If $G$ acts such that $|\mbox{fix}(g)|\in \{0,p\}$ for $g \ne 1$ and $N \unlhd G$. Then every element outside of $N$ fixes at most $p$ $N$-orbits.

Question

Let $G$ be a transitive permutation group acting on $\Omega$ such that each nontrivial element either fixes no point or exactly $p$ points for some prime $p$. Also assume that for $g \notin N_G(G_{\alpha})$ we have $$ G_{\alpha} \cap G_{\alpha}^g = 1 $$ and denote by $\overline \alpha$ the set of $p$ fixed points of $G_{\alpha}$. Suppose further we have a normal subgroup $N \unlhd G$ of order coprime to $p$. The orbits of $N$ form a system of blocks, and $G$ acts naturally on them.

Then why we have that:

No element of $G$ outside $N$ fixes more than $p$ orbits of $N$.

I thought part of this statement involves that the kernel of the action is precisely $N$ (in this case for the elements outside of $N$ it would be just left to show that they only fix $p$ orbits of $N$), but as I learned here in the forum, this is neither true in the general situation, nor in a more restricted situation of the above assumptions. This is a statement from a published paper, but maybe they forgot to state that $p$ must be odd, as the counterexample in the other post uses $p = 2$; but nevertheless I am not able to prove it or see why it should be true. So I am asking you for help?!

Answer 1

Let $\Delta$ be an orbit of $N$. We claim that no element of $g$ that fixes $\Delta$ can fix more than one point of $\Delta$. If it did then, by arguments we have seen before, all $p$ of its fixed points would lie in $\Delta$. But that would imply $p$ divide $|\Delta|$, which is not possible because $N$ is a $p'$-group.

Now let $g \in G \setminus N$ and suppsoe that $g$ fixes $\Delta$. Then by multiplying $g$ by an element of $N$, we may assume that $g$ fixes a point $\alpha \in \Delta$. So by the above, $\alpha$ is the unique fixed point of $g$ and all its powers in $\Delta$ and hence all of its orbits on $\Delta$ must have the same length $|g|$, which must divide $|\Delta|-1$. So $g$ fixes a (unique) point in each orbit of $N$ that it fixes, and hence it cannot fix more than $p$ such orbits.

I have the feeling that all of these arguments have been used in answers to previous queries.

Answer 2

Just as a note and addition. I guess I have an alternative proof. By googling I found the result that for a Frobenius group each normal subgroup either contains the kernel, or is striclty contained in the kernel. By this, if $\Delta^g = \Delta$ and seeing that $G_{\alpha}N$ for $\alpha \in N$ is a Frobenius group (we have a point such that every element fixing that point fixes no other point, and some nontrivial element fixes that point), I guess we must have $K \le N$ if $K$ denotes its Frobenius kernel. As $N$ is normal in $G_{\alpha}N$ and if $N_{\alpha} \ne 1$ the case $N \le K$ is not possible, and if $N_{\alpha} = 1$ we would have $|N| = |K|$, hence $N = K$. So that if $\Delta^g = \Delta$ we have $g \in G_{\alpha}N$ and if $g$ fixes no point this would imply $g \in K \le N$, giving that for each element $g$ outside of $N$ we have $\Delta^g = \Delta$ $\Leftrightarrow$ $g$ fixes exactly one point in $\Delta$.

So it seems to hold that for arbitrary $g \notin N$ (and not just the ones of the form $gn^{-1}$ which are contained in some point stabilizer) we have that $\Delta^g = \Delta$, then $g$ fixes a point in $\Delta$, see my last comment under Derek's answer.