# 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[/itex] 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 $c_0 > 4$.