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 $A$.∎