![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
sniffnoyFollowing up to this entry, I want to report: Yes, we can indeed construct a group of order pp+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 Fp with 1's on the diagonal such that the first superdiagonal is an arithmetic progression.
But, OK, how can you do it with only pp+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 (Cp²×Cpp-2)⋊Cp?
Let's say A = Cp²×Cpp-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 Mp=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 + ... + Mp-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 MT instead. Take the ring Z[ζp], where ζp is a primitive p'th root of unity. As an abelian group this has a free basis 1, ζp-1, (ζp-1)2, ..., (ζp-1)p-2. Let's reverse that, so (ζp-1)p-2 is the first basis element and 1 is the last. Now consider the additive endomorphism that is multiplication by (1-p)ζp. 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 vp(ζp-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[ζp], 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[ζp]/((ζp-1)p), and then our automorphism is just multiplication by ζp, 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 pp+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
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 Fp with 1's on the diagonal such that the first superdiagonal is an arithmetic progression.
But, OK, how can you do it with only pp+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 (Cp²×Cpp-2)⋊Cp?
Let's say A = Cp²×Cpp-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 Mp=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 + ... + Mp-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 MT instead. Take the ring Z[ζp], where ζp is a primitive p'th root of unity. As an abelian group this has a free basis 1, ζp-1, (ζp-1)2, ..., (ζp-1)p-2. Let's reverse that, so (ζp-1)p-2 is the first basis element and 1 is the last. Now consider the additive endomorphism that is multiplication by (1-p)ζp. 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 vp(ζp-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[ζp], 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[ζp]/((ζp-1)p), and then our automorphism is just multiplication by ζp, 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 pp+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
