Making a group with many but not all elements of order p

[sniffnoy]

Following up to this entry, I want to report: Yes, we can indeed construct a group of order p^p+1 where the fraction of elements of order p is 1-1/p+1/p². Thanks to Hunter and to Julian for figuring out an important part of the construction.

Although, first I should point out David Speyer's construction from the comments, which does achieve the desired fraction but uses a larger order to do so; it has order p^{(p choose 2) + 2}. The group consists of all (p+1)×(p+1) upper-triangular matrices over F_p with 1's on the diagonal such that the first superdiagonal is an arithmetic progression.

But, OK, how can you do it with only p^p+1 elements?

EDIT: Actually Julian came up with an even simpler description; skip to the end if you just want to see that.

Well, I was inspired by the p=3 case. In the p=3 case, there's a unique group of order 81 that achieves the fraction 7/9, and it has the form (C₉×C₃)⋊C₃. And of course for p=2 our group D₈ has the form C₄⋊C₂. So, I decided to try to generalize this. I wondered, can we do it as a group of the form (C_p²×C_p^p-2)⋊C_p?

Let's say A = C_p²×C_p^p-2 and G is our group. We need to specify some automorphism of A of order p. But that's not enough; in order to get it to work, we need that all the elements of G\A have order p, which is not something that happens automatically.

So, what do we actually need here? Let's consider A additively; then our automorphism is given by a matrix M. We can consider this as a matrix over Z, if we want to be lazy, or over Z/(p²), if we want to be slightly less lazy, but really it's a sort of mixed matrix; the first row consists of elements of Z/(p²), while the other rows can be taken to be over Z/(p). Moreover, in the first row, all the off-diagonal elements must be divisible by p. (And also the diagonal element in the first row had better not be divisible by p, but that's not exactly a tough restriction and also it's redundant with the additional requirement I'm about to tack on.)

Now, we said it has to be of order p, right? So we need M^p=1. Except as mentioned above, that's not strong enough; in order to get that all elements of G\A have order p, what we want is the stronger condition 1 + M + M² + ... + M^p-1 = 0. Except, what do I mean by these matrix equations? Equal mod what? Well, they need to be equal mod p² in the first row, and equal mod p in the rest.

So the problem remains to construct such a matrix! Whether considered as a mixed matrix like I described above or as a matrix over Z or whatever. This is the problem as I posed it to various people I know the other day. (Figuring I'd take it to MathOverflow if I didn't get an answer, but I did.)

Hunter came up with a way of doing it first, and his way is definitely more direct conceptually, but I'm going to go with Julian's construction because it's really nice and also easier to carry out; no need to go finding eigenvectors or extending things to free bases. Note that this construction (both Hunter's and Julian's, actually) only work if p>2, but of course we already know how to handle the p=2 case (you take the 1x1 matrix (-1) and get the group D₈).

Julian's construction is as follows: Let's construct the transpose M^T instead. Take the ring Z[ζ_p], where ζ_p is a primitive p'th root of unity. As an abelian group this has a free basis 1, ζ_p-1, (ζ_p-1)², ..., (ζ_p-1)^p-2. Let's reverse that, so (ζ_p-1)^p-2 is the first basis element and 1 is the last. Now consider the additive endomorphism that is multiplication by (1-p)ζ_p. Then the matrix of this, with respect to the basis I just gave, satisfies the transposed version of all the requirements. (It's a matrix over Z of course but you reduce it down.) So you transpose it and you get M.

I'll skip his proof, I don't think it should be too hard to rederive. :) (If you just use the fact that v_p(ζ_p-1) = 1/(p-1).)

So there you go! Take M as above, take the semidirect product, and you've got your group. (Julian also asked an interesting question: Since he found a way to construct M by means of Z[ζ_p], can the group G be naturally expressed in terms of this ring somehow?)

EDIT: Yes it can, at least somewhat. We can write A as Z[ζ_p]/((ζ_p-1)^p), and then our automorphism is just multiplication by ζ_p, and then we construct the semidirect product with that. No need to construct any fancy matrix, hooray! And unlike the other two constructions, this one still works when p=2. :)

Anyway, that question at least is answered: You can achieve 1-1/p+1/p², and you can do it with only p^p+1 elements. Unfortunately that was the easy half, of a watered-down version, of a special case, of my original question! But hey, like I said, I'm not really planning on tackling any of the hard versions, so I'll take it. :)

-Harry

Comments

[sniffnoy]

Also, Hunter's construction (again, without proof), since this should probably be written down too:

Let H(x) = x^p-1 - (p-1)x^p-2 + (p-1)²x^p-3 - ... + (p-1)^p-1. Let N be the companion matrix for H(x) (considered as an endomorphism of Z^p-1). Then N has 1 as an eigenvalue mod p, i.e. fixes some nonzero vector mod p; let v be a primitive lift of this to Z. Extend v to a free basis of Z^p-1. Take N in this basis. And then once again finally transpose.

Also Julian's construction is apparently equivalent to this he says, and he got it by adapting Hunter's construction? I should look at this more and try to understand this equivalence, I guess...

(Edit: Note this one also only works for p>2.)

[(Anonymous)]

Good to hear! This problem has also wormed its way into my head. Here is the most important thing I've learned.

Let G be a p-group where not all g have order p. The Hughes subgroup, H_p(G), is the subgroup generated by the elements not of order p. Clearly, the probability of g^p=1 is always at least 1-[G : H_p(G)]. For p=2 or 3, we always have [G:H_p(G)] = 1 or p, and Hughes conjectured that this would hold for all primes.

But Hughes conjecture was false! Wall constructed at 5-group where H_5 had index 25, and Havas and Vaughan-Lee achieved p^2 for p=7, 11, 13, 17 and 19. So these give cases where the probability of g^p=1 is at least 1-p^{-2}.

Havas and Vaughan-Lee also reduce the question of whether p^3 can be achieved to a finite computation for fixed p. For p=5, they show p^3 is NOT achievable, but p=7 was beyond 2009's computational power.

The MR review of Havas and Vaughan-Lee is pretty readable https://www.ams.org/mathscinet-getitem?mr=2531223

Best,
David Speyer

[sniffnoy]

Whoa! So it is possible to do better than 1-1/p+1/p² after all! Just not for p=2, and possibly not for p=3.

Still, the fact that p³ is not achievable as a Hughes index for p=5 is to me still suggestive of a gap still existing. Maybe doesn't have any simple formula though.

But, hey, I did only plot things for p=2 and p=3, so for all I know well-ordering (assuming it happens at all) could break down for higher p... sounds unlikely, but...

[(Anonymous)]

For p=3, the bound 1-1/3+1/3^2 = 7/9 is correct. See https://www.ams.org/mathscinet-getitem?mr=409643

[sniffnoy]

Oh nice! This seems to restrict to 3-groups, though, instead of considering it for all finite groups... still very suggestive obviously.

[(Anonymous)]

Oh, sorry, I should have also linked https://doi.org/10.1017/S0305004100052865 -- if G is not a p-group, the number of solutions to g^p=1 in G is at most p/(p+1). (Note that p/(p+1) is less that 1-1/p+1/p^2.) So 7/9=0.77777 is optimal for g^3 in 3-groups and 3/4=0.75 is optimal in non 3-groups. I think A_4 achieves the latter.

[sniffnoy]

Oh, nice. So we really can restrict to p-groups, then.

[sniffnoy]

There is I guess one remaining question though, regarding both this and the 3/4 bound for p=2: Can we generalize beyond finite groups to compact groups? The 5/8 bound for commuting pairs generalizes straightforwardly to compact groups, for instance. Maybe I should try actually reading these papers...

(It would certainly be quite something if well-ordering held for finite groups, but failed for compact groups! Well, OK, I guess that would be pretty believable, but still...)

Edit: I guess this quick proof of a weaker 7/8 bound for p=2 certainly applies to compact groups just as well. Haven't yet checked the actual proof of 3/4...

Edit again: Darn, the one I can find easily relies on finiteness. There could be others though.

[sniffnoy]

Oh wait, here's a simple argument for the p=2 case that doesn't rely on finiteness.

Suppose that the fraction of involutions is greater than 3/4. Fix an involution g. Then picking a random x, the probability that x is an involution is greater than 3/4, and the probability that gx is an involution is greater than 3/4, so the probability that both are involutions is greater than 1/2. But this means the fraction of elements commuting with g is greater than 1/2, so g is central. So all involutions are central; and that's more than half the elements, so the group is abelian. But now you have an abelian group with over half the elements being involutions, so all the elements are involutions. Done.