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Created exponent of a group.
[On edit: Ah, I’m being dumb, see below!]
The note of concern over how to interpret “least” makes no difference, in that using either the ordinary ordering or the divisibility ordering will yield the same exponent, for the same reason that they will yield the same least common multiple of any naturals. (But perhaps the purpose of the note is precisely to note that there is no difference between these?)
(If the purpose of this note is in fact to quibble about the Right Way of looking at things, I would say the Right Way to think of orders of group elements in general is in terms of the induced ideal on the integers (the order of $g$ being the kernel of the mapping $n \mapsto g^n$); the exponent of group $G$ (i.e., the order of the generic element $id_G$ within the group $G^{|G|}$) then amounts to the intersection of the ideals corresponding to the orders of the elements of $G$. Of course, this ends up being just another way of talking about least common multiples of naturals, by the correspondence of integer ideals and natural numbers).
$0$ is maximal in the divisibility ordering, and minimal in the ordinary ordering.
Something about that note seems to bother you. I suggest you merely think of it as an internal justification for why I’m adopting the convention about $0$ as a possible exponent, and what the rule is, and why it’s reasonable.
Ah, you’re right about 0; I wasn’t thinking about that (even though the note explicitly pointed it out!). With that in mind, I sheepishly retract my entire previous post. :)
[I wasn’t actually bothered by the note; just trying to understand what its importance was. Which I now do, and agree with you for including! Sorry about the confusion.]
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