# 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 $(\sigma \cdot B)\cap B=\mathrm{\varnothing}$, 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 $\mathrm{\varnothing}$, $A$, 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 ${B}_{\mathrm{1}}\mathrm{,}{B}_{\mathrm{2}}\mathrm{\subset}A$ are blocks of $G$, then $B\mathrm{=}{B}_{\mathrm{1}}\mathrm{\cap}{B}_{\mathrm{2}}$ 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 $(\sigma \cdot B)\cap B\ne \mathrm{\varnothing}$, 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 $\sigma \cdot {B}_{i}\cap {B}_{i}\ne \mathrm{\varnothing}$ 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 $H\mathrm{\subset}G$ are permutation groups on $A$, $B\mathrm{\subset}A$ is a block of $H$, and $\sigma \mathrm{\in}G$, then $\sigma \mathrm{\cdot}B$ is a block of $\sigma \mathit{}H\mathit{}{\sigma}^{\mathrm{-}\mathrm{1}}$.

###### Proof.

Choose $\tau \in H$ and assume that

$$((\sigma \tau {\sigma}^{-1})\sigma \cdot B)\cap \sigma \cdot B\ne \mathrm{\varnothing}$$ |

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

$$(\tau \cdot B)\cap B\ne \mathrm{\varnothing}$$ |

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 $G$, $\sigma \mathrm{\in}G$, then $\sigma \mathrm{\cdot}B$ 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 |
---|---|

Canonical name | BlocksOfPermutationGroups |

Date of creation | 2013-03-22 17:19:05 |

Last modified on | 2013-03-22 17:19:05 |

Owner | rm50 (10146) |

Last modified by | rm50 (10146) |

Numerical id | 15 |

Author | rm50 (10146) |

Entry type | Topic |

Classification | msc 20B05 |

Defines | trivial block |

Defines | block |

Defines | block system |

Defines | conjugate block |