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I am reading the following theorem:

Let $G$ is a group acting on a set $\Omega$ transitively and let $B\neq\emptyset $ be a block of $G$. Then $|B|$ divides $|\Omega|$.

From the first step till the proof ends, I see the transitively is being used and it is really necessary in this theorem. Is there any counter example showing that omitting transitively doesn't lead us to desire conclusion? Thanks.

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    @DerekHolt: From the time you kindly pointing me out the hint, atlast I can find one counter example. May I have see that for sure? I found G=<(1 2)><(3 4 5)> is not acting on 5 letters transitively and $B=\{1,2\}$ is a non trivial block of this action.2012-06-22

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Going for a minimal counterexample. Let $G$ be the group of permutations generated by the 2-cycle $(12)$. As we can view $G$ as a subgroup of the symmetric group $S_3$, it acts (intransitively) in the set $\Omega=\{1,2,3\}$. The set $\{1,2\}$ is a block of $G$, but its size is not a factor of $|\Omega|$.

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    @Babak, you found that counterexample yourself, so you could post it as an answer! I posted this as a CW just in case your brain gets stuck in a loop - happens to all of us I think :-)2012-06-22