Orbital maximin is bounded below by constant fraction of number of ordered pairs of distinct elements for solvable groups

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Statement

Let S be a set of size n, with n \ge 2. Consider the orbital maximin problem for the action of solvable groups on S, i.e., we want to find the maximum possible value of the smallest orbital size under a solvable group action on S. The total number of ordered pairs of distinct elements is n(n - 1). The claim is that the following equivalent statements are true:

  1. There is a constant c such that the orbital maximin is \ge n(n-1)/c for all n.
  2. There is a constant c_0 such that, for all sufficiently large n, the orbital maximin is \ge n(n-1)/c_0.

Proof

We prove (2).

The general construction: additive partitioning

Consider an expression for n as a sum of primes:

n = p_1 + p_2 + \dots + p_r

Consider now the group:

G := GA(1,p_1) \times GA(1,p_2) \times \dots GA(1,p_r)

where GA(1,p_i) is the general affine group (also called the affine general linear group) over the prime field with p_i elements. In other words, it is the semidirect product of the additive group of the field and the multiplicative group of the field. Note that each GA(1,p_i) is a metabelian group, hence so is G.

We consider now the following action of G on a set S of size n. First, divide S into subsets T_i of size p_i each. Identify each T_i with the underlying set of the field of p_i elements. Now, the action of an element of G on the piece T_i is the action of its i^{th} coordinate viewed as an element of the affine group acting on the field.

The orbits of elements are the sets T_i, and there are two kinds of orbitals:

  • The orbitals where both elements are in the same T_i: These have size p_i(p_i - 1) because the action of the general affine group is a doubly transitive group action.
  • The orbitals where the two elements are in different T_is, say T_i and T_j: These have size p_ip_j.

The minimum of all these turns out to be p(p - 1) where p is the minimum of the p_is. Therefore, if we can guarantee a partition into primes where each of the primes is \le n/K for some constant K, we will be done.

Result from additive number theory

The result now follows from Haselgrove's strengthening of Vinogradov's theorem, which states that any odd integer n can be divided into three primes that are (n/3) + o(n), and an easy corollary which shows that any even integer n can be divided into three primes that are n/4 + o(n). This would show that any constant $c_0 > 16</math> will work. Also, we see that our choice of G has polycyclic breadth at most eight.

If strong versions of Goldbach's conjecture are true, we can push it down to any constant c_0 > 4.