# block structure of a subgroup

Let $$H \leq S_N$$ be a transitive permutation group. A system of blocks $$\Sigma = \{\Delta\}$$ for $$H$$ is a partition of $$[N]$$ such that either $$\Delta^h = \Delta$$ or $$\Delta \cap \Delta^h = \emptyset$$. For primitive $$H$$, the only system of blocks are the trivial ones $$\Delta = \{e\}$$ and $$\Delta = [N]$$. It is also known that there is a one-to-one mapping between the system of blocks of $$H$$ and overgroups of the stabilizer $$H_{n}$$ for $$n \in [N]$$.

My question is this: what can we say about the system of blocks for a larger group $$G > H$$, $$G \leq S_N$$? We know the relationship between the stabilizers, that is $$H_n < G_n$$. I assume the blocks are generally not preserved (?) what more can we say?

• why is that? can you elaborate? – self-educator Aug 2 at 19:22
• ...or If H is a proper subgroup of G, can G and H still have exactly same block systems? – self-educator Aug 2 at 20:16
• Yes it's possible for $G$ and $H$ to have the same block systems. – Derek Holt Aug 2 at 21:08

There's not a lot to say here, but we can say a couple of things. We'll assume $$H\le G\le S_n$$ and $$H$$ and $$G$$ are both transitive. A block system is defined by a choice of a single block $$\Delta\ni 1$$, so let $$B_H$$ be the set of blocks for $$H$$ containing $$1$$ and $$B_G$$ be the set of blocks for $$G$$ containing $$1$$.

Immediately if $$\Delta^g=\Delta$$ or $$\Delta^g\cap\Delta=\emptyset$$ for all $$g\in G$$ then certainly this holds for all $$g\in H$$. So a block for $$G$$ is a block for $$H$$ which we can write as $$B_G\le B_H$$.

The inverse equality clearly doesn't always hold (e.g. $$H$$ any imprimitive group and $$G=S_n$$). One can show however that blocks of $$G$$ containing $$1$$ are in one-to-one correspondence with subgroups of $$G$$ containing the point stabiliser $$G_1$$ (specifically $$\Delta$$ corresponds to $$G_\Delta$$ the setwise stabiliser of $$\Delta$$). I claim that a block $$\Delta$$ of $$H$$ containing $$1$$ is a block for $$G$$ if and only if $$G_\Delta\ge G_1$$. This means $$B_G=\{\Delta\in B_H|G_1\le G_\Delta\}$$ - that is a block of $$G$$ containing $$1$$ is any block of $$H$$ containing $$1$$ fixed by $$G_1$$.

Proof of Claim:

"Prove $$\Delta$$ is a block of $$G$$ implies $$G_1\le G_\Delta$$" is a standard exercise, so I will only do the reverse implication.

Assume $$G_\Delta\ge G_1$$ where $$\Delta$$ is a block for $$H$$.

Let $$g\in G$$ and suppose $$x\in\Delta\cap\Delta^g$$.

Relabelling if necessary we may assume $$x=1$$.

As $$1\in\Delta^g$$ there is some $$y\in\Delta$$ with $$y^g=1$$.

As $$H$$ is transitive there is some $$h\in H$$ with $$1^h=y$$ so $$1^{hg}=1$$.

That is $$hg\le G_1\le G_\Delta$$.

As $$\Delta$$ is a block for $$H$$, $$h\le H_\Delta\le G_\Delta$$.

Hence $$g=h^{-1}(hg)\le G_\Delta$$, so $$\Delta^g=\Delta$$ completing the proof.