Group Action Approaches in Erdos Quotient Set Problem

Let ${\mathbb{F}}_q$ denote the finite field of $q$ elements. For $E\subset {\mathbb{F}}_q^d$, denote the distance set $\mathrm{\Delta }(E)=\{\mathrm{\parallel }x-y{\mathrm{\parallel }}^2:=(x_1-y_1{)}^2+\cdots +(x_d-y_d{)}^2:(x,y)\in E^2\}$. The Erdos quotient set problem was introduced in \cite{Iosevich_2019} where it was shown that for even $d\ge 2$ that if $\mathrm{\mid }E\mathrm{\mid }\subset {\mathbb{F}}_q^2$ such that $\mathrm{\mid }E\mathrm{\mid }>>q^{d\mathrm{/}2}$, then $\frac{\mathrm{\Delta }(E)}{\mathrm{\Delta }(E)}:=\{\frac{s}{t}:s,t\in \mathrm{\Delta }(E),t≠0\}={\mathbb{F}}_q^d$. The proof of the latter result is quite sophisticated and in \cite{pham2023group}, a simple proof using a group-action approach was obtained for the case of $q\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$ when $d=2$. In the $q\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$ setting, for each $r\in ({\mathbb{F}}_q{)}^2$, \cite{pham2023group} showed if $E\subset {\mathbb{F}}_q$, then $V(r):=\mathrm{\#}\{(a,b,c,d)\in E^2:\frac{\mathrm{\parallel }a-b{\mathrm{\parallel }}^2}{\mathrm{\parallel }c-d{\mathrm{\parallel }}^2}=r\}>>\frac{\mathrm{\mid }E{\mathrm{\mid }}^4}{q}$. In this work we use group action techniques in the $q\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$ setting, for $d=2$ and improve the results of \cite{pham2023group} by removing the assumption on $r\in ({\mathbb{F}}_q{)}^2$. Specifically we show if $d=2$ and $q\equiv 3\mathrm{m}\mathrm{o}\mathrm{d}4$, then for each $r\in {\mathbb{F}}_q^{\ast }$,$V(r)\ge \frac{\mathrm{\mid }E{\mathrm{\mid }}^4}{2q}$if $\mathrm{\mid }E\mathrm{\mid }\ge \sqrt{2}q$ for all $r\in {\mathbb{F}}_q$. Finally, we improve the main result of \cite{bhowmik2023near} using our proof techniques from our quotient set results.