About blocks on permutation groups I have the following situation:
$G$ is a permutation group, transitive on $\Omega$, and $B$ is a non-trivial block.
(for every $x\in G$, a block $B$ verifies either $B\cdot x=B$ or $B\cdot x\cap B=\emptyset$)
Then I have to prove that $\Sigma=\lbrace{B\cdot x\;/\;x\in G}\rbrace$ is a partition of $\Omega$. And in particular, that $|B|$ divides $|\Omega|$.

What I have:
Assuming that $\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$, I have proved that this union must be disjoint (I just supposed $B\cdot x\cap B\cdot y\neq\emptyset$, and I concluded that $B\cdot x=B\cdot y$ using the definition of block).
Also, as  $\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$, then $|\Omega|=|B\cdot x||\Sigma|=|B||\Sigma|$, so  $|B|$ divides $|\Omega|$. (is this correct?)
Now, my problem is that I don't see how to justify the union I was assuming to be true:$\Omega=\displaystyle\cup_{(B\cdot x) \in\Sigma}(B\cdot x)$
My teacher told me that I should use the transitivity of $G$ on $\Omega$, but I still can't see how to justify this step.
Thanks a lot for any help.

Fix $\gamma \in B$ and let $\omega \in \Omega$. Since $G$ acts transitively on $\Omega$, there exists $g \in G$ such that $\gamma g = \omega$. Now $B.g$ contains $\omega$. So $\Omega = \bigcup_{g \in G} (B.g)$.
You've shown that two 'translates' $B.g$ and $B.h$ are either equal or disjoint. So it follows that there exist $x_1, \ldots, x_m \in G$ such that $\Omega = B.x_1 \cup \ldots B.x_m$, where the union is disjoint. Since $|B.x_i| = |B|$ for all $i$, we get $|\Omega| = m|B|$.
