# Finding a counter Example

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.

• Have you actually tried to construct any non-transitive group action? – Erick Wong Jun 22 '12 at 7:24
• @ErickWong: I can consider $V=\{id,(1,2),(3,4),(1,2)(3,4)\}|\{1,2,3,4\}$ which is not transitive. There are two orbits which could be blocks also, $B_1=\{1,2\}$ and $B_2=\{3,4\}$. – Mikasa Jun 22 '12 at 7:41
• Try an intransitive example in which the orbits have different lengths! – Derek Holt Jun 22 '12 at 8:56
• @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. – Mikasa Jun 22 '12 at 19:34

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 $|\Omega|$.
• If you want such a group action that $G$ does not fix any point of $\Omega$, you can try the subgroup of $S_5$ generated by $(123)(45)$. – Jyrki Lahtonen Jun 22 '12 at 12:51