On the intersection spectrum of \({\text {PSL}}_2(q)\)

Given a group G and a subgroup \(H \le G\) $H\le G$ , a set \(\mathcal {F}\subset G\) $\mathcal{F}\subset G$ is called H-intersecting if for any \(g,g' \in \mathcal {F}\) $g,g^{\prime }\in \mathcal{F}$ , there exists \(xH \in G/H\) $xH\in G/H$ such that \(gxH=g'xH\) $gxH=g^{\prime }xH$ . The intersection density of the action of G on G/H by (left) multiplication is the rational number \(\rho (G,H)\) $\rho (G,H)$ , equal to the maximum ratio \(\frac{|\mathcal {F}|}{|H|}\) $\frac{|\mathcal{F}|}{|H|}$ , where \(\mathcal {F} \subset G\) $\mathcal{F}\subset G$ runs through all H-intersecting sets of G. The intersection spectrum of the group G is then defined to be the set \(\begin{aligned} \sigma (G) := \left\{ \rho (G,H) : H\le G \right\} . \end{aligned}\) $$\begin{array}{c}\hfill \sigma (G):=\left\{\rho ,(,G,,,H,),:,H,\le ,G\right\}.\end{array}$$ It was shown by Bardestani and Mallahi-Karai (J Algebraic Combin, 42(1):111–128, 2015) that if \(\sigma (G) = \{1\}\) $\sigma (G)=\{1\}$ , then G is necessarily solvable. The natural question that arises is, therefore, which rational numbers larger than 1 belong to \(\sigma (G)\) $\sigma (G)$ , whenever G is non-solvable. In this paper, we study the intersection spectrum of the linear group \({\text {PSL}}_2(q)\) ${\text{PSL}}_2(q)$ . It is shown that \(2 \in \sigma \left( {\text {PSL}}_2(q)\right) \) $2\in \sigma \left({\text{PSL}}_2,(q)\right)$ , for any prime power \(q\equiv 3 \pmod 4\) $q\equiv 3(mod4)$ . Moreover, when \(q\equiv 1 \pmod 4\) $q\equiv 1(mod4)$ , it is proved that \(\rho ({\text {PSL}}_2(q),H)=1\) $\rho ({\text{PSL}}_2(q),H)=1$ , for any odd index subgroup H (containing \({\mathbb {F}}_q\) ${\mathbb{F}}_q$ ) of the Borel subgroup (isomorphic to \({\mathbb {F}}_q\rtimes {\mathbb {Z}}_{\frac{q-1}{2}}\) ${\mathbb{F}}_q⋊{\mathbb{Z}}_{\frac{q-1}{2}}$ ) consisting of all upper triangular matrices.