Let $G$ be an infinite transitive permutation group acting on a set $\Omega$. Is $\text{fix}_\Omega(G_\alpha)$ a block of imprimitivity for $G$ in $\Omega$?
$G_\alpha$ is the set of elements of $G$ that fix $\alpha \in \Omega$
$\text{fix}_\Omega(G_\alpha)$ is the set of points in $\Omega$ fixed by $G_\alpha$.
A block of imprimitivity for $G$ in $\Omega$ is a set $B\subseteq \Omega$ such that for any $g \in G$, $B^g=B$ or $B^g \cap B = \emptyset$
I can show this in the finite case where $\text{fix}_\Omega(G_\alpha) = \alpha^{N_G(G_\alpha)}$, the orbit of $\alpha$ under the normalizer of $G_\alpha$, but in the infinite case I'm not sure if this is still true.
Thank you for your help.
This is false in general, because there are infinite groups $G$ containing a subgroup $H$ and an element $y$ such that $K := y^{-1}Hy$ is a proper subgroup of $H$. An example is the Baumslag-Solitar group $\langle x,y \mid y^{-1}xy = x^2 \rangle$ with $H = \langle x \rangle$. (This can also be defined as a $2 \times 2$ group of matrices over ${\mathbb Q}$.)
Consider the transitive permutation action by right multiplication on the right cosets of $H$ in $G$. Let $\alpha=H$ and $\beta=Hy$. Then $G_\alpha=H$ and $G_\beta = y^{-1}Hy=K$, so $G_\beta$ is a proper subgroup of $G_\alpha$. Hence ${\rm Fix}(G_\alpha)$ is a proper subset of ${\rm Fix}(G_\beta)$ (it contains $\alpha$ but not $\beta$). But $y^{-1}$ maps ${\rm Fix}(G_\beta)$ to ${\rm Fix}(G_\alpha)$, so ${\rm Fix}(G_\beta)$ cannot be a block.