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Suppose $G$ is a finite, non-regular, faithful, transitive group such that each nontrival element has at most two fixed points. Suppose $S$ is a Sylow-$2$-subgroup of $G$ and $\alpha \in \Omega$. If $|\Omega|$ is even and $G_{\alpha}$ has even order, we want to show that if $S$ is not dihedral or semidihedral, than $G$ has a normal subgroup of index $2$ that is a Frobenious group. By $\mbox{fix}_{\Gamma}(x) = \{ \gamma \in \Gamma : \gamma^x = \gamma \}$ we denote the fixed points of $x$ in $\Gamma \subseteq \Omega$.

Suppose that $|\Omega|$ is even and $G_{\alpha}$ has even order. Then there exists a Sylow $2$-subgroup $P$ of $G_{\alpha}$. Choose some Sylow $2$-subgroup $S$ of $G$ such that $P \le S$, then $P \le S \cap G_{\alpha} = S_{\alpha}$, hence $S_{\alpha} \ne 1$, and $S \nleq G_{\alpha}$ as $|\Omega| = |G : G_{\alpha}|$ is even, and so $S$ could not be a Sylow $2$-subgroup of $G_{\alpha}$.

[ Now the proof shows that $S$ is dihedral or semidihedral or that there exists some $\beta \in \Omega$ such that $\alpha \ne \beta, S_{\alpha} = S_{\beta}$ has index $2$ in $S$ and all elements in $S \setminus S_{\alpha}$ interchanges $\alpha$ and $\beta$. As $S_{\alpha}$ already has two fixed points, this subgroup must have regular orbits on the remaining points of $\Omega$. It follows that $\{\alpha,\beta\}$ is the unique $S$-orbit of length $2$ and all other orbits have length $|S|$. In particular $|S| \ge 4$ and together with the hypothesis about $G$ this implies $|\Omega| \ge 6$. Now the rest of the proof wants to show that $G$ has a normal subgroup of index $2$ that is a Frobenius group, and this is the part that follows. ]

Let $x \in S \setminus S_{\alpha}$. If we view $x$ as an element of the symmetric group on $\Omega$, then there are two possibilities: Either $x$ induces a single cycle on $\Omega \setminus \{ \alpha,\beta \}$ or an even number of cycles of $2$-power length. In the first case $x$ has order $|S|$ and therefore $S$ is cyclic. Then Burnside's $p$-Complement Theorem yields that $G$ has a normal $2$-complement and in particular $G$ has a subgroup $N$ of index $2$. In the second case it follows that, viewed as a permutation on $\Omega \setminus \{\alpha,\beta\}$, the element $x$ is an even permutation. But $x$ interchanges $\alpha$ und $\beta$ and therefore $x$ is an odd permutation in its action on $\Omega$. It follows that $G$ is not contained in the alternating group on $\Omega$, so again $G$ has a normal subgroup $N = A_{\Omega} \cap G$ of index $2$. We note that $G = N \cdot \langle x \rangle$ (in the first case because $N$ contains a complement to $\langle x \rangle$, in the second case by $x \notin N$ and $\langle x \rangle N / N$ must have index one in $G$), whence $G_{\alpha} \le N$. In particular $N$ does not act regularly and it has two orbits of equal size on $\Omega$ that are interchanged by $x$. We denote the $N$-orbit that contains $\alpha$ by $\Gamma$ and the orbits containing $\beta$ by $\Lambda$.

Suppose now that $y \in G_{\alpha}, y \ne 1$ is such that $|\mbox{fix}_{\Gamma}(y)| = 2$. Then as $y$ could have at most two fixed points we have $|\mbox{fix}_{\Lambda}(y)| = 0$. By considering the orbits of $\langle y \rangle$ we can choose $a,b \in \mathbb N_0$ such that $$ |\Gamma| = |\mbox{fix}_{\Lambda}(y)| + a \cdot o(y) = 2 + a\cdot o(y) $$ and $$ |\Gamma| = \Lambda| = b \cdot o(y). $$ Then we deduce that $|\Gamma| \equiv 0 \pmod{o(y)}$ and $|\Gamma| \equiv 2 \pmod{o(y)}$ and hence as $0 \equiv 2 \pmod{o(y)}$ that $o(y) = 2$. But then $\alpha$ and $\beta$ are the unique fixed points of $y$. This is impossible. We conclude that nontrivial elements of $N$ fix at most one point of $\Gamma$ and thus $N$ is a Frobenius group. $\square$

The following points in this proof I do not understand:

1) In the third paragraph: "Either $x$ induces a single cycle on $\Omega \setminus \{ \alpha,\beta \}$ or an even number of cycles of $2$-power length."

As $x$ has to move every element from $\Omega\setminus \{\alpha,\beta\}$ I see that we might get a single cycle, but why in the other case we must have an even number of cycles in a cycle decomposition? I see that they must have $2$-power length as their order must divide some power of $2$, but I do not see why there must be an even number of such cycles?

2) "In particular $N$ does not act regularly and it has two orbits of equal size on $\Omega$ that are interchanged by $x$."

Why we must have two orbits of equal size? I do not see that...

3) "[... ] that $o(y) = 2$. But then $\alpha$ and $\beta$ are the unique fixed points of $y$. This is impossible."

Why does order two implies that we could not have two fixed points?

I would be really glad if someone could answer my questions regarding the above proof!

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  • $\begingroup$ I don't understand 1) either. In fact I don't even understand why $S$ is cyclic when $x$ has a single orbit on $\Omega \setminus \{\alpha,\beta\}$. I suspect that you might have left out something important in the part of the proof that you have described in the second paragraph of your proof. As it stands, it is difficult to see how to use the fact that $S$ is neither dihedral nor semidihedral. $\endgroup$ – Derek Holt Oct 22 '15 at 17:46
  • $\begingroup$ Thanks for your feedback! Yes I forget something, namely that by the hypothesis about the fixed points, $S$ acts regular on $\Omega \setminus \{\alpha,\beta\}$ (I have edited the part to give these arguments in detail), and therefore $|S|$ equals the size of each such orbit, and as other fixed points are excluded and the only orbit of size two is $\{\alpha,\beta\}$ the other orbits have size $\ge 4$. This implies that if $x$ has a single orbit, $|S|$ equals the size of that orbit, and as $o(x)$ equals the size of the single orbit $S$ must be cyclic in this case. Is it more understandable now? $\endgroup$ – StefanH Oct 22 '15 at 18:00
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    $\begingroup$ I think the subdivision of cases in 1) is slgihtly wrong. The first case is when $x$ induces a single cycle on each of its orbits on $\Omega \setminus \{\alpha,\beta\}$. In that case $o(x)= |S|$, so $S$ is cyclic. Since the orbits of $S$ on $\Omega \setminus \{\alpha,\beta\}$ all have the same size, which must be a power of $2$, and the orbits of $x$ on $\Omega \setminus \{\alpha,\beta\}$ also all have the same size, in the other situation, $x$ must have an even number of orbits on each of the orbits of $S$ on $\Omega \setminus \{\alpha,\beta\}$. $\endgroup$ – Derek Holt Oct 22 '15 at 20:35
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    $\begingroup$ For 2), since $N$ has index $2$ in a transitive group, it must either be transitive itself or have two orbits of equal length. Since all elements of $S \setminus S_\alpha$ lie outside of $N$, we must have $S_\alpha \le N$. So $N$ cannot be transitive, or else its order wopuld be divisible by $2|S_\alpha|$ and it would contain a Sylow $2$-subgroup of $G$ which it clearly cannot. $\endgroup$ – Derek Holt Oct 22 '15 at 21:54
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    $\begingroup$ In the $2$-complement situation, $N'$ is the unique subgroup of index $2$ in $G$, so we must have $N=N'$. That's my very last comment! $\endgroup$ – Derek Holt Oct 23 '15 at 11:48

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