Properties of point stabilizers of $PSL(2, q)$

Let $G$ be a transitive, non-regular finite permutation group such that each non-trivial element fixing some point fixes exactly two points. Suppose that $G \cong PSL(2, q), q > 5$ and $H = G_{\alpha}$ for some point $\alpha \in \Omega$.

Assume $q = p^n$.

i) If $p \ne 2$ and $|H|$ is even, then $H$ contains an involution $u$ and $N_G(H)$ contains the centralizer of $u$.

ii) If $p = 2$ and $|H|$ is even, then $N_G(H)$ contains a Sylow $2$-subgroup $Q$ of order $2^n$.

How to see these facts?

Maybe they are related to the subgroups of $PSL(2, p^n)$, which are classified according to Dickson (given here as formulated in B. Huppert: Endliche Gruppen I):

(Dickson) The group $PSL(2, q)$ with $q = p^n$ has only the following subgroups:

(1) elementary abelian $p$-groups;

(2) cyclic groups of order $k$ for each divisor of $k$ of $q \pm 1$ if $q$ is even and each divisor $k$ of $\frac{q \pm 1}{2}$ if $q$ is odd;

(3) dihedral groups of order $2k$ for each divisor of $k$ of $(q \pm 1) / 2$ if $q$ is odd, and dihedral groups of order $2k$ for each divisor $k$ of $q \pm 1$ if $q$ is even (i.e. $k$ as in (2) above);

(4) alternating groups $A_4$ for $p > 2$ or $p = 2$ and $n \equiv 0 \pmod{2}$;

(5) symmetric groups $S_4$ if $p^{2n} - 1 \equiv 0 \pmod{16}$;

(6) alternating groups $A_5$ for $p = 5$ or $p^{2n} - 1 \equiv 0 \pmod{5}$;

(7) semidirect products of elementary abelian groups of order $p^m$ with cyclic groups of order $t$, where $t \mid p^m - 1$ and $t \mid p^n - 1$;

(8) groups $PSL(2, p^m)$ for $m \mid n$ and $PGL(2, p^m)$ for $2m \mid n$.

• What is $\Omega$? The natural set that $PSL(2,q)$ acts on is the $q+1$ points of a projective line, but here I believe it acts (sharply) triply transitively, so that is definitely not the action you are talking about. – Morgan Rodgers Oct 21 '15 at 11:23
• There is no restriction on the action other then the one said, i.e. about the number of fixed points. So the situation is we have an action of $G$ on some set $\Omega$ with the fixed point restriction, and we know that $G \cong PSL(2,q), q > 5$. That action is not necessarily the natural action of $PSL(2,q)$. – StefanH Oct 21 '15 at 11:34
• What source does this problem come from, and is it copied exactly as it appears? (I would guess that this can be proven without resorting to the classification of the subgroups). – Morgan Rodgers Oct 21 '15 at 12:57
• It is from a paper by O. Pretzel, A. Schleiermacher: On Permutation Groups in Which Non-Trivial Elements have $p$ Fixed Points or None. No I guess the classification might not be necessary, it is just an idea of me that it might play a role, so I added it. – StefanH Oct 21 '15 at 13:12
• I don't completely see how to prove, but I think it is based on the fact that if $g \in N_{G}(H)$, then $g$ permutes the orbits of $H$. I'm guessing that if $p \neq 2$ then you can show that the other fixed point of every element in $H$ is the same point $\alpha^{\prime}$ (which proves i) ). – Morgan Rodgers Oct 21 '15 at 13:38

Now after almost two months I guess I know the answer, and @Morgan Rodgers was right in his comment and his observation is the key. First I show an observation on nonregular groups acting such that every nontrivial element has at most one fixed point, and conclude from this that $H$ has two unique fixed points, which implies i).

Let $G$ act on $\Omega$ faithfully such that each nontrivial element fixes at most one point and there exists at least one that fixes some point, then consider its orbits on $\Omega$. There exists some orbit on which $G$ acts nonregular, and we show that this is the only on. For suppose we have another orbit on which it does, then it acts on both as a Frobenius group. Now the possible Frobenius complements (which are the point stabilizers in this case) form a unique conjugacy class, hence the point stabilizers on both orbits are conjugate, which shows that the elements in them fix at least two points. But this is not possible, hence on every other orbit $G$ acts regular. The result that the Frobenius complements form a unique conjugacy class (which implies that the Frobenius representation as its action on a subgroups is essentially unique) is quite deep, so I give a more elementary proof of the above in the end.

Okay, now let $G$ be a group acting on $\Omega$ faithfully and transitively on $\Omega$ such that each element fixing some point has exactly $2$ fixed points. Let $\alpha \in \Omega$ and consider $G_{\alpha}$, then this group acts on $\Omega \setminus \{\alpha\}$ as described in the previous paragraph, i.e. every nontrivial element has at most one fixed point, and some (in fact every) element must fix some other point. Hence we have a unique orbit $\Delta$ on which $G_{\alpha}$ acts non-regularly. If $\Delta$ has more than one point, then $G_{\alpha}$ acts as a Frobenius group on $\Delta$ and the elements of its kernel fix no point of $\Delta$, and also none of $\Omega \setminus \{\alpha\}$, hence these elements have just the single fixed point $\alpha$ which is not possible. Hence $\Delta = \{\beta\}$, and this shows that $G_{\alpha}$ has a unique other fixed point $\beta$ and $G_{\alpha} = G_{\beta}$.

Observe that for point stabilizers in general we have $$g \in N_G(G_{\alpha}) \Leftrightarrow \mbox{g permutes the fixed points of G_{\alpha}}.$$

Now for i). If $u \in H$ is an involution, then the elements from $C_G(u)$ permute the fixed points of $u$, but as written above these are exactly the fixed points of $H$, hence with the above observation we have $C_G(u) \le N_G(H)$.

Okay. Now for ii). As shown above, and as $N_G(G_{\alpha})$ acts on the two fixed points $\alpha,\beta$ of $G_{\alpha}$ we have $|N_G(G_{\alpha}) : G_{\alpha}| \le 2$ by orbit-stabilizer. Now by transitivity we have some $g \in G$ with $\alpha^g = \beta$. Then $G_{\beta^g} = G_{\beta}^g = G_{\alpha}^g = G_{\alpha^g} = G_{\beta}$, but this forces $\beta^g = \alpha$ as the elements from $G_{\alpha}$ could not fix any other $\beta^g \notin \{\alpha,\beta\}$ and $\beta^g = \beta$ would yield $\alpha^g = \alpha$. Hence $g$ interchanges $\alpha$ and $\beta$, so $g \in N_G(G_{\alpha})$ and this shows $|N_G(G_{\alpha}) : G_{\alpha}| = 2$.

We want to show that $N_G(H)$ with $H = G_{\alpha}$ contains a full Sylow $2$-subgroup. Now let $S$ be a Sylow $2$-subgroup with $S \cap H = S_{\alpha} \ne 1$, as $|H|$ is even we could find such a Sylow subgroup. Then $|S|$ is the highest $2$-power dividing $|G|$. If we could show that $|S : S_{\alpha}| \le 2$ then with the above it would follow that $N_G(H)$ is divided by the highest $2$-power in $|G|$, hence the claim.

Suppose that $|S : S_{\alpha}| = 2^m$ and assume $m \ge 2$. Set $\Delta = \alpha^S$. Then $S_{\alpha} = S_{\beta}$ and $S_{\alpha}$ acts semi-regularly on $\Delta \setminus \{\alpha, \beta\}$, hence by an orbit decomposition $$2^m = |\Delta| = 2 + k\cdot 2^l$$ for some $k$ and $|S_{\alpha}| = 2^l$. By assumption $2^m \ge 4$ and $n \ge 1$ and hence $2^m = 2\cdot (1 + k \cdot 2^{l-1})$, which implies that $k\cdot 2^{l-1}$ must be odd, i.e. $l = 1$. As $p = 2$ we have $SL(2, p^n) \cong PSL(2, p^n)$, and we know that the Sylow $2$-subgroups of $SL(2, p^n)$ are generalized quaterionen subgroups (see for example B. Huppert, Endliche Gruppen, page 196, Satz 8.10). As $|S_{\alpha}| = 2$ it contains the unique central involution, hence $C_S(S_{\alpha}) = S$ has order $\ge 8$. But $$|C_S(S_{\alpha}) : S_{\alpha}| \le |N_S(S_{\alpha}) : S_{\alpha}| \le 2$$ with similar arguments as above, which gives $C_S(S_{\alpha}) \le 4$, a contradiction. Hence we must have $m \le 1$, which gives the claim.

Note that just for ii) we need $G = PSL(2, q)$, i.e. the special structure of $G$. And also note that the action of $G$ in this case is not the natural action of the projective linear group on the set of projective points (or lines in this case), as this action has elements fixing just one point (namely the maps $x \mapsto x + a$ which just fix $\infty$), which is excluded by the assumptions about the action.

Appendix: Here I give a more elementary proof of the observations made in the 2nd paragraph, without using uniqueness results about Frobenius representations.

Suppose the situation is as written there, and $G$ acts on two orbits $\Delta, \Gamma$ nonregular. Then it acts on both as a Frobenius group, hence on $\Delta$ with $\alpha \in \Delta$ we have $$|G : G_{\alpha}|(|G_{\alpha}| - 1)$$ nontrivial elements fixing some point, and analogously on $\Gamma$ with $\beta \in \Gamma$ we have $|G : G_{\beta}|(|G_{\beta}| - 1)$ such elements. So we count the nontrivial elements fixing points in both orbits and find \begin{align*} |G : G_{\alpha}|(|G_{\alpha}| - 1) + |G : G_{\alpha}|(|G_{\alpha}| - 1) & = 2|G| - |G : G_{\alpha}| - |G:G_{\beta}| \\ & \ge 2|G| - |G|/2 - |G|/2 \\ & = |G| \end{align*} that we have more than $|G|$ of them, hence by pigeonhole we have some element of $G$ which fixes a point on both orbits, which is excluded by assumptions about the action of $G$ on $\Omega$.

• Nice work, sorry I never got back to you with a deeper explanation. – Morgan Rodgers Dec 18 '15 at 20:47
• Thank you! You are welcome, finally I found an explanation and now everything is clear to me, and your comment brought me on the right track. If you or anyone else has other, or shorter arguments, I would be glad to hear them :) – StefanH Dec 18 '15 at 20:57