I am noting a simple problem about a permutation group from "Permutation Group" By J.Dixon, its answer and my attempt to understand it in details:
Q: A primitive permutation group $G(≠1$) is transitive.
A: If $G$ is an intransitive group ($≠1$), then it has an orbit of length at least $2$. This orbit is a nontrivial block for $G$.
This is clear that any intransitive group ($≠1$) can possess an orbit $B$ of length at least $2$. Let $G$ is acting on a set $\Omega$. Since $∅≠B≠\Omega$ and it has at least two elements, it is enough to show that $B$ is a block. If for example $B$={$\alpha$,$\beta$} then I shuold check $B^g∩B=∅$ or $B^g=B$ for any $g\in G$. $B^g$={$\alpha^g$,$\beta^g$} and if $g\in G_{\alpha,\beta}$ then $B^g=B$ clearly. If $g∉G_{\alpha,\beta}$ then we get $B^g∩B=∅$.
Honestly, I cannot go for the rest. If my approach is not wrong, please help me to complete the answer. Thanks