# For a group acting such that each nontrivial element has at most two fixed point, size of orbit of single $2$-power order element

Let $G$ be a nonregular, transitive permutation group on $\Omega$ such that each nontrivial element has at most two fixed points. Suppose $S \in \mbox{Syl}_2(G)$ and that we have $\alpha, \beta \in \Omega$ distinct such that $S_{\alpha} = S_{\beta}$ has index $2$ in $S$ and every element in $S \setminus S_{\alpha}$ interchanges $\alpha$ and $\beta$ (this follows as $S$ normalizes $S_{\alpha}$). Now assume $\{\alpha,\beta\}$ is the unique $S$-orbit of length two, and all other $S$-orbits have length $|S|$.

Now take some $x \in S \setminus S_{\alpha}$ and look what it does on the $S$-orbits on $\Omega \setminus \{\alpha,\beta\}$.

Why does every orbit of $X := \langle x \rangle$ on $\Omega \setminus \{\alpha, \beta\}$ has the same size?

These orbits of $X$ (or $x$ for short) are cycles if we view $x$ as an element of the symmetric group on $\Omega$.

There must be one very short and easy way to see it; but I do not see it???

Suppose some $x^k$ has a fixed point, then as $x^{2k} \in S_{\alpha} = S_{\beta}$ (as $S / S_{\alpha}$ has order two) has this fixed point too by hypothesis we must have $x^{2k} = 1$. This shows $|X_{\omega}| \le 2$, and if $o(x) = 2^m$, then only $x^{2^{m-1}}$ is a candidate to fix some point. This is all I got, so the orbits have either size $|X|$ or $|X| / 2$, but I do not get that they all have equal size?

• Since all other $S$-orbits have length $|S|$, $S$ is acting regularly on each orbit, so all all orbits of $X$ have length $|X|$. – Derek Holt Dec 2 '15 at 22:57
• Oh yes, $X_{\omega} = S_{\omega} \cap X \le S_{\omega} = 1$, that is so simple; I somehow confused that with $C_{\Omega}(U) = \{\alpha \in \Omega : U \subseteq G_{\alpha} \}$ where $C_{\Omega}(U) \subseteq C_{\Omega}(V)$ for $V \le U$ yesterday. – StefanH Dec 3 '15 at 10:02