Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


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.


share|cite|improve this question
Hint: You have so far used that the group is closed under group operations, but you haven't yet used the existence of inverses. – joriki Mar 30 '11 at 19:06

You are misinterpreting the definition of block. You are interpreting it to mean that


  • For all $x\in G$, $B^x = B$; or
  • For all $x\in G$, $B^x\cap B = \emptyset$.

This is incorrect (for one thing, the second clause can never happen: if you take $x=e$, the identity, then $B^x = B$).

The reason this is incorrect is that "For all $x$ (P or Q)" is not the same thing as "(For all $x$ P) or (For all $x$ Q)". For example, "Every human in history is either still alive, or already dead" is not the same thing as "Every human in history is still alive, or every human in history is already dead."

The correct interpretation is:

$B$ is a block for $G$ if and only if for all $x\in $G$,

  • Either $B^x=B$; or
  • $B^x\cap B = \emptyset$.

That is, for each $x\in G$, you will either have $B^x=B$, or you will have $B^x\cap B =\emptyset$. The same $B$ can have one behavior with respect to one $x$, and the other behavior with respect to another $x$.

For example, take $A$ to be the set of vertices of a square, and let $G$ be the cyclic group of order $4$. Let $G$ act on $A$ by "rotations of the square". So, if the square has vertices $(\pm 1,\pm 1)$, then the generator of $g$ maps $(1,1)$ to $(-1,1)$, $(-1,1)$ to $(-1,-1)$, $(-1,-1)$ to $(1,-1)$, and $(1,-1)$ to $(1,1)$; with $g^2$, $g^3$, and $g^4=e$ acting in the obvious way.

Take $B=\{(1,1),(-1,-1)\}$. Then $B$ is a block: because if you take $x=e$ or $x=g^2$, then $B^x = B$; and if you take $x=g$ or $x=g^3$, then $B^x = \{(1,-1), (-1,1)\}$, and then $B^x\cap B = \emptyset$. So for some elements of $G$ you have $B^x = B$, and for the other elements of $G$ you have $B^x\cap B = \emptyset$.

Now that this is cleared up: you want to show that $C$ is a partition of $A$, and that each element of $C$ is a block of $G$.

To show that each element of $C$ is a block of $G$, take an element of $C$, $B^x$ for some fixed $x\in G$. You want to show that for every $g\in G$, either $(B^x)^g = B^x$, or else $(B^x)^g\cap B^x = \emptyset$. Hint: Try to bring things back to $B$, which you know to be a block.

Added. I fear my hint above led you in the wrong direction.

Suppose that $(B^x)^g\cap B^x\neq\emptyset$; we want to show that $(B^x)^g = B^x$. Let $a\in (B^x)^g\cap B^x$. That means that there exists a $b\in B$ such that $a = b^x$ (since $a\in B^x$), and there exists $c\in B^x$ such that $a = c^g$ (since $a\in (B^x)^g$). Since $c\in B^x$, there exists $b'\in B$ such that $c=b^x$, so $a = c^g = (b'^x)^g = b'^{xg}$.

Now consider $a^{x^{-1}}$. We have $a^{x^{-1}} = (b^x)^{x^{-1}} = b^{xx^{-1}} = b^1 = b\in B$; and $a^{x^{-1}} = (b'^{xg})^{x^{-1}} = (b')^{xgx^{-1}}\in B^{xgx^{-1}}$. So $a^{x^{-1}}\in B\cap B^{xgx^{-1}}$. So $B\cap B^{xgx^{-1}}\neq\emptyset$. What does that tell you about $B$ and $B^{xgx^{-1}}$? And what does that tell you about $B^x$ and $(B^x)^g$?

Or, if you are comfortable working with the sets and the induced action of $G$ on subsets of $A$, you can argue as follows:

Note that $a\in B^x\cap B^y$ if and only if $a^{y^{-1}}\in (B^x)^{y^{-1}}\cap (B^y)^{y^{-1}}$, if and only if $a^{y^{-1}}\in B^{xy^{-1}}\cap B$. In other words, $B^x=B^y$ if and only if $B^{xy^{-1}} = B$, and $B^x\cap B^y=\emptyset$ if and only if $B^{xy^{-1}} \cap B=\emptyset$. Go from there.

To show that $C$ is a partition, you need to show that two things happen: that $A = \mathop{\cup}\limits_{B^x\in C}B^x$, and that if $B^x$ and $B^y$ are in $C$, then either $B^x=B^y$, or else $B^x\cap B^y = \emptyset$. The second part should be fairly straightforward (it's the same argument as to show that the elements of $C$ are blocks for $G$); the first part should use the transitivity of $G$ on $A$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.