Primitivity implies transitivity? 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
A: Suppose $G$ acts on a set $\Omega$.  Then the orbits of $G$ form a partition of $\Omega$, and each orbit is a block of $G$; in fact, each orbit $B$ is a minimal fixed block, so that $B \cap B^g=B, \forall g \in G$. If $G \ne 1$, then there is an orbit $B$ of length at least 2, and in addition if $G$ is intransitive, then $|B| < |\Omega|$, so that $B$ is a nontrivial block.  Thus, every intransitive group $G \ne 1$ has a nontrivial block.  
Given a nontrivial block $B$, if $G$ is transitive, then $\Sigma:=\{B^g: g \in G\}$ is a partition of $\Omega$, and $G$ acts on $\Sigma$.  As the authors of the text mention in p. 12, we can sometimes obtain useful information about $G$ by considering this action.  If the group is intransitive, the resulting $\{B^g: g \in G\}$ is not a partition of $\Omega$.  
Every intransitive group $G \ne 1$ has a nontrivial block and hence (by definition of primitivity) cannot be primitive.  Thus, if the group is intransitive, there is no question as to whether it is primitive or imprimitive.
