# Show that the order of an element g is well-defined

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.

• Which definition of the order of an element do you have? Apr 24 '19 at 14:00
• "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. Apr 24 '19 at 14:00
• 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. Apr 24 '19 at 14:06
• @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. Apr 24 '19 at 14:13
• 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.. Apr 24 '19 at 14:54

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 $$\operatorname{ord}g=k$$.
• 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... Apr 24 '19 at 15:31
• 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. Apr 24 '19 at 17:29
• Okay, I understand now. Apr 29 '19 at 8:27