Transitive action, blocks, primitive action, maximal subgroups.

Let act transitively on a finite set . A 'block' in is a non-empty subset of such that for all either or (where ).

This action is called 'primitive' if the only blocks in are trivial ones: the sets of size and itself.

Prove that:

The action(transitive) of on is primitive if and only if for each is a maximal subgroup of . ( stabilizer of in )

Here is what i have(with a little help from my friend):

Define . Its easy to see that . Moreover if then .

I came to know that there exists a bijection between the blocks containing and the subgroups of containing .I couldn't prove this. Help needed.

Re: Transitive action, blocks, primitive action, maximal subgroups.

First we show that if for each , is a maximal subgroup, then is primitive. In particular, we prove the contrapositive that if is not a maximal subgroup for some then is not primitive. Let be a chain of proper subgroups. We claim that is a block. It's easy to see that if then . On the other hand, let , and suppose towards a contradiction that . Then there are such that and hence . But then , giving us , a contradiction. So is indeed a block. Furthermore, is nontrivial: for if then and are distinct elements of . But , because if it were then we would have for some and hence , which would contradict the fact that . Thus the first half of the proof is complete.

For the converse, let be a nontrivial block, and let be distinct. Then . We claim that both subgroup relations are proper: For let . Then there is with by the transitivity of , which means . So is a proper subgroup of . Also by the transitivity of there is with , which means . Since is a block with then it must be that , and hence . So is a proper subgroup of . We conclude that is not maximal, and this completes the second half of the proof.

Re: Transitive action, blocks, primitive action, maximal subgroups.

I was just thinking about how to show the stronger result that there is a bijection between blocks containing and the subgroups of containing ...

Let as the set of blocks in containing , and let be the set of subgroups of containing . Define by , for each . We claim that is injective: let be distinct blocks containing . Then (without loss of generality) there is . By the transitivity of there is with . So . However ; since is a block this means and hence . Thus , and it follows that is injective.

Now we show that is surjective: Let be a chain of subgroups of . We showed in the previous post that is a block, and obviously it contains . We claim that . Clearly for each we have ; so . Now let . Then , which means there is with . So , and thus . We conclude , and it follows that is surjective, indeed, bijective.

Re: Transitive action, blocks, primitive action, maximal subgroups.

thank you! that was brilliant.