Let a finite group $G$ act on a set $\Omega$. For some $\alpha \in \Omega$, denote by $G_{\alpha} = \{ g \in G : \alpha^g = \alpha \}$ the stabiliser of $\alpha$ in $G$. I want to show that:
(i) if $p$ is an odd prime dividing $G_{\alpha}$ and $P$ is a $p$-Sylow-subgroup of $G_{\alpha}$ which has at most two fixed points (i.e. elements $\alpha \in \Omega$ such that $\alpha^g = \alpha$ for all $g \in P$) then $P$ is a $p$-Sylow-subgroup of $G$.
I have some partial results, namely:
1) if $x \in G$ has exactly one fixed point (i.e. an element $\alpha$ with $\alpha^x = \alpha$), then $C_G(x) \le G$,
2) if $x \in G$ has exactly two fixed points $\alpha, \beta$, then every element of odd order from $C_G(x)$ is contained in $G_{\alpha}\cap G_{\beta}$.
But here I am stuck and do not know how to proceed... any help?