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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?

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    $\begingroup$ Looks good. Minor point: in both cases you might want to say you assume $m<n$ so that $n-m>0$. $\endgroup$
    – rogerl
    Commented Jun 6, 2020 at 0:35
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    $\begingroup$ 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. $\endgroup$
    – Koro
    Commented Jun 6, 2020 at 0:36
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    $\begingroup$ @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? $\endgroup$
    – Koro
    Commented Jun 6, 2020 at 0:39
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    $\begingroup$ @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. $\endgroup$
    – rogerl
    Commented Jun 6, 2020 at 0:41
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    $\begingroup$ @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. $\endgroup$
    – Koro
    Commented Jun 6, 2020 at 0:44

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