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Suppose $G$ is a group and let $g∈G$, explain why the order of $g$ is well-defined, while the definition of the order is the following:

The smallest positive r such that $g^r=e$, if no such r is found then we say g has infinite order.

My Question: What strategy should I adopt to check the well-definedness? I know we are essentially checking if the output is unique or not.

My Attempt: Suppose the order of g is finite then consider the set $\{r>0:g^r=e\}$. We know this set is non-empty since $g^k=e$ for some $k$. Then by Well-Ordering Principle there exists smallest such $r$ and hence the order is well-defined.

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  • $\begingroup$ Which definition of the order of an element do you have? $\endgroup$
    – Bernard
    Apr 24, 2019 at 14:00
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    $\begingroup$ "Well-defined" is usually used when there is a choice to be made. For example, when working with cosets the representatives represent a "choice". A question might be "prove that the order of an element of $G/N$ is well defined", and you prove that if $gN=hN$ and if $g^n\in N$ then $h^n\in N$. So, at least to me, this question makes no sense in its current form. $\endgroup$
    – user1729
    Apr 24, 2019 at 14:00
  • $\begingroup$ Usually the order of an element $g$ is defined as smallest positive integer $r$ with $g^r=e$ together with the clausule that $g$ has infinite order if no such integer exists. A check on well-definedness is not justified, since no choice is made. It is as if you are asked to prove that $\mathbb N$ is well-ordered. $\endgroup$
    – drhab
    Apr 24, 2019 at 14:06
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    $\begingroup$ @Bernard Wait, is it the case that I need to check if such function cannot assign two values at once? As in, Suppose the orders of g are $m$, $n$ respectively then $m-n$ qualifies for such definition, too? I really have no idea what I am checking to be fair. $\endgroup$
    – JustWandering
    Apr 24, 2019 at 14:13
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    $\begingroup$ If $m-n>0$, this means $m$ was not the smallest positive integer such that … Again, the smallest positive integer in a set of positive integers is unique, as the order on $\mathbf N$ is a total order.. $\endgroup$
    – Bernard
    Apr 24, 2019 at 14:54

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Your approach is good, but not correct. You cannt say “Suppose the order of $g$ is finite”, since this already assumes that the order exists. You can consider two cases:

  1. $(\forall k\in\mathbb N):g^k\neq e$: then, by definition, $\operatorname{ord}f=\infty$.
  2. $g^k=e$ for some natural $k$. Then, by the Well-Ordering Principle, there is a smallest such $k$, in which case $\operatorname{ord}g=k$.
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    $\begingroup$ It would be nice if you could address the comments to the question, which are asking why well-definedness is necessary here. So far as I understand it, it is not necessary and the question makes no sense... $\endgroup$
    – user1729
    Apr 24, 2019 at 15:31
  • $\begingroup$ Actually, I don't understand the spirit of the comments. Yes, it is easy to prove what the OP tried to prove just by mentioning the Well-Ordering Principle. Which the OP did, by the way. If all the rest was fine, I would not have answered; I would have just posted a comment saying that the proof was fine. But it was not fine and there was something to be proved, no matter how easy the proof is. $\endgroup$
    – José Carlos Santos
    Apr 24, 2019 at 17:29
  • $\begingroup$ Okay, I understand now. $\endgroup$
    – user1729
    Apr 29, 2019 at 8:27

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