# If $P\unlhd G$ is semiregular and $G$ such that $|\mbox{fix}(g)| \le 2$ for $g \ne 1$. Each $g \ne 1$ has at most one fixed point in each $P$-orbit

Let $G$ be a finite group acting nonregular and transitive on $\Omega$ such that each nontrivial element has at most two fixed points and $|\Omega| \ge 4$.

Suppose that $N$ is a proper normal $2'$-subgroup of $G$, and that $\alpha \in \Omega$. Suppose that $p \in \pi(N)$ is such that $P := O_p(N) \ne 1$. Then every $x \in G_{\alpha}, x \ne 1$ acts fixed point freely on $P$. Denote by $\overline \Omega$ the set of $P$-orbits on $\Omega$ and $\overline G := G / P$. Suppose $|\overline \Omega| > 2$ and that $P$ acts semiregular (i.e. no element has fixed points).

Then $|\overline \Omega| \ge 4$. Further if $u^x = u$ for $x \in G_{\alpha}, u \in P$ both nontrivial, then $u$ permutes the fixed points of $x$. As $u$ itself and each of its powers has no fixed points but odd order $p$, we would have $p > 2$ fixed point (the orbit of $u$ on some fixed point), which is not possible, hence the action is fixed point freely. Now suppose $\overline g \in \overline G$ is nontrivial and fixes $\omega^P$. As $\overline G_{\omega^P} = G_{\omega}P/P$ we may suppose that $g \in G_{\omega}$. As $P$ is semi-regular, it acts regular on each of its $P$-orbits. It follows that $g$ acts on $\omega$ in the same way as on $P$, hence with a unique fixed point. This means that $g$ has a fixed point in every $P$-orbit that it stabilizes.

I have some questions:

1) This means that $g$ has a fixed point in every $P$-orbit that it stabilizes.

Suppose $g \in G_{\omega}$ stabilizes another $P$-orbit $\alpha^P$, then $\overline g \in \overline G_{\overline \alpha} = G_{\alpha} P / P$, hence $g \in G_{\alpha}P$. But to apply a similar reasoning as for the $P$-orbit $\omega^P$ we would need $g \in G_{\alpha}$, but we just know $gu^{-1} \in G_{\alpha}$ for some $u \in P$. So how could they conclude that $g$ acts on $\alpha^P$ in the same way as on $P$?

2) Why $|\overline \Omega| = 3$ is not possible?

I have no idea how to see that.

• Since $g$ fixes $\omega$, it acts fixed point freely (by conjugation) on $P$, so it has order coprime to $p$. So, since the orbits of $P$ have order a power of $p$, if $g$ fixes such an orbit in $P$ then it must fix a point in that orbit. I don't think you have provided enough information to deduce $|\overline{\Omega}| \ne 3$ - I can think of examples in which it is. Do you know that $|\Omega|$ is even for example? – Derek Holt Dec 6 '15 at 21:55
• Thanks for your comment, but I do not understand that: "So, since the orbits of $P$ have order a power of $p$, if $g$ fixes such an orbit in $P$ then it must fix a point in that orbit" - I can think of examples where $g$ acts on a set of size coprime to its order (of prime size) but has no fixed points in that set, for example $g = (1 ~ 2 ~ 3)(4 ~ 5)$. Have I overlooked some assumption? I do not see where I have overlooked information, the question is related to Lemma 5.2 from this paper: conway1.mathematik.uni-halle.de/~waldecker/…, – StefanH Dec 7 '15 at 10:18
• Yes you have overlooked an assumption (in the first sentence of the post). – Derek Holt Dec 7 '15 at 11:05
• All cycles of $g$ apart from $\omega$ and at most one other fixed point must have the same length. – Derek Holt Dec 7 '15 at 11:54
• Isn't it clear now that the nontrivial orbit lengths of $g$ divide $|P|-1$? – Derek Holt Dec 7 '15 at 12:51