I am currently practicing group theory and came across the following task:
If $G$ is a transitive permutation group then $\mathrm{fix}(G_\alpha)$ is a block
Unfortunately, the solution shown here is unclear to me, is there an alternative solution? I know for example that all stabilizers have to be conjugated and that $G=G_{\alpha}N$ for $N$ nontrivial normal subgroup holds. Any help is greatly appreciated!
First we have know that $N_{G}(G_{\alpha})$ acts transitively in $\operatorname{fix}(G_{\alpha})$ as you can check. Define $\Delta=\operatorname{fix(G_{\alpha})}$, now if
$$\exists x\in G, \Delta^{x}\cap\Delta\neq\emptyset,$$
there must exist $\beta\in\Delta$ such that $\beta^{x}\in\Delta$. By the transitivity of $N_{G}(G_{\alpha})$, there are $n_{1},n_{2}\in N_{G}(G_{\alpha})$, such that
\begin{align*}
\beta&=\alpha^{n_{1}}\\
\alpha^{n_{1}x}=&\beta^{x}=\alpha^{n_{2}}.
\end{align*}
And we can deduce
\begin{align*}
&n_{2}x n_{1}\in G_{\alpha}\\
\Rightarrow &n_{2}x n_{1}\in N_{G}(G_{\alpha})\\
\Rightarrow &x\in N_{G}(G_{\alpha})
\end{align*}
Now $\forall \delta\in\Delta$, we have
$$
(\delta^{x})^{G_{\alpha}}=\delta^{xG_{\alpha}}
=\delta^{G_{\alpha}x}=\delta^{x},
$$
so $\Delta^{x}\in\Delta,\Delta^{x}\subseteq\Delta$, hence $\Delta^{x}=\Delta$. Now for any $x\in G$, either $\Delta^{x}=\Delta$ or $\Delta^{x}\cap\Delta=\emptyset$, means $\Delta=\operatorname{fix}(G_{\alpha})$ is a block. (When |fix| is finite.)
