$. Am I right?
– Koro
Commented Jun 6, 2020 at 0:39
• @Koro Correct, it's not required. But if you don't make that assumption, then in the first part for example saying that $|x|\le n-m$ requires a little more care. It just makes the whole proof pedagogically simpler. Commented Jun 6, 2020 at 0:41
• @rogerl,thanks. I always wondered that. It confused me a lot in proving that every permutation( finite order) is either cyclic or union of disjoint cycles.
– Koro
Commented Jun 6, 2020 at 0:44