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I need help with next statement :

Let $G$ be a group acting transitively on a set $A$. A nonempty subset $B$ of $A$ is called a block for $G$ if for each $x\in G$ either $B^x=B$ or $B^x \cap B=\emptyset$.

Suppose that $G$ acts transitively on $A$ and that $B$ is a block for $G$. Put $C=\{B^x\mid x\in G\}$.

Next part is the problem :

Then the sets in $C$ form a partition of $A$ and each element of $C$ is a block for $G$.

Trying to prove that every element of $C$ is a block, when I assume that $B$ is a kind of block that for each $x\in G$, $B^x=B$, proof is trivial I guess? But if $B$ is a kind of block that for each $x\in G$, $B^x\cap B=\emptyset$, I don't know how to prove that.

Thanks!

Thank You for answer. So, if I got it right, to prove that $B^x$ is block I have to prove that for every g $\in$ G $(B^x)^g=B^x$ or $(B^x)^g\cap(B^x)=\emptyset $. Because B is block, I will have four possible situations :

  1. $(B^x)^g=B$ and $B^x=B$

  2. $(B^x)^g=B$ and $B^x \cap B = \emptyset$

  3. $(B^x)^g \cap B = \emptyset$ and $B^x = B$

  4. $(B^x)^g \cap B = \emptyset$ and $B^x \cap B = \emptyset$

Things are trivial in first trhree situations I guess, but I don't know what to do in fourth. I don't know how to conclude that either $(B^x)^g=B^x$ or $(B^x)^g\cap(B^x)=\emptyset $ based on 4. assumption.

Thanks!

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    Hint: You have so far used that the group is closed under group operations, but you haven't yet used the existence of inverses.2011-03-30

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