This is a rather specific post, but I hope nevertheless someone can help me. I am refering to a specific paper, namely

K. Mayaard, R. Waldecker, Transitive permutation groups where nontrivial elements have at most two fixed points

I have a special question on three statements and their interconnections. I will explain it in a way that hopefully it isn't necessary to look up the paper itself.

The statements are (I number them according to the paper):

Lemma 2.11 (b) Let $\omega \in \Omega$. Suppose that $S \in \mbox{Syl}_2(G)$ is such that $S_{\omega} \ne 1$. Then $S$ is dihedral or semidihedral and $|S_{\omega}| = 2$ or $G_{\omega}$ contains a subgroup of index at most $2$ of $S$. In the second case, if $S \nleq G_{\omega}$, then there exists $\delta \in \Omega$ such that $\omega \ne \delta$, $S_{\omega} = S_{\delta}$ and some element in $S$ interchanges $\omega$ and $\delta$.

Let me note that the last formulation in the second case means $|S : S \cap G_{\omega}| \le 2$. In the proof several "higher level" arguments enter, but basically it is just a distinction by cases, namely if we set $|S : S_{\omega}| = 2^m, |S_{\omega}| = 2^n$, then if $m \ge 2$ this implies $n = 1$, hence $|S_{\omega}| = 2$ which in turn implies that $S$ is semidihedral or dihedral. If $m = 1$ then the second case follows.

Let me note at this point, as I see the formulation (and proof) of this Lemma, it is not excluded in the second case that $S$ is dihedral or semidihedral, i.e. from a logical point of view to me it looks perfectly valid to have $S$ dihedral or semidihedral but $|S_{\omega}| > 2$ or $|S : S_{\omega}| \le 2$.

Now lets turn to a second lemma:

Lemma 2.12. Let $S \in \mbox{Syl}_2(G)$ and $\alpha \in \Omega$. Then one of the following holds:

(1) $G_{\alpha}$ has odd order.

(2) $S$ is dihedral or semidihedral and $|S_{\alpha}| = 2$.

(3) $|S| \ge 4$, there is a unique $S$-orbit on $\Omega$ of length $2$, and all other $S$-orbits have length $|S|$. In this case $G$ has a normal subgroup of index $2$ that is a Frobenius group.

(4) $|\Omega|$ is odd.

So now my question basically regard the interpretation of this Lemma (which arose in certain difficulties in following some arguments in it):

Interpretation (a): For an arbitrary but fixed $S \in \mbox{Syl}_2(G)$ and $\alpha \in \Omega$ we have one of the cases (1)-(4). [$\forall S \forall \alpha$]

Interpretation (b): For an arbitrary but fixed $S \in \mbox{Syl}_2(G)$ there exists some $\alpha \in \Omega$ such that one of the cases (1) - (4) holds. [$\forall S \exists \alpha$]

Interpretation (c): There exists some $S \in \mbox{Syl}_2(G)$ and some $\alpha \in \Omega$ such that one of the cases (1) - (4) hold. [$\exists S \exists \alpha$]

Of course, after a first reading interpretation (a) seems to be the natural one, but led me explain why this does not seem to make sense. First simply how the proof starts

Suppose that neither (1) nor (4) holds. Then with Sylow's Theorem we may suppose that $S_{\alpha} \ne 1$, but $S\nleq G_{\alpha}$.

The assumption $S_{\alpha} \ne 1$ seems questionable with respect to interpretation (a). If we look at some $G_{\alpha}$, then we know that there exists $S \in \mbox{Syl}_2(G)$ such that $S_{\alpha} \in \mbox{Syl}_2(G)$ (or the weaker property $S_{\alpha} \ne 1$) but not that a fixed $S \in \mbox{Syl}_2(G)$ must have nontrivial intersection with $G_{\alpha}$. This is strongly supported by later arguments, namely the proof of case (3) where it is indeed shown that $S_{\delta} = 1$ for $\delta\in \Omega\setminus\{\alpha,\beta\}$. So this too supports interpretaion (c).

The proof proceed to prove part (3). There is also a minor point that confused me, where I even wrote on MSE (see here), namely that $S$ acts semi-regular on $\Omega \setminus \{\alpha,\beta\}$. But this could be repaired by the implicit assumption that we are not in case (2) but only under the interpretaion (b) or (c).

Namely we know that $S_{\alpha}$ acts semi-regular on $\Omega\setminus\{\alpha,\beta\}$. Now suppose $S$ does it not and as $|S : S_{\alpha}| = 2$ this implies $|S_{\delta}| = 2$. Now with similar argument as in the proof of Lemma 2.11 (b) we could show that $S$ must be semidihedral or dihedral, so this would contradict that we are not in case (2) under interpretation (b) or (c), but not under interpretation (a). As it would totally valid for $S$ to be dihedral or semidiheral, but $|S_{\alpha}| \ne 2$, so this does not contradicts case (2) then.

Also at the end of the proof there is a line like "$y \in G_{\alpha}$ [..] $o(y) = $2 [...] then $y \in T$, but then $\alpha$ and $\beta$ are the unique fixed points."

Here is definitely a typo, this $T$ could be $S$, this would then correspond to the conclusion that $\alpha$ and $\beta$ are the unique fixed points. But by Sylow theorems as I see it, we just have $y \in S^g$ for some $g \in G$, set $T := S^g$ might be an intended interpretation. So lets stick to that. Then with $y \in G_{\alpha}$ we have $T_{\alpha} \ne 1$. But also as $S_{\alpha} = S_{\beta}$ both have index two in $S$, we have two subgroups in $T$ of index two that fix $\alpha^g, \beta^g$. We could apply Lemma 2.11 (b) to $T$ and $\alpha$, this gives that either $T$ is semidihedral or dihedral and $|T_{\alpha}|=2$, or the other case. I guess that the other case could be handled as the constructed normal complement is unique, but the case that $|T_{\alpha}|=2$ and $T$ is semidihedral or dihedral I just see that it is no problem if we use interpretation (b) or (c) of the lemma, but not under interpretation (a).

Okay, now suppose we abandon interpetation (a). But this is not conform with how it is used in Theorem 2.23. This Theorem aims at a classification of the groups in question by using Lemma 2.12. The arguments goes like this:

If we have a Sylow $2$-subgroup $S$ which is semidihedral or dihedral, then point stabilizers have twice odd order.

But this conlusion about the stabilizers having twice odd order is not valid under interpretation (b) or (c). As this allows $|S_{\alpha}| = 2$, but $S \cap G_{\alpha}$ need not be a Sylow $2$-subgroup in general, we could have that four divides $|G_{\alpha}|$. This conclusion is just valid if we have for all $S \in \mbox{Syl}_2(G)$ and $\alpha\in \Omega$ that $|S_{\alpha}| \le 2$.

One last possiblity is that if case (2) of Lemma 2.12. holds, then this implies for all $\alpha\in \Omega$ that $|S_{\alpha}| \le 2$; but I do not see that. Also as $S$ does not act transitive, if $\delta^g = \gamma$ we just have $S_{\delta}^g = (S \cap G_{\delta})^g = S^g \cap G_{\delta^g} = (S^g)_{\gamma}$, but the elements mapping the Sylow subgroups onto each other and the points could be different, so as I see it we could not conclude that all point stabilizers have equal size like in transitive actions.

Okay, my question regards Lemma 2.12 and its interpretation? What interpretation is the right one. Interpretation (a) does not work well with its proof (at least as I see it), but interpretation (b) and (c) does not work well with how it is used in Theorem 2.23. So what is the right interpretation, or is interpretaion (a) the right one and I have overlooked something in the application of the Sylow theorems??? But then I hope you can explain it to me, I have thought about this quite a long time but cannot see it...


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