3
$\begingroup$

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).

$\endgroup$

1 Answer 1

4
$\begingroup$

${\rm PSL}(2,q)$ 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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$ Commented Feb 27, 2016 at 21:00
  • $\begingroup$ I mean their natural actions of course. $\endgroup$ Commented Feb 27, 2016 at 23:06
  • $\begingroup$ Yes, that's right! Thank you. $\endgroup$ Commented Feb 28, 2016 at 13:47

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.