3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ I am not sure what you mean by "the latter". I think that $|H| = |M|, 2^s, 2^s \pm 1$ or $1$. $\endgroup$
    – Derek Holt
    Oct 21, 2015 at 22:23
  • $\begingroup$ @DerekHolt For example if $r=4$ and $s=2$ then for all $g \in G$ the subgroup $M^g \cap M$ is non-trivial. It seems true whenever $r=2s$ but I have not seen any proof of that. $\endgroup$
    – Amin
    Oct 22, 2015 at 11:12
  • $\begingroup$ You should edit the post to correct it. You have $2^r$ instead of $2^s$ in the possible orders of $|H|$, and I think you should say that an intersection of order $1$ is not possible when $r=2s$. I will think about the proof! Unfortunately we still have $|H|^2 < |G|$ in that case (although not by very much). $\endgroup$
    – Derek Holt
    Oct 22, 2015 at 15:52

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you very much. I think a similar argument shows that the converse is also true, i.e. $H \cap K = 1$ for some $g \in G$ iff $r \ne 2s$. $\endgroup$
    – Amin
    Oct 22, 2015 at 20:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.