# 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 $$\operatorname{ord}g=k$$.