# Maximal subgroups of almost simple groups with socle $PSL(2, q)$

Let $G$ be an almost simple group with socle $PSL(2,q)$ where $q=p^f>3$ is the $f$th power of some odd prime $p$, and $M$ a maximal subgroup of $G$. By http://arxiv.org/abs/math/0703685, except for some small cases, $M\cap PSL(2,q)$ is maximal in $PSL(2,q)$ whose subgroups are explicitly known. Denote by $M(G)$ the maximal subgroup of $G$ such that $M(G)\cap PSL(2,q)=D_{q+1}$.

1. How many conjugacy classes are there for $M(G)$ in $G$?

2. Does anyone know more about the structure of $M(G)$ as an abstract group than $M(G)$ is just an extension of $D_{q+1}$ by $Out(G)$? Expecially for $G=P\Sigma L(2,q)$ and $P\Gamma L(2,q)$ when $M(P\Sigma L(2,q))$ and $M(P\Gamma L(2,q))$ are shown to be $N_G(D_{q+1})$ and $N_G(D_{2(q+1)})$ respectively in http://arxiv.org/abs/math/0703685.

3. I'm working with the transitivity of the natural action of $M(G)$ on $PG(1,q)$. It is not hard to prove that $M(G)$ is transitive on $PG(1,q)$ for $G\geq PGL(2,q)$, and $M(PSL(2,q))$ is transitive on $PG(1,q)$ iff $q\equiv3\pmod{4}$. Based on computation results, I think that $M(P\Sigma L(2,q))$ is transitive on $PG(1,q)$ iff $p\equiv3\pmod{4}$. Then how to prove this and what about the remaining cases of $G$?

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## 1 Answer

Qn 1. There is a unique conjugacy class of subgroups of ${\rm PSL}_2(q)$ isomorphic to $D_{q+1}$, and $M(G)$ is the normaliser of this subgroup in $G$, so there must be a unique class of $M(G)$ in $G$.

Qn 2. Let $q=p^e$. For $G = {\rm P \Sigma L}_2(q)$, $M(G)$ is a metacyclic group with presentation

$\langle x,y \mid x^{(q+1)/2}=y^{2e}=1, y^{-1}xy=x^p \rangle$,

which intersects ${\rm PSL}_2(q)$ in the subgroup $\langle x,y^e \rangle$.

The easiest way to see this is to consider the image of the group in ${\rm P \Sigma L}_2(q^2)$, bearing in mind that ${\rm P \Sigma L}_2(q)$ is the quotient of a subgroup of ${\rm P \Sigma L}_2(q^2)$ by a central element of order 2.

Qn 3. Again let's consider $G = {\rm P \Sigma L}_2(q)$, and we may as well assume that $q \equiv 1 \mod 4$, since otherwise $M(G)$ is certainly transitive. The element $y \in M(G)$ (coming from the above presentation) either has two fixed points on ${\rm PG}_1(q)$, in which case $M(G)$ will have two orbits, or it interchanges two points from the two orbits of the subgroup $\langle x \rangle$, in which case $M(G)$ is transitive.

I haven't worked out every detail, but it is possible to figure out what is happening by considering the setwise stabilizer of two points in $G$. This has the presentation

$\langle x,y,z \mid x^{(q-1)/2}=y^e=z^2=1, y^{-1}xy=x^p, yz=zy, z^{-1}xz=x^{-1} \rangle$,

where $x$ and $y$ fix the two points, and $z$ interchanges them. I believe that when $p \equiv 1 \mod 4$, there are elements of the form $yx^i$ having order $2e$, and none of the form $zyx^i$, and the situation is reversed when $p \equiv 3 \mod 4$. This would confirm your conjecture that $M(G)$ is transitive if and only if $p \equiv 3 \mod 4$.

It is not hard to use the results for ${\rm PGL}_2(q)$ and ${\rm P \Sigma L}_2(q)$ to work out what happens for other subgroups $G$.

By the way, you might get a better response to research level questions if you ask them on the MathOverflow site.

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