# blocks of permutation groups

Throughout this article, $A$ is a set and $G$ is a permutation group on $A$.

A block is a subset $B$ of $A$ such that for each $\sigma\in G$, either $\sigma\cdot B=B$ or $(\sigma\cdot B)\cap B=\emptyset$, where $\sigma\cdot B=\{\sigma(b)\ \mid\ b\in B\}$. In other words, if $\sigma\cdot B$ intersects $B$, then $\sigma\cdot B=B$.

Note that for any such permutation group, each of $\emptyset$, $A$, and every element of $A$ is a block. These are called trivial blocks.

It is obvious that if $H\subset G$ are permutation groups on $A$, then any block of $G$ is also a block of $H$.

Blocks are closed under finite intersection:

###### Theorem.

If $B_{1},B_{2}\subset A$ are blocks of $G$, then $B=B_{1}\cap B_{2}$ is a block of $G$.

###### Proof.

Choose $\sigma\in G$. Note that $\sigma\cdot(B_{1}\cap B_{2})=(\sigma\cdot B_{1})\cap(\sigma\cdot B_{2})$. Thus if $(\sigma\cdot B)\cap B\neq\emptyset$, then

 $(\sigma\cdot B)\cap B=(\sigma\cdot(B_{1}\cap B_{2}))\cap(B_{1}\cap B_{2})=(% \sigma\cdot B_{1}\cap B_{1})\cap(\sigma\cdot B_{2}\cap B_{2})$

is nonempty, and thus $\sigma\cdot B_{i}\cap B_{i}\neq\emptyset$ for $i=1,2$. But $B_{1}$ and $B_{2}$ are blocks, so that $\sigma\cdot B_{i}=B_{i}$ for $i=1,2$. Thus

 $\sigma\cdot B=\sigma\cdot(B_{1}\cap B_{2})=(\sigma\cdot B_{1})\cap(\sigma\cdot B% _{2})=B_{1}\cap B_{2}=B$

and $B$ is a block. ∎

We show, as a corollary to the following theorem, that blocks themselves are permuted by the action of the group.

###### Theorem.

If $H\subset G$ are permutation groups on $A$, $B\subset A$ is a block of $H$, and $\sigma\in G$, then $\sigma\cdot B$ is a block of $\sigma H\sigma^{-1}$.

###### Proof.

Choose $\tau\in H$ and assume that

 $((\sigma\tau\sigma^{-1})\sigma\cdot B)\cap\sigma\cdot B\neq\emptyset$

Then, applying $\sigma^{-1}$ to this equation, we see that

 $(\tau\cdot B)\cap B\neq\emptyset$

But $B$ is a block of $H$, so $\tau\cdot B=B$. Multiplying by $\sigma$, we see that

 $\sigma\cdot(\tau\cdot B)=\sigma\cdot B$

and thus

 $(\sigma\tau\sigma^{-1})\sigma\cdot B=\sigma\cdot B$

and the result follows. ∎

###### Corollary.

If $B$ is a block of $G$, $\sigma\in G$, then $\sigma\cdot B$ is also a block of $G$.

###### Proof.

Set $G=H$ in the above theorem. ∎

###### Definition.

If $B$ is a block of $G$, $\sigma\in G$, then $B$ and $\sigma\cdot B$ are conjugate blocks. The set of all blocks conjugate to a given block is a block system.

It is clear from the fact that $B$ is a block that conjugate blocks are either equal or disjoint, so the action of $G$ permutes the blocks of $G$. Then if $G$ acts transitively on $A$, the union of any nontrivial block and its conjugates is $A$.

###### Theorem.

If $G$ is finite and $G$ acts transitively on $A$, then the size of a nonempty block divides the order of $G$.

###### Proof.

Since $G$ acts transitively, $A$ is finite as well. All conjugates of the block have the same size; since the action is transitive, the union of the block and all its conjugates is $A$. Thus the size of the block divides the size of $A$. Finally, by the orbit-stabilizer theorem, the order of $G$ is divisible by the size of $A$. ∎

Title blocks of permutation groups BlocksOfPermutationGroups 2013-03-22 17:19:05 2013-03-22 17:19:05 rm50 (10146) rm50 (10146) 15 rm50 (10146) Topic msc 20B05 trivial block block block system conjugate block