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

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 $$T_i$$s, 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 $$p_i$$s. 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 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 $$c_0 > 4$$.