More on probabilities in groups
Jul. 9th, 2020 02:46 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo the past few weeks I've gotten sucked back into thinking about probabilities of equations being satisfied in groups after deciding to get in touch with Sean Eberhard. (You may notice I've added a bit to that page.) I know, I know, I should be thinking about integer complexity, but, y'know, sometimes something else calls.
By the way, a question I forgot to ask about on the original page is whether these sets are closed. That question's been added now.
Here are some things I've learned -- some of which has been added to my website page, some of which has not:
1. I asked what happens if we allow for compact groups instead of just finite groups. Well, for the case of [a,b], i.e. commuting probabilities, what happens is nothing! (Aside from the inclusion of 0.) Eberhard shows in this blog post that given any compact group, with commuting probability greater than 0, there's a finite group with the same commuting probability.
2. For ap, p an odd prime, the set is definitely not closed... at least not for finite groups. The point (p-1)/p is a limit point, but not in the set. However, it is in the set if we allow more general compact groups.
This raises the question of whether, for a given word, the set for compact groups might be equal to the closure of the set for finite groups. (Remember, in the case of [a,b], it's conjectured that the set for finite groups might be already closed, aside from 0.)
If the set for [a,b] is closed, it would have to be for a pretty crazy reason. Like, for [a,b], the limit point 1/2 is approached by 1/2 + 1/22n+1; the latter is realized by a group G which consists of the product of n copies of D8, but with their centers identified. But 1/2 is instead realized by S3. Yeah...
3. If you do want to consider compact groups, connected compact groups will get you nowhere. In a connected compact group, regardless of the word, you'll only ever get 0 or 1.
Unfortunately, that doesn't mean you can just pass to G modulo the connected component of the identity, and thereby only consider profinite groups; taking this quotient may change the probability. (E.g., consider if G is S1⋊C2, the generalized dihedral group over the circle, and your word is a²; the correct probability is 1/2, but if you mod out you'll get a probability of 1.)
Still, it does simplify things somewhat, as it does mean you only need consider each way of assigning a connected component to each variable, and seeing whether that gets you 0 or 1; you can then average that over the connected components.
4. A thing I didn't make clear last time -- I mentioned that the 5/8-gap argument for [a,b] is simple enough that it obviously applies to compact groups; I didn't mention that the same is obviously true of the 3/4-gap argument for a². (Unfortunately, the only 7/9-gap arguments I can find for a³ seem specific to finite groups.) But way more is true!
In their paper Groups with automorphisms inverting most elements, Liebeck and MacHale give a classification of finite groups with over half the elements being involutions; in particular, the probabilities you get this way are 1/2 + 1/(2n). So, that's our top ω for a², or ω+1 if you include 1/2 itself.
But what the authors don't note is that -- as far as I can tell -- their proofs work just as well (with only a little modification) for compact groups as well! Their classification goes through with only minimal modification, and you get the same probabilities! (Basically: In Type I*, A can now be infinite, although [A:CA(x)] must still be finite, and in the other types, Z can now be infinite. If we're concerned with counting involutions, then for the other types that means E can now be infinite.)
The most significant thing to note about the necessary modifications, FWIW, is that for Theorem 3.5, which is pretty key, you don't need to assume that H is finite index; rather, you can conclude it! If H is infinite index, you get a contradiction. Similarly with 3.6; you don't need to assume that the qi are finite, you can conclude it. Note that you'll need Zorn's Lemma to prove the existence of a maximal abelian subgroup.
Unfortunately, it's probably going to be really hard for anyone to get that into the literature...
5. If we're looking at top ω's, then for [a,b], this is known as well, plus a little more. In his paper What is the probability that two elements of a finite group commute?, David Rusin finds that, for finite groups, the top probabilities are 1/2 + 1/22n+1 (as mentioned above), then 1/2, then 7/16, then 11/27, then 2/5, then 25/64, then 3/8, then 5/14, then 11/32. So, that's the top ω+8.
Except... he doesn't actually include 5/14 (realized by D14); I added that in. This doesn't seem to be a fundamental mistake, just a minor omission in case 2; he failed to realize that setting p=7, n=2 yields 5/14 > 11/32; he claimed that p=7 always yields something less than 11/32, which is false.
Dunno if anyone's corrected that in the literature; doubt it though.
Rusin's methods seem specific to finite groups, but by Eberhard's work above we know that doesn't matter!
6. Eberhard's proof that the probabilities for [a,b] are reverse-well-ordered really doesn't seem to imply that the order type is ωω, even though it seems like it should. :-/ (The main problem seems to be that the mapping p↦q need not be injective.)
There's more I could say, but I think this is a good place to stop.
-Harry
By the way, a question I forgot to ask about on the original page is whether these sets are closed. That question's been added now.
Here are some things I've learned -- some of which has been added to my website page, some of which has not:
1. I asked what happens if we allow for compact groups instead of just finite groups. Well, for the case of [a,b], i.e. commuting probabilities, what happens is nothing! (Aside from the inclusion of 0.) Eberhard shows in this blog post that given any compact group, with commuting probability greater than 0, there's a finite group with the same commuting probability.
2. For ap, p an odd prime, the set is definitely not closed... at least not for finite groups. The point (p-1)/p is a limit point, but not in the set. However, it is in the set if we allow more general compact groups.
This raises the question of whether, for a given word, the set for compact groups might be equal to the closure of the set for finite groups. (Remember, in the case of [a,b], it's conjectured that the set for finite groups might be already closed, aside from 0.)
If the set for [a,b] is closed, it would have to be for a pretty crazy reason. Like, for [a,b], the limit point 1/2 is approached by 1/2 + 1/22n+1; the latter is realized by a group G which consists of the product of n copies of D8, but with their centers identified. But 1/2 is instead realized by S3. Yeah...
3. If you do want to consider compact groups, connected compact groups will get you nowhere. In a connected compact group, regardless of the word, you'll only ever get 0 or 1.
Unfortunately, that doesn't mean you can just pass to G modulo the connected component of the identity, and thereby only consider profinite groups; taking this quotient may change the probability. (E.g., consider if G is S1⋊C2, the generalized dihedral group over the circle, and your word is a²; the correct probability is 1/2, but if you mod out you'll get a probability of 1.)
Still, it does simplify things somewhat, as it does mean you only need consider each way of assigning a connected component to each variable, and seeing whether that gets you 0 or 1; you can then average that over the connected components.
4. A thing I didn't make clear last time -- I mentioned that the 5/8-gap argument for [a,b] is simple enough that it obviously applies to compact groups; I didn't mention that the same is obviously true of the 3/4-gap argument for a². (Unfortunately, the only 7/9-gap arguments I can find for a³ seem specific to finite groups.) But way more is true!
In their paper Groups with automorphisms inverting most elements, Liebeck and MacHale give a classification of finite groups with over half the elements being involutions; in particular, the probabilities you get this way are 1/2 + 1/(2n). So, that's our top ω for a², or ω+1 if you include 1/2 itself.
But what the authors don't note is that -- as far as I can tell -- their proofs work just as well (with only a little modification) for compact groups as well! Their classification goes through with only minimal modification, and you get the same probabilities! (Basically: In Type I*, A can now be infinite, although [A:CA(x)] must still be finite, and in the other types, Z can now be infinite. If we're concerned with counting involutions, then for the other types that means E can now be infinite.)
The most significant thing to note about the necessary modifications, FWIW, is that for Theorem 3.5, which is pretty key, you don't need to assume that H is finite index; rather, you can conclude it! If H is infinite index, you get a contradiction. Similarly with 3.6; you don't need to assume that the qi are finite, you can conclude it. Note that you'll need Zorn's Lemma to prove the existence of a maximal abelian subgroup.
Unfortunately, it's probably going to be really hard for anyone to get that into the literature...
5. If we're looking at top ω's, then for [a,b], this is known as well, plus a little more. In his paper What is the probability that two elements of a finite group commute?, David Rusin finds that, for finite groups, the top probabilities are 1/2 + 1/22n+1 (as mentioned above), then 1/2, then 7/16, then 11/27, then 2/5, then 25/64, then 3/8, then 5/14, then 11/32. So, that's the top ω+8.
Except... he doesn't actually include 5/14 (realized by D14); I added that in. This doesn't seem to be a fundamental mistake, just a minor omission in case 2; he failed to realize that setting p=7, n=2 yields 5/14 > 11/32; he claimed that p=7 always yields something less than 11/32, which is false.
Dunno if anyone's corrected that in the literature; doubt it though.
Rusin's methods seem specific to finite groups, but by Eberhard's work above we know that doesn't matter!
6. Eberhard's proof that the probabilities for [a,b] are reverse-well-ordered really doesn't seem to imply that the order type is ωω, even though it seems like it should. :-/ (The main problem seems to be that the mapping p↦q need not be injective.)
There's more I could say, but I think this is a good place to stop.
-Harry
