# If $G$ acts such that each non-trivial element either has no fixed point or exactly two, then there exists fixed-point free involutory map on $\Omega$

If a finite group $G$ acts on a set $\Omega$ non-regulary (i.e. there is some element fixing some point) and each element having some fixed point has exactly $n$ fixed points, then we say the group is of type $(0,n)$. I know:

Lemma: If $G$ acts faithfully as a group of type $(0,1)$ on a set $\Omega$ then $G$ has precisely one non-regular orbit (i.e. an orbit on which $G$ acts non-regular).

Now from this follows:

If $G$ acts on a set $\Omega$ as a group of type $(0,2)$, there is an involutory map, $\alpha \mapsto \alpha^*$, of $\Omega$ onto itself, not fixing any point, such that $\alpha$ and $\alpha^*$ have equal stabilizers.

Proof: Let $\alpha \in \Omega$. Then $G_{\alpha}$ acts on $\Omega \setminus \{ \alpha\}$ as a group of type $(0,1)$. Thus there is precisely one orbit $\Delta \subseteq \Omega\setminus \{\alpha\}$ on which $G_{\alpha}$ acts non-regularly. If $\Delta$ has more than one point $G_{\alpha}$ acts on $\Delta$ as a Frobenius group and the elements of its kernel fix no point of $\Delta$ and hence none of $\Omega \setminus \{\alpha\}$. $\square$

I do not get the proof, there is no map given. I conjecture that we can get a map by selecting some $\alpha^* \in \Delta$ from the unique orbit on which $G_{\alpha}$ acts non-regular, but I fail to verify its properties. We have to verify two properties, i) that the points stabilizers of $\alpha$ and $\alpha^*$ are equal, and ii) that the map is involutory.

For i) let $\alpha \in \Omega$ and let $\Delta$ be as in the proof, then we must have that for each $\beta \in \Delta$ $$G_{\alpha} = G_{\beta}.$$ The facts I know:

1) as $G_{\alpha}$ acts non-regular on $\Delta$, we have $1 \ne (G_{\alpha})_{\beta} = G_{\alpha}\cap G_{\beta}$ for $\beta \in \Delta$,

2) as $G$ has type $(0,2)$ for each $g \in G_{\gamma}$ there must exists some $\delta$ with $g \in G_{\delta}$.

Applying 2) to the situation above, we must have $G_{\alpha} \cap G_{\beta} = 1$ for each $\beta \notin \Delta, \beta \ne \alpha$, and $G_{\alpha}\cap G_{\beta} \ne 1$ for each $\beta \in \Delta$. So that if $h \in G_{\alpha}$ the other $\gamma$ with $\gamma^h = \gamma$ must be in $\Delta$. But for different $h \in G_{\alpha}$ these $\gamma$ might be different, what would be desirable would be that $\gamma = \alpha^*$, then we would have $G_{\alpha} \le G_{\alpha^*}$ and using the involutory property of the map the other inclusion would follow. But despite verifying the properties of the map I also have no idea for showing that the map is an involution?

So any hints? Maybe the map I am considering is the wrong one..

Remark: I can supply the proof of the first Lemma if wanted.

• The final sentence of the proof is a contradiction to the assumption that $\Delta$ has more than one point. So $\Delta$ must have exactly one point, $\alpha^*$. That gives you the map. – Derek Holt Oct 20 '15 at 8:44
• Ah okay I see, then the element in the kernel would have exactly one fixed point, and for a Frobenius group the kernel is non-trivial. – StefanH Oct 20 '15 at 9:49