Can we make groups with many-but-not-all elements of order p?

[sniffnoy]

So I've recently been wondering again about this question (yes, this was originally a post here, but I decided it deserved a page on my website). Just mostly idly wondering about it, not seriously studying it; I'm no group theorist, and I don't really have time to think about such a problem what with everything else that needs doing (both in terms of old math that needs getting back to or writing up, and non-math stuff).

Anyway. As mentioned in the linked page, the simplest place to start would be words of the from aⁿ, so we're counting all elements of order dividing n. Or really the simplest place to start would be words of the form a^p, for p prime, so you're just counting elements of order p (together with the identity). (Or maybe prime power order if you want to be a little more general.)

Also as mentioned, for p=2, it's known that there really is a gap between ¾ and 1, with ¾ being achievable as D₈; I'm using the convention here that the dihedral group with 2n elements is D_2n, not D_n.

What I'm wondering is, what's the best known fraction (less than 1) for other p (or maybe other n)? Even if we can't prove a gap, it would still be nice to see how high we can find, to maybe get some idea of what the best might be (assuming that there really is a best like I'm hypothesizing).

Like, sticking to primes first -- for p=2, the best is ¾, that's a theorem. What about for p=3? I can again get ¾ by looking at A₄; I don't know offhand how to do better than that. And for p=5, uh, no clue. Well, I mean, obviously one can get ½ for any n (take C_2n), but I'm considering that baseline -- can we do better?

Since for p=2 you get a good fraction by looking at D₈, i.e., C₄⋊C₂, you might think more generally to try looking at C_p²⋊C_p; however, this doesn't work so well when p≠2 -- you don't get lots of elements of order p like you do when p=2. Like so many other differences between 2 and the other primes, this one once again basically comes down to the fact that 2 is the only prime p for which p does not divide (p choose 2). (I really think this is an underappreciated point, it really does seem to underlie so many of the differences, but I've never seen anyone else point this out!)

If we allow more general n then of course for any even n we can get ¾ by looking at D_4n. But mostly I don't want to look at more general n because this quickly gets kind of silly. I guess powers of primes might be kind of noteworthy? In which case that's how to get ¾ for arbitrary powers of 2? But like let's say you look at powers of 3 -- well then A₄ still does the job, you don't need to change the example at all! This sort of thing is part of why I say more general n is kind of silly. I guess more generally this gets you anything divisible by 3 but not 2, so we've now covered all n divisible by either 2 or 3. And of course in the trivial case of n=1 the best you can do is the baseline ½. But this is kind of why I said more general n is kind of silly.

Anyway, I dunno, that's what I've got. Anyone know anything about this?

ADDENDUM: OK, doing some quick computations with GAP here's what I get for some small primes based on groups of size less than 256 (haven't attempted to determine what groups witness these; also man I really do not know GAP, using it is a pain):

p=2: 3/4 (as we know)
p=3: 7/9
p=5: 9/11
p=7: 7/8
p=11: 21/23
p=13: 1/2
p=17: 1/2
p=19: 1/2

(EDIT: See the comments for some updates here)

Surprising, if you ask me! My offhand suspicion is that those ½'s at the end are spurious, an artifact of me not using large enough groups, but who knows? And it's not at all clear what pattern might be going on in the first few...

FURTHER ADDENDUM: Wow, after plotting the results from above for p=2 on a graph and doing the same for p=3... yeah, I'd totally bet on well-ordering, for this simple case at least. But I don't think I can go further and actually attempt to discern the order type from the graph, as I'd like to.

(Like I said I have no intent of trying to prove this. Maybe I will run some more numerical experiments to gather more empirical evidence, but that's it.)

YET MORE: You know what, just see the comments for more regarding the numbers.

-Harry

Comments

[(Anonymous)]

I can make a group with probability (p^2-p+1)/p^2, though it has more than p^{p+1} elements: We start with an auxilliary construction: Let G be the group of (p+1)x(p+1) upper triangular matrices with entries in F_p and 1's on the diagonal. If g is in this group, then the only potential nonzero entry of g^p is in the upper right, and that entry is \prod_{i=1}^p g_{i(i+1)}.

The probability of that product being 0 is (1-1/p)^p or about 1/e, so not so good. But consider the subgroup h where we now require that the p numbers g_{i(i+1)} form an arithmetic progression. (In other words, g_12 - g_23 = g_23 - g_34 = ... = g_{(p-1)p} - g_{p(p+1)}.) The chance that a p term arithmetic progression has no zeroes is (p^2-p+1)/p^2, just like you want.

My guess is that your groups are the result of passing to some smaller subgroup of G which imposes the same condition on the g_{i(i+1)} and also imposes some other random condition on entries further into the matrix.

[(Anonymous)]

Here is a wide range of ideas that don't improve on 1-p^{-1}+p^{-2}. Let A be a finite F_p algebra with a two sided ideal m such that A/m = F_p. Then the set of elements in A of the form 1+m forms a p-group. This is nice to work with, because (1+x)^p = 1+x^p, so we can think about p-th powers in m. The group I gave above to realize 1-p^{-1}+p^{-2} is of this kind (using a subring of upper triangular matrices) and so are several other examples I found using the same bound. Unfortunately, I have learned from this question https://mathoverflow.net/questions/325378/ that a nonzero degree p homogenous polynomial on an F_p-vector space is nonzero with probability at least 1-p^{-1}+p^{-2} . So, if x^p is not identically zero, then it must nonzero with at least this probability. (The map x^p takes values in an F_p vector space, not in F_p, but just take one component of it to get the bound.)

I think, though, that I can use this idea to that the probability of g^p=1 can fill an interval between 0 and 1-p^{-1}+p^{-2}. (David S.)

[sniffnoy]

Interesting! So you think there is no gap? If so, that'd answer not just this narrow question of elements of order p, but the original broad question!

(I guess the obvious question here is, if indeed you can prove this way that there is no gap, what causes the proof to break down when p=2? Or for whatever small primes it breaks down for; like I said I plotted the p=3 case and it was pretty suggestive...)

[sniffnoy]

Oh, that's really nice! So that has order p^{(p choose 2)+2}, OK.

Meanwhile it turns out that it actually is doable with order p^p+1, Hunter and I found a way, I'll post it later once I've actually checked it more. :P

Edit: OK, posted in a new entry: https://sniffnoy.dreamwidth.org/548721.html

[sniffnoy]

Also I just noticed you flipped some things, but not in any ways that matter. Like, that 1-1/p+1/p² is the probability that the progression has a zero, not that it has no zeroes, but that's fine because that's what we want.