PlanetMath: blocks of permutation groups

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blocks of permutation groups

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

A block is a subset of 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 , then $\sigma\cdot B = B$ .

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

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

Blocks are closed under finite intersection:

Theorem 1 If $B_1, B_2\subset A$ are blocks of , then $B=B_1\cap B_2$ is a block of .

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

$\displaystyle (\sigma\cdot B)\cap B = (\sigma\cdot (B_1\cap B_2))\cap(B_1\cap B_2)=(\sigma\cdot B_1 \cap B_1)\cap (\sigma\cdot B_2\cap B_2)$

is nonempty, and thus $\sigma\cdot B_i \cap B_i\neq\emptyset$ for

. But

and

are blocks, so that $\sigma\cdot B_i=B_i$ for

. Thus

$\displaystyle \sigma\cdot B = \sigma\cdot (B_1\cap B_2) = (\sigma\cdot B_1)\cap (\sigma\cdot B_2)=B_1\cap B_2=B$

and

is a block. $\qedsymbol$

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

Theorem 2 If $H\subset G$ are permutation groups on , $B\subset A$ is a block of , and $\sigma\in G$ , then $\sigma\cdot B$ is a block of $\sigma H\sigma^{-1}$ .

Proof. Choose $\tau\in H$ and assume that

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

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

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

But

is a block of

, so $\tau\cdot B=B$ . Multiplying by $\sigma$ , we see that

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

and thus

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

and the result follows. $\qedsymbol$

Corollary 1 If is a block of , $\sigma\in G$ , then $\sigma\cdot B$ is also a block of .

Proof. Set

in the above theorem. $\qedsymbol$

Definition 1 If

is a block of

, $\sigma\in G$ , then

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 is a block that conjugate blocks are either equal or disjoint, so the action of permutes the blocks of . Then if acts transitively on , the union of any nontrivial block and its conjugates is .