Let $p$ be a prime and $G$ a faithful non-regular transitive finite group acting on $\Omega$ with $|\Omega| > p$ such that some element fixes no point, and that each nontrivial element fixing some point fixes exactly $p$ points. Also assume that for $g \notin N_G(G_{\alpha})$ we have $$ G_{\alpha}^g \cap G_{\alpha} = 1 $$ and $p$ divides the order of $G_{\alpha}$. Let $\overline \alpha := \mbox{fix}(G_{\alpha})$ be the set of points fixed by $G_{\alpha}$ and assume that a Sylow $p$-subgroup $S$ of $G$ fixes $\overline \alpha$ and acts semi-regulary on $\Omega \setminus \overline \alpha$.

a) Show $N_G(S) \subseteq N_G(G_{\alpha})$ and $N_G(S)' \subseteq G_{\alpha}$,

b) $G$ has a normal subgroup $F$ of index $p$ [Hint: Grün's Theorem],

c) Show that $F$ has $p$ orbits and acts as a Frobenious group on these orbits.

Some facts I know:

i) $S$ has order at least $p^2$.

As $S$ acts semi-regulary on $\Omega \setminus \overline \alpha$ for each orbit $\Delta \subseteq \Omega \setminus \overline \alpha$ of $S$ we have $|\Delta| = |S : S_{\beta}| = |S|$ with $\beta \in \Delta$. Let $k$ denote the number of these orbits, and as $S$ fixes(1) $\overline \alpha$ $$ |\Omega| = |\overline \alpha| + k\cdot |S| = p + k \cdot |S| $$ and hence $p$ divides $|\Omega|$. As $G$ acts transitive $|\Omega| = |G : G_{\alpha}|$. So $p$ divides the index of $G_{\alpha}$ in $G$ and the order of $G_{\alpha}$, therefore $S$ has order at least $p^2$ as a Sylow $p$-subgroup.

ii) $G_{\alpha}$ has index $p$ in its normalizer.

Let $k = |N_G(G_{\alpha}) : G_{\alpha}|$. Then if $g \in N_G(G_{\alpha})$ we have $G_{\alpha} = G_{\alpha}^g = G_{\alpha^g}$, so that $G_{\alpha}$ fixes all points $\alpha^g$ with $g \in N_G(G_{\alpha})$, and the orbit of the normalizer has size $|N_G(G_{\alpha}) : G_{\alpha}|$. So at least $k$ points are fixed by $G_{\alpha}$. Further as $\alpha^g = \alpha$ implies $G_{\alpha} = G_{\alpha}^g$ exactly $k$ points are fixed by $G_{\alpha}$.

By $h \in N_G(G_{\alpha}^g) \Leftrightarrow (G_{\alpha}^g)^h = G_{\alpha}^g \Leftrightarrow G_{\alpha}^{ghg^{-1}} = G_{\alpha} \Leftrightarrow ghg^{-1} \in N_G(G_{\alpha}) \Leftrightarrow h \in N_G(G_{\alpha})^g$ the normalizers of conjugates to $G_{\alpha}$ are isomorphic, hence $|N_G(G_{\alpha}^g) : G_{\alpha}^g| = k$ and we see that each conjugate of $G_{\alpha}$ fixes also exactly $k$ points.

Let $g \in G$ and denote by $l$ the number of conjugates of $G_{\alpha}$ which contain $g$. By the above the number of points fixed by $g$ is precisely $kl$.

But as the conjugates intersect trivially, if $g \ne 1$ and $g$ fixes some point then $l = 1$, and so $g$ fixes exactly $k$ points, i.e. the number of points fixed equals the index, and by supposition the number of points equals $p$, hence $k = p$.

So this is all I got, hoping you can help me to solve the points a), b) and c)!

(1) I guess this means $\overline \alpha^S = \overline \alpha$, i.e. $\overline \alpha$ is a union of orbits under $S$, and not necessarily that $S$ fixes each point from $\overline \alpha$. But note that either $S$ fixes each point from $\overline \alpha$, or if one point is not fixed, then $\overline \alpha$ must be an orbit of $S$ as $|\overline \alpha| = p$ and the size of each orbit must divide $|S|$. But maybe the case that $S \le G_{\alpha}$ could be excluded somehow..

  • $\begingroup$ So $S$ has one orbit of length $p$, namely $\overline{\alpha}$ and the remainder have length $|S| \ge p^2$. Since $N_G(S)$ permutes the orbits of $S$, it must fix the single orbit $\overline{\alpha}$ of length $p$. But any element fixing $\overline{\alpha}$ normalizes $G_\alpha$, so $N_G(S) \le N_G(G_\alpha)$. Since any element fixing a point of $\overline{\alpha}$ fixies all such points, the induced action of $N_G(S)$ on $\overline{\alpha}$ is cyclic of order $p$ and hence $N_G(S)'$ fixes all poitns of $\overline{\alpha}$. That's a). Can you remind me what Grün's Theorem says? $\endgroup$ – Derek Holt Oct 28 '15 at 15:46
  • $\begingroup$ It is stated here: groupprops.subwiki.org/wiki/… $\endgroup$ – StefanH Oct 28 '15 at 18:29

I showed how to prove a) in the comments. Here is an outline proof of b) and c).

Grün's Theorem says that $G' \cap S = S_0 := \langle N_G(S)' \cap S, T' \cap S \mid T \in {\rm Syl}_p(G) \rangle$. We shall show that $S_0 \le G_\alpha$. We proved $N_G(S)' \le G_\alpha$ in a). Since $S$ acts semiregularly on $\Omega \setminus \{ \overline{\alpha} \}$, the elements in $S \setminus S_\alpha$ are fixed-point-free. Since $S' \le G_\alpha$, for any $T \in {\rm Syl}_p(G)$, the elements of of $T'$ are not fixed-point-free, and hence $S \cap T' \le G_\alpha$. Hence $G' \cap S = S_0 \le G_\alpha$. Now, since $|S:S_\alpha| = p$ and $G/G'$ is abelian, $G$ must have a normal subgroup $F$ of index $P$ with $F \cap S = S_\alpha$. That proves b).

Now, since $|G_\alpha/F_\alpha|$ is not divisible by $p$, but $|FG_\alpha/F|=1$ or $p$, we must have $FG_\alpha=F$, so $G_\alpha \le F$. For $\alpha \ne \beta \in \overline{\alpha}$, there is an element of $g \in S \setminus S_\alpha$ with $\alpha^g \in \beta$ and, since $G_\alpha \le F$, it follows that no element of $F$ can map $\alpha$ to $\beta$. So the $p$ points in $\overline{\alpha}$ lie in distinct orbits of $F$, and hence $F$ has $p$ orbits, with elements of $F_\alpha$ fixing one point in each orbit. So the action on each orbit is a Frobenius group.

  • $\begingroup$ Thanks for your detailed solution! But two points of your second paragraph I do not get, i) Why does $|S : S_{\alpha}| = p$ and $G/G'$ abelian implies the existence of $F$, and ii) I see that as the Sylow subgroups are conjugate, that $T'$ has to lie in some $G_{\alpha^g}$, but I do not see why it has to lie in $G_{\alpha}$? $\endgroup$ – StefanH Oct 30 '15 at 12:53
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    $\begingroup$ $|S:S_\alpha|=p$ from the Orbit-Stabilizer Theorem, since $|\alpha^S|=|\overline{\alpha}|=p$. Since $S \cap G' \le S_\alpha$, we have $S_\alpha \in {\rm Syl}_p(G'G_\alpha)$, so $|G:G'G_\alpha|$ is divisible by $p$, and hence (since $G/G'$ is abelian) $G'G_\alpha$ is contained in a subgroup $F$ of index $p$ in $G$. I didn't say that $T' \le G_\alpha$, I said $S \cap T' \le G_\alpha$ and I explained why! $\endgroup$ – Derek Holt Oct 30 '15 at 13:17
  • $\begingroup$ Okay, because we can find in $G/G'G_{\alpha}$ a subgroup of index $p$. You said the elements from $S\setminus S_{\alpha}$ are fixed-point-free, so as $T'$ is not we have $T'\cap S \le S_{\alpha} \le G_{\alpha}$. In your last paragraph you said that $|G_{\alpha}/F_{\alpha}|$ is not divisible by $p$, why that? I know that $|G:F_{\alpha}| = |G:G_{\alpha}||G_{\alpha}:F_{\alpha}|$, but I do not see that it might imply that the last factor might not be divisble by $p$? $\endgroup$ – StefanH Oct 30 '15 at 14:11
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    $\begingroup$ Because $F_\alpha$ contains $S_\alpha$, which is a Sylow $p$-subgroup of $G_\alpha$. $\endgroup$ – Derek Holt Oct 30 '15 at 14:18
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    $\begingroup$ Sorry! That was nonsense. I meant no element of $F$ can map $\alpha$ to $\beta$ (I have edited it now), because the set of elements of $G$ that map $\alpha$ to $\beta$ is the coset $G_\alpha g \le F g$, which is disjoint from $F$, since $g \not\in F$. $\endgroup$ – Derek Holt Oct 30 '15 at 15:00

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