I am looking for a finite, transitive and nonregular permutation group $G$ acting on $\Omega$, such that every nontrivial element fixes at most two points and such that

i) the point stabilizers $G_{\alpha}$ have even order,

ii) $G$ has even degree, i.e. $|\Omega|$ is even,

iii) the Sylow $2$-subgroups are dihedral or semidihedral

iv) $|S_{\alpha}| > 2$ for some Sylow $2$-subgroup $S$ and $\alpha \in \Omega$.

The condition i) and ii) [and also iv)] imply that four divides $|G|$. For example if $G = \mathcal S_4$ in its natural action, then i), ii) and iii) are fulfilled (see for example here), but as for example for $\alpha = 1$ we have $G_{\alpha} = \langle (234), (23), (34), (24) \rangle$ and $S = \langle (1234), (24) \rangle$, then $S \cap G_{\alpha} = \{ (), (24) \}$, and by symmetry considerations we see that every Sylow $2$-subgroup intersects with the stabilizers in a subgroup of order two, so this is not an example.

So okay do you know any examples? (Remark: I added the GAP-tag, maybe some clever GAP-user knows how to use a computer program to find an example).


${\rm PSL}(2,q)$ with $q \equiv 1 \bmod 8$ or ${\rm PGL}(2,q)$ with $ q \equiv 1 \bmod 4$ are examples with dihedral Sylow $2$-subgroups. So ${\rm PGL}(2,5)$ is the smallest example.

For some $q$, such as $q=9$, there is a related example ${\rm PSL}(2,q^2).\langle \tau \rangle$, where $\tau$ induces a product of a field and a diagonal isomorphism, with semidihedral Sylow $2$-subgroups.

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  • $\begingroup$ Thanks, but could you provide more information on the specific action (with the fixed points restrictions)? For example $PGL(2,5) \cong S_5$; and I also tried $S_5$ but its natural action does not work (not even degree, but also some nontrivial element has more then 2 fixed points) and similar if $S_5$ acts on all $2$-point subsets then we have $|S_{\alpha}| > 2$, but again we find elements fixing three points. So in your examples I do not see immediately that they also fulfill all the restrictions. $\endgroup$ – StefanH Feb 27 '16 at 21:00
  • $\begingroup$ I mean their natural actions of course. $\endgroup$ – Derek Holt Feb 27 '16 at 23:06
  • $\begingroup$ Yes, that's right! Thank you. $\endgroup$ – StefanH Feb 28 '16 at 13:47

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