Motivated by the work of Cohn, Umans and collaborators on bounding the complexity of matrix multiplication, a student and I have been investigating the following problem. For subsets $X$ and $Y$ of a group $G$, let $X \cdot Y = \{ xy \mid x \in X, y \in Y \}$.
We would like to find the size of the smallest group with three distinct Sylow $p$-subgroups $P_{1}, P_{2}$ and $P_{3}$ such that \begin{equation}\label{Syl} (P_{1}\cdot P_{2}) \cap P_{3} = \{1_{G}\} \,. \end{equation}
Any group having this property has at least $p+1$ Sylow $p$-subgroups, and so has order greater than $p^{2}$, while $\mathrm{PSL}_{2}(p)$ is an explicit example of a group with the required property. So $p^{2} < |G| < p^{3}$, we are interested in whether there exists a stronger bound of the form $|G|< p^{3-\epsilon}$ for some absolute $\epsilon > 0$ (not tending to $0$ with $p$).
We tried restricting to Frobenius groups $G = C_{q} \rtimes C_{p}$ where $q = kp+ 1$ is prime. Computational evidence suggests that for $p > 5$ such groups exist with $q < p^{2}$. Realising $G$ as a group of linear polynomials with coefficients in $\mathbb{F}_{q}$ we can reformulate the problem of estimating the intersection of point stabilisers $(G_{0} \cdot G_{1}) \cap G_{t}$ as counting the number of solutions of the polynomial $x^{k} + (t-1)y^{k} - t = 0$, for some $t \in \mathbb{F}_{q}$. The Hasse-Weil bound gives a lower bound for $q$ in terms of $p$, but these are weaker than we would like. If we assume that the fixed points of elements in $G_{0} \cdot G_{1}$ are distributed close to uniformly at random in $G$, then we have a heuristic argument that a Frobenius group satisfies the displayed equation only when $q = \Omega(p^{2}/\log(p))$.
Questions:
1)For a given prime $p$, what is the size of smallest Frobenius group $C_{q} \rtimes C_{p}$ satisfying the displayed equation?
2) For a given prime $p$, what is the size of the smallest group satisfying the displayed equation?
