Back to groups for a bit
Jan. 1st, 2022 02:56 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoySo I was talking to Julian yesterday and as a result I decided to go back and look at groups some more. I figured I'd focus on words of length 4 or less, as well as ak for k composite (since primes have already been looked at so much) and not divisible by any primes larger than 3 (since GAP is unlikely to give me the real maxima for such numbers).
For words of length 4 or less... well, it turns out there just aren't that many of them that aren't in some way trivial or reducible to a shorter one. Other than a4, which I'll cover below, there's aba-1b-1 (i.e. the commutator, already well-studied), and abab-1, and... that's it. You may say, wait, what about a²b², but nope, that's equivalent to the latter of these.
So, uh, as for abab-1, it also has a gap of 5/8, achieved by D8, and the proof is just an easy elaboration on the usual 5/8 proof. And then as mentioned a²b² is equivalent, although I didn't realize this at first and initially proved that it has a gap of 5/8 by a separate means.
OK, so what about ak for k small, composite, and 3-smooth? Here of course I didn't prove anything, but I decided to run the numbers with GAP and see what I'd get.
Let's start with the least surprising one; for k=6, I got an empirical gap of 8/9, achieved by a group that could be described as ((C9×C3)⋊C3)⋊C2. (Note: I didn't bother checking exactly how C2 acts on the rest.) Note that ((C9×C3)⋊C3) is the group that gets you the 7/9 gap for k=3. (Or at least, I assume it's that same group! I'll admit I didn't actually check that it's the same semidirect product.) So, 8/9 as the fraction is a bit surprising, but that group makes some sense... it's 3ish and it's 2ish, you know?
OK, but what about k=4? Here I *also* got an empirical gap of 8/9! Yes, that is a 9 in the denominator! The group that achieves this is a group that could be described as (C3×C3)⋊Q8 -- and make sure you use the right one of those, there are multiple groups matching that description and not all of them have this property! So, we got a factor of 3 seemingly out of nowhere, and not even any powers of 2 in the denominator...
This same group will also yield 8/9 for k=8, and for k any larger power of 2, as there are no elements of order 8.
So what about when I tried k=9? This got me an empirical gap of 17/19, which I really have to wonder if it's right, due to the involvement of a prime larger than 3, which as mentioned above I'm skeptical of. (But that just means I'm skeptical that this is the exact right number... not that it'll be something weird! If it were something non-weird, I would have expected to see that appear as better. :P ) Here, that fraction is achieved by C19⋊C9. Huh!
I guess it is worth noting that in both these cases, of k=8 and k=9, we have a semidirect product where the expected p-part is the quotient, and the normal subgroup is something else entirely. I would not have much confidence in such a pattern holding up, but it's worth paying some attention to I think...
So uh yeah. I guess this is part of why even for groups this is a hard problem that is unlikely to be solved anytime soon!
I'll see if I can quickly prove that 8/9 gap for k=4, but my guess is that if nobody else has so far (and I don't think anyone has), then I won't be able to either, as, uh, I'm not much of a group theorist...
-Harry
For words of length 4 or less... well, it turns out there just aren't that many of them that aren't in some way trivial or reducible to a shorter one. Other than a4, which I'll cover below, there's aba-1b-1 (i.e. the commutator, already well-studied), and abab-1, and... that's it. You may say, wait, what about a²b², but nope, that's equivalent to the latter of these.
So, uh, as for abab-1, it also has a gap of 5/8, achieved by D8, and the proof is just an easy elaboration on the usual 5/8 proof. And then as mentioned a²b² is equivalent, although I didn't realize this at first and initially proved that it has a gap of 5/8 by a separate means.
OK, so what about ak for k small, composite, and 3-smooth? Here of course I didn't prove anything, but I decided to run the numbers with GAP and see what I'd get.
Let's start with the least surprising one; for k=6, I got an empirical gap of 8/9, achieved by a group that could be described as ((C9×C3)⋊C3)⋊C2. (Note: I didn't bother checking exactly how C2 acts on the rest.) Note that ((C9×C3)⋊C3) is the group that gets you the 7/9 gap for k=3. (Or at least, I assume it's that same group! I'll admit I didn't actually check that it's the same semidirect product.) So, 8/9 as the fraction is a bit surprising, but that group makes some sense... it's 3ish and it's 2ish, you know?
OK, but what about k=4? Here I *also* got an empirical gap of 8/9! Yes, that is a 9 in the denominator! The group that achieves this is a group that could be described as (C3×C3)⋊Q8 -- and make sure you use the right one of those, there are multiple groups matching that description and not all of them have this property! So, we got a factor of 3 seemingly out of nowhere, and not even any powers of 2 in the denominator...
This same group will also yield 8/9 for k=8, and for k any larger power of 2, as there are no elements of order 8.
So what about when I tried k=9? This got me an empirical gap of 17/19, which I really have to wonder if it's right, due to the involvement of a prime larger than 3, which as mentioned above I'm skeptical of. (But that just means I'm skeptical that this is the exact right number... not that it'll be something weird! If it were something non-weird, I would have expected to see that appear as better. :P ) Here, that fraction is achieved by C19⋊C9. Huh!
I guess it is worth noting that in both these cases, of k=8 and k=9, we have a semidirect product where the expected p-part is the quotient, and the normal subgroup is something else entirely. I would not have much confidence in such a pattern holding up, but it's worth paying some attention to I think...
So uh yeah. I guess this is part of why even for groups this is a hard problem that is unlikely to be solved anytime soon!
I'll see if I can quickly prove that 8/9 gap for k=4, but my guess is that if nobody else has so far (and I don't think anyone has), then I won't be able to either, as, uh, I'm not much of a group theorist...
-Harry
