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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 $u$.

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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 $2m \mid n$.

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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 $\Omega$.

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