Intersection of maximal subgroups of PSL(2,q)

Question

Let $G := PSL(2,2^r)$, and let $M$ be a maximal subgroup of $G$ isomorphic to $PSL(2,2^s)$. I need to compute $H := M \cap M^g$ for $g \in G-M$. It seems to me that $|H|$ must be $2^r, 2^r\pm 1$ or $1$ (the latter may not hold for some values of $r$ and $s$). I think this is not difficult to prove, but the result sounds to be well-known. It would be appreciated if you introduce any references.

Answer 1

As a partial answer, here is a sketch of a counting argument to show that, for $r=2s$, two subgroups $H,K \cong {\rm PSL}(2,s)$ cannot intersect trivially. Suppose for a contradiction that $H \cap K=1$.

The total number of conjugates of $H$ in $G$ is $|G:H| = (2^r+1)2^s$. Of these, the $|H| = (2^r-1)2^s$ conjugates of $K$ under elements of $H$ are all distinct (otherwise some nontrivial element of $H$ would normalize $K$, but $K$ is self-normalizing), and they all intersect $H$ trivially.

Now $H$ has $2^{s-1}(2^s+1)$ cyclic subgroups of order $2^s-1$. Let $C$ be one of these. Then $C$ is centralized in $G$ by a cyclic subgroup of order $2^s+1$ that intersects $H$ trivially (since $C$ is self-centralizing in $H$). The $2^s$ conjugates of $H$ under the nontrivial elements of $D$ are all distinct, and their intersection with $H$ contains $C$. The proper subgroups of $H$ containing $C$ have orders $2^s-1$, $2(2^s-1)$ and $2^s(2^s-1)$. An intersection $H \cap H^d$ of order $2^s(2^s-1)$ is not possible (I'll leave the proof of that to you), so the intersections $H \cap H^d$ for $d \in D$ all contain the unique subgroup $C$ of order $2^s-1$.

Since there are $2^{s-1}(2^s+1)$ subgroups of $H$ of order $2^s-1$, there are at least $2^{2s-1}(2^s+1)$ conjugates of $H$ in $G$ that intersect $H$ in a subgroup of order $2^s-1$ or $2(2^s-1)$ and these, together with the $(2^r-1)2^s$ that intersect $H$ trivially, come to more than the total number of conjugates of $H$ in $G$, so we have a contradiction.