Let $G \subseteq \mathbb F\_p^\*$ be a subgroup. Then $G$ is called almost trivial if $G \cap (2-G)$ consists of the element 1.


Then I am wondering how big $G$ can be in terms of $p$. If $G$ is a random set containing 1 then according to the birthday paradox one expects that $G$ and $2-G$ have nontrivial intersection as soon as $\#G \gg \sqrt p$.



> 
> I am wondering if maybe one could prove that all almost trivial subgroups of $\mathbb
>  F\_p^\*$ have size at most $\tilde O(\sqrt p)$.
> 
> 
> 


Or if a $\sqrt p$ up to log factors bound is out of reach, I was wondering if there is $1/2 < \epsilon < 1$ and some constant $c$ such that for all almost trivial groups we have $\#G < cp^\epsilon$.


I did some numeric experimentation by finding the largest almost trivial subgroup of $\mathbb F\_p^\*$ for all p < 80000. And the largest ratio that I could find for $\frac {\#G}{\sqrt p}$ was $\frac {884}{\sqrt {41549}} = 4.3368\ldots$ because $\mathbb F\_{41549}$ contains an almost trivial subgroup of order 884, so a square root (possibly up to log factors) bound seams reasonable.


P.S. The notion of being almost trivial is based on the notion of almost rational in the paper [Almost rational torsion points on semistable elliptic curves by
Frank Calegari](https://doi.org/10.1155/S1073792801000253).



Subgroups of the multiplicative group of a finite field satisfying a certain additive property
If we let $S$ be the set of characters of $\mathbb F\_p^\times$ trivial on $G$ then $$\sum\_{\chi \in S} \chi(g) = \begin{cases} \frac{p-1}{|G|} & g\in G \\ 0 & g\notin G \end{cases}$$


so $$\sum\_{\chi\_1,\chi\_2\in S} \sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g) = 0$$ if $G$ is almost trivial.


Now if $\chi\_1 =\chi\_2=1$ then $\sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g)=p-3$ and otherwise, by the bound for Jacobi sums, $\left| \sum\_{ g \in \mathbb F\_p \setminus \{0,1,2\} } \chi\_1(g) \chi\_2(2-g) \right| \leq \sqrt{p}+1 $.


Thus, if $G$ is almost trivial then $p-3 \leq ( ((p-1)/|G|)^2-1) (\sqrt{p}+1)$, or $p \leq c^2 p^{5/2} / |G|^2$ for a constant $c$, meaning $|G| \leq c p^{3/4}$, answering the weaker form of your question.


