Let G be a transitive permutation group on the finite set A. a 'block' is a non-empty subset B of A such that for all either or . (here is the set ).
It can be easily shown that if B is a block and are all distinct images of B under the elements of G, then these form a partition of A.
Can the same be proved if G is not transitive on the finite set A.
Yes, if I understand your definition of a `block' correctly. I mean, let , where is the orbit of . Then clearly , because ...does that make sense?...Basically, every element of G takes everything in A and sticks it somewhere within the orbit. So the orbit is partitioned nicely (because G acts transitively on the orbit), and we can just take `everything else' to be another partition...
Although in the question it was given that G acts transitively on a, i still am assuming that G does not act transitively on A. Then is non-empty. is this correct??
Then . is this correct too??
If these two are correct then i will be able to understand your post.
THANKS IN ADVANCE.
yes swlabr. i can see why they are correct.
I infer that your solution uses the fact that G acts transitively on A. does this mean that the distinct images of the block B under the elements of G do not necessarily form a partition of A if G does not act transitively on A.
I have quoted the solution you have given.
you have defined .
I agree with the fact that , because this follows from the definition of .
But i don't agree with . According to me this is only true if G acts transitively on A. i say this because if G does not act transitively on A then is non-empty( as in this case ), and there must be some element such that .
May be i am just being a 'block'head here but i am only a newbie at group theory so please help this poor guy.