3
$\begingroup$

Let $G$ be a finite permutation group on $\Omega$ acting nonregular and transitive such that each nontrivial element fixes at most two points of $\Omega$.

Suppose that for $\alpha \in \Omega$ the stabilizer $G_{\alpha}$ has even order and $|\Omega|$ is even too. Let $S \in \mbox{Syl}_2(G)$ and suppose $S_{\alpha} \ne 1$, but $S \nleq G_{\alpha}$ and that there exists $\beta \in \Omega$ distinct form $\alpha$ such that $S_{\alpha} = S_{\beta}$ with $|S : S_{\alpha}| = 2$ and some element of $S$ interchanges $\alpha$ and $\beta$ (note that by normality of $S_{\alpha}$ this holds for each $S \setminus S_{\alpha}$).

As $S_{\alpha}$ already has two fixed points, this subgroup must have regular orbits (i.e. acts semiregular) 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 $|S|$.

I do not see why all other $S$-orbits have length $|S|$? As $S_{\alpha}$ is normal in $S$ the $S_{\alpha}$ orbits form a system of blocks for each $S$-orbit and hence have all equal size on each $S$-orbit, so if $\gamma \notin \{\alpha,\beta\}$ then $$ |\gamma^S| = |S : S_{\gamma}| = k \cdot |S_{\alpha} : S_{\alpha}\cap S_{\gamma}| = k\cdot |S_{\alpha}| $$ for some $k$. Now $|S| = 2|S_{\alpha}|$, so that $k \in \{1,2\}$. What I want is $k = 2$ (which is equivalent to $S_{\gamma} = 1$, i.e. $S$ itself acts regular on the orbit $\gamma^S$). If $k = 1$ we must have $|S_{\gamma}| = 2$ and as $S_{\alpha} \cap S_{\gamma} = 1$ we have $S = S_{\gamma}S_{\alpha}$. Further $\gamma^S = \gamma^{S_{\alpha}}$, so that $S_{\alpha}$ already acts transitive on the orbit. But I do not see where this all might lead to a contradiction, thereby excluding the case $k = 1$?

Do you see why $|\gamma^S| = |S|$? Can you help me seeing it!?

$\endgroup$
  • 1
    $\begingroup$ What about $G \cong A_4$ with $|\Omega|=6$, $|G_\alpha|=2$. Unless I have overlooked something, that satisfies the hypotheses. Then $S$ has order $4$ and has three orbits each of length $2$. $\endgroup$ – Derek Holt Nov 10 '15 at 22:34
  • $\begingroup$ Why $|G_{\alpha}| = 2$? A point stabilizer of $A_4$ has order three, for example for $\alpha = 4$ we have $G_4 = \langle (1 ~ 2 ~ 3) \rangle$, or not? $\endgroup$ – StefanH Nov 10 '15 at 22:51
  • $\begingroup$ Derek is considering a different action of $A_4$, on the cosets of the group generated by a double transposition. (Equivalently, acting on the 6 unordered pairs of the natural $4$-set.) This group is imprimitive with three blocks of size 2, and the Sylow $2$-subgroup is normal, with these blocks as orbits. It does seem like a counter-example. $\endgroup$ – verret Nov 11 '15 at 6:07
  • $\begingroup$ $S_4$ acting on the $6$ cosets of $\langle (1,2),(3,4) \rangle$ is also a counterexample. Here $|S|=8$, and $S$ has orbits of lengths $2$ and $4$. $\endgroup$ – Derek Holt Nov 11 '15 at 9:28
  • $\begingroup$ Okay, this is truly an error in the paper. Thank you all! And Derek, I would prefer if you put your counterexample in an answer so I can award the bounty... $\endgroup$ – StefanH Nov 11 '15 at 10:39
3
+50
$\begingroup$

The transitive action of $A_4$ of degree $6$ on the cosets of the subgroup $\langle (1,2)(3,4) \rangle$ is a counterexample. Here we have $|S_\alpha|=2$, $|S|=4$, and $S$ has three orbits of zise $2$. You could take $$G = \langle (1, 2)(5, 6), (1, 3, 5)(2, 4, 6) \rangle$$ with $$S = \{ 1,\,(1,2)(3,4),\,(1,2)(5,6),\,(3,4)(5,6) \}.$$

Another counterexample, which contains this one, is $S_4$ acting on the $6$ cosets of $\langle (1,2),(3,4) \rangle$. Here $|S|=8$, and $S$ has orbits of sizes $2$ and $4$. This is the intersection of $S_2 \wr S_3$ with $A_6$.

I don't know whether there are any larger counterexamples - possibly not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.