Suppose that $G$ is a finite permutation group acting transitively and non-regularly on $\Omega$. Also suppose that each non-trivial element has at most two fixed points and $|\Omega| \ge 4$.

Let me note that for a group acting that way, if $1 \ne X \le G_{\alpha}$, then as $N_G(X)$ acts on the set of fixed points of $X$, because if $X^g = X$ then $X\le G_{\alpha^g}$, and as $X$ has at most two fixed point we have that $|N_G(X) \cap G_{\alpha} : N_G(X)| \le 2$ by the orbit-stabilizer theorem. Also if $p \in \pi(G_{\alpha})$ and $p$ is odd, then $G_{\alpha}$ contains a full Sylow $p$-subgroup of $G$, for if $P$ contains a Sylow $p$-subgroup of $G_{\alpha}$, then $$|G:G_{\alpha}| \equiv \mbox{number of fixed points on }\alpha^G \pmod{p}$$ and hence is not divisible by $p$.

These facts are used in the following arguments, for which I ask this question:

Let $G \cong PSL(2, q)$ with $q = r^m \ge 13$. Suppose $q \equiv 1 \pmod{4}$. Let $\alpha \in \Omega$ and set $H := G_{\alpha} \ne 1$. Suppose $r$ divides $|H|$. Then $H$ contains a Sylow $r$-subgroup of $G$ and hence a subgroup of index at most two of its normalizer. This means that $H$ contains a subgroup of order $\frac{q-1}{4}$. As $q \ge 7$ and as $q-1$ is divisible by $4$, we know that $q-1 \ge 8$. Hence $H$ contains a subgroup of order $(q-1)/2$ of $G$. Now by looking at the subgroup structure of $G$ we find that $H$ is the normalizer of a Sylow $r$-subgroup of $G$ whence $|\Omega| = q + 1$. Also if we let $\Omega$ denote the set of cosets of the normalizer of a Sylow $r$-subgroup in $G$, then $\Omega$ has $q+1$ elements and acts in the way described above.

How does the fact that $H$ contains a subgroup of index at most two of a normalizer of a Sylow $r$-subgroup implies that $H$ contains a subgroup order $(q-1)/4$? And further why does the fact that $H$ contains a subgroup of order $(q-1)/2$ implies it must be the normalizer of a Sylow $r$-subgroup, and why does this implies that $|\Omega| = q + 1$?

The subgroup structure of $PSL(2,q)$ I have posted here, but it could also be seen in this paper.

  • $\begingroup$ As it says, all of these things follow from the subgroup structure of ${\rm PSL}(2,q)$. Oddly enough, you haven't asked about the part that I do not find clear, which is the claim that $H$ contains a subgroup of order $(q-1)/2$. $\endgroup$ – Derek Holt Nov 4 '15 at 9:03
  • $\begingroup$ I do not see the connection. Can you say what subgroups are exactly involved? For the point you mentioned, here the fact that the normalizers of subgroups have at most index $2$ is involved, so it has something to to with that, but I am not 100% sure about that either, hoping it becomes clearer as my other questions get answered... $\endgroup$ – StefanH Nov 4 '15 at 9:41
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    $\begingroup$ The normalizer of a Sylow $r$-subgroup of $G$ has order $q(q-1)/2$, hence the fact that $H$ has a subgroup of order at least $q(q-1)/4$. There are now only two possibilities for $H$, $H$ is the full Sylow normalizer of order $q(q-1)/2$ or the unique subgroup of index 2, which has order $q(q-1)/4$. I think in the second case there would elements fixing more than two points, but I am surpised that no details are given in the proof. $\endgroup$ – Derek Holt Nov 4 '15 at 9:47
  • $\begingroup$ Thanks for your comment! There is not much more said in the original source, if you want you can take a look there, its Lemma 3.11 conway1.mathematik.uni-halle.de/~waldecker/… $\endgroup$ – StefanH Nov 4 '15 at 10:09
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    $\begingroup$ I see! They are applying Lemma 2.8 to the subgroup of $G_\alpha$ of order $(q-1)/4$. Its normalizer in $G$ is the dihedral maximal subgroup of order $(q-1)$, so we get at least $(q-1)/2$ in $G_\alpha$. $\endgroup$ – Derek Holt Nov 4 '15 at 10:14

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