Are probabilities in groups well-ordered?

So, this is speculation without having done my homework first -- I haven't really tried to look at the existing literature here. If you know of any, please let me know! Because perhaps way more is known on this than I know.

Anyway, this is based on combining ideas from two papers: firstly, Commuting Probabilities of Finite Groups, by Sean Eberhard (thanks to Juan Arias de Reyna for pointing this out to me), and Word-Induced Measures on Compact Groups, by John Wiltshire-Gordon and Gene Kopp.

So, first, let's summarize the first of these papers. If you have a finite group, you can consider the probability that a randomly chosen ordered pair of elements from it commutes. You could then consider the set of all probabilities obtained this way; this is some set of rational numbers between 0 and 1.

In this paper, Eberhard shows that this set is in fact reverse well-ordered! Moreover, this later paper of Browning shows its (reverse) order type is ω^ω, and that it's closed in (0,1]. Here's my questions:

What if we allowed compact groups instead of just finite groups? Obviously this would allow 0 as well; would it allow anything else? Would it just be the closure of the finite case, say? (So the order type would be either ω^ω+1 or ω^ω²+1.) That would be nice, certainly! (Edit: Turns out Sean Eberhard has answered this one too! And aside from zero, it's the same as for finite groups.)
The big question: What if we tried using other words? That is to say: We pick a word from some free group F_n, and consider the probability that a randomly chosen n-tuple from G satisfies that word. Would we still get (reverse) well-ordering? What can be said about the order type and the closure? (And obviously we can try this one with compact groups too.)

It's this last question that's the most interesting to me. Say our word is just "a", in F₁. Then obviously for finite groups we just get ω, or ω+1 for compact groups. But what about other words in F₁? I.e., counting elements of order dividing some fixed k? That sounds like a good place to start before tackling the general problem. For the particular case of the word a², i.e. counting elements of order at most 2, it is at least known that there's a gap between ¾ and 1... but is that the start of a well-ordering? (For a³, it's similarly known that there's a gap between 7/9 and 1, by combining this and this; thanks to David Speyer for pointing this out to me.)

In particular it likely makes sense to focus on a^p for p prime, so you're just counting elements of order p together with the identity. As already mentioned there's at least a starting gap proven for p=2. And having run some numbers with p=2 and p=3... I gotta say they look reverse well-ordered to me. So I'd bet on this simple case being true, at least. (I had thought moreover that the largest probability less than 1 would be 1-1/p+1/p², as occurs with p=2 and p=3 as mentioned above, but this turns out not to be the case; thanks to David Speyer for pointing this out to me. Apparently it's an open problem whether a gap exists for a^p in general.))

Note that both for aba^-1b^-1 and for a², the proof of a gap is simple enough that it works for compact groups too, not just finite groups. Unfortunately the existing gap proof for a³ seems to rely heavily on finiteness. (There are proofs out there for a² that rely on finiteness, but you can do it without it.) (Note that a trivial case is the F₂ word "ab", which will again get you ω. More generally this will work for any word a₁...a_n∈F_n for n>0, where the a_i are the free generators. But that's kind of a trivial case. (I suppose an even more trivial case would be n=0, where you get 1 as your order type. But...))

(Another trivial variant of the problem is if you restrict to abelian groups, because then your probabilities will always be of the form 1/n (or 0 if you allow compact), since the satisfying tuples will form a subgroup of the Cartesian product. But that doesn't say very much...)

Also a fourth question worth asking: Are these sets closed? For aba^-1b^-1 this is now proven; could it be true more generally? Well, it actually can't if we restrict to finite groups; it's easy to see that a^p can't be closed for p an odd prime (since (p-1)/p is a limit point but not in the set). However, could it be true if we allow compact groups? Then that limit point appears. Not something I'm going to work on, though... (of course, maybe someone's already done it and I don't know!)

(Updates: See blog entries here and here for more on this, and this page for other algebraic structures.)

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