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.

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:


*

*$(\forall k\in\mathbb N):g^k\neq e$: then, by definition, $\operatorname{ord}f=\infty$.

*$g^k=e$ for some natural $k$. Then, by the Well-Ordering Principle, there is a smallest such $k$, in which case $\operatorname{ord}g=k$.

