# Thread: Blocks in the finite set A when acted on by G.

1. ## Blocks in the finite set A when acted on by G.

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 $\displaystyle \sigma \text \in G$ either $\displaystyle \displaystyle \sigma (B)=B$ or $\displaystyle \sigma(B) \cap B=\phi$. (here $\displaystyle \sigma(B)$ is the set $\displaystyle \left\{\sigma(b)|b \in B \right\}$).

It can be easily shown that if B is a block and $\displaystyle${\sigma}_1 (B),\text{} ${\sigma}_2 (B), \ldots, \text{}${\sigma}_n (B)$are all distinct images of B under the elements of G, then these form a partition of A. Can the same be proved if G is not transitive on the finite set A. 2. Originally Posted by abhishekkgp Let G be a transitive permutation group on the finite set A. a 'block' is a non-empty subset B of A such that for all$\displaystyle \sigma \text \in G$either$\displaystyle \displaystyle \sigma (B)=B$or$\displaystyle \sigma(B) \cap B=\phi$. (here$\displaystyle \sigma(B)$is the set$\displaystyle \left\{\sigma(b)|b \in B \right\}$). It can be easily shown that if B is a block and$\displaystyle ${\sigma}_1 (B),\text{}${\sigma}_2 (B), \ldots, \text{} ${\sigma}_n (B)$ 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.
Your terminology is unfamiliar to me. Are you saying that $\displaystyle G$ acts on $\displaystyle A$ transitively? Is that what your first sentence means?

3. Originally Posted by abhishekkgp
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 $\displaystyle \sigma \text \in G$ either $\displaystyle \displaystyle \sigma (B)=B$ or $\displaystyle \sigma(B) \cap B=\phi$. (here $\displaystyle \sigma(B)$ is the set $\displaystyle \left\{\sigma(b)|b \in B \right\}$).

It can be easily shown that if B is a block and $\displaystyle${\sigma}_1 (B),\text{} ${\sigma}_2 (B), \ldots, \text{}${\sigma}_n (B)$are all distinct images of B under the elements of G, then these form a partition of A. Can the same be proved if G is not transitive on the finite set A. Yes, if I understand your definition of a block' correctly. I mean, let$\displaystyle A_0 = A\setminus Gx$, where$\displaystyle Gx$is the orbit of$\displaystyle G$. Then clearly$\displaystyle \sigma(A_0)\cap A_0 = \emptyset$, because$\displaystyle A_0 \cap Gx=\emptyset$...does that make sense?...Basically, every element of G takes everything in A and sticks it somewhere within the orbit. So the orbit is partitioned nicely (because G acts transitively on the orbit), and we can just take everything else' to be another partition... 4. Originally Posted by Drexel28 Your terminology is unfamiliar to me. Are you saying that$\displaystyle G$acts on$\displaystyle A$transitively? Is that what your first sentence means? yes, i mean the same. I have taken the question from the text on abstract algebra by dummit and foote. 5. Originally Posted by Swlabr Yes, if I understand your definition of a block' correctly. I mean, let$\displaystyle A_0 = A\setminus Gx$, where$\displaystyle Gx$is the orbit of$\displaystyle G$. Then clearly$\displaystyle \sigma(A_0)\cap A_0 = \emptyset$, because$\displaystyle A_0 \cap Gx=\emptyset$...does that make sense?...Basically, every element of G takes everything in A and sticks it somewhere within the orbit. So the orbit is partitioned nicely (because G acts transitively on the orbit), and we can just take everything else' to be another partition... Although in the question it was given that G acts transitively on a, i still am assuming that G does not act transitively on A. Then$\displaystyle $A_0$ is non-empty. is this correct??
Then $\displaystyle \sigma ($A_0) \cap $A_0 \neq \phi$. is this correct too??

If these two are correct then i will be able to understand your post.

6. Originally Posted by abhishekkgp
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 $\displaystyle$A_0$is non-empty. is this correct?? Then$\displaystyle \sigma ($A_0) \cap$A_0 \neq \phi$. is this correct too?? If these two are correct then i will be able to understand your post. THANKS IN ADVANCE. Yes, these are correct. Can you see why they are? 7. Originally Posted by Swlabr Yes, these are correct. Can you see why they are? yes swlabr. i can see why they are correct. I infer that your solution uses the fact that G acts transitively on A. does this mean that the distinct images of the block B$\displaystyle ${\sigma}_1 (B),\text{}${\sigma}_2 (B), \ldots, \text{} ${\sigma}_n (B)$ under the elements of G do not necessarily form a partition of A if G does not act transitively on A.

8. Originally Posted by abhishekkgp
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 $\displaystyle${\sigma}_1 (B),\text{} ${\sigma}_2 (B), \ldots, \text{}${\sigma}_n (B)$under the elements of G do not necessarily form a partition of A if G does not act transitively on A. No no, I use the fact that G acts transitively on$\displaystyle G_x = A\setminus A_0$. So your blocks are the partitions of G_x along with A_0. 9. Originally Posted by Swlabr Yes, if I understand your definition of a block' correctly. I mean, let$\displaystyle A_0 = A\setminus Gx$, where$\displaystyle Gx$is the orbit of$\displaystyle G$. Then clearly$\displaystyle \sigma(A_0)\cap A_0 = \emptyset$, because$\displaystyle A_0 \cap Gx=\emptyset$...does that make sense?...Basically, every element of G takes everything in A and sticks it somewhere within the orbit. So the orbit is partitioned nicely (because G acts transitively on the orbit), and we can just take everything else' to be another partition... I have quoted the solution you have given. you have defined$\displaystyle $A_0 = A \setminus Gx$.
I agree with the fact that $\displaystyle$A_0 \cap Gx=\emptyset$, because this follows from the definition of$\displaystyle $A_0$.
But i don't agree with $\displaystyle \sigma($A_0) \cap $A_0=\emptyset$. 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 $\displaystyle$A_0$is non-empty( as in this case$\displaystyle Gx \subset A$), and there must be some element$\displaystyle a \in $A_0$ such that $\displaystyle \sigma(a) \notin Gx \text{ } \forall \text{ } \sigma \in G$.

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.

10. Originally Posted by abhishekkgp
I have quoted the solution you have given.
you have defined $\displaystyle$A_0 = A \setminus Gx$. I agree with the fact that$\displaystyle $A_0 \cap Gx=\emptyset$, because this follows from the definition of $\displaystyle$A_0$. But i don't agree with$\displaystyle \sigma($A_0) \cap$A_0=\emptyset$. According to me this is only true if G acts transitively on A. i say this because if G does not act transitively on A then$\displaystyle $A_0$ is non-empty( as in this case $\displaystyle Gx \subset A$), and there must be some element $\displaystyle a \in$A_0$such that$\displaystyle \sigma(a) \notin Gx \text{ } \forall \text{ } \sigma \in G $. May be i am just being a 'block'head here but i am only a newbie at group theory so please help this poor guy. What do you mean by$\displaystyle \sigma(a)$? Do you mean$\displaystyle \sigma(a)=\{a\cdot g: g \in G\}$. As in,$\displaystyle \sigma(a)$is the orbit of$\displaystyle a$. This is, I believe, what is happening. Then by definition,$\displaystyle \sigma(a) \in Gx$for all$\displaystyle a \in A$, and so$\displaystyle \sigma(a) \not\in A_0$for all$\displaystyle a \in A$, as required... 11. Originally Posted by Swlabr What do you mean by$\displaystyle \sigma(a)$? Do you mean$\displaystyle \sigma(a)=\{a\cdot g: g \in G\}$. As in,$\displaystyle \sigma(a)$is the orbit of$\displaystyle a$. This is, I believe, what is happening. Then by definition,$\displaystyle \sigma(a) \in Gx$for all$\displaystyle a \in A$, and so$\displaystyle \sigma(a) \not\in A_0$for all$\displaystyle a \in A$, as required... no no.. when i write$\displaystyle \sigma(a)$i mean$\displaystyle \sigma \cdot a$. the orbit of a will be represented by$\displaystyle Ga$. I wrote$\displaystyle \sigma \cdot a$as$\displaystyle \sigma(a)$because in the question i had mentioned that G is permutation group on the finite set A. now do you understand what i am trying to say?? for example let$\displaystyle S_4$act on {1,2,3,4}. then for$\displaystyle \sigma=\left\{2,3,1,4\right\} \in S_4$we write$\displaystyle \sigma(2)=\sigma \cdot 2=3$12. Originally Posted by abhishekkgp no no.. when i write$\displaystyle \sigma(a)$i mean$\displaystyle \sigma \cdot a$. the orbit of a will be represented by$\displaystyle Ga$. I wrote$\displaystyle \sigma \cdot a$as$\displaystyle \sigma(a)$because in the question i had mentioned that G is permutation group on the finite set A. now do you understand what i am trying to say?? for example let$\displaystyle S_4$act on {1,2,3,4}. then for$\displaystyle \sigma=\left\{2,3,1,4\right\} \in S_4$we write$\displaystyle \sigma(2)=\sigma \cdot 2=3$Oh, I think I understand what you mean now. So just take your blocks to be the individual orbits. These partition your set in the desired fashion. (e.g.$\displaystyle S_4 = \langle (1\:2\:3\:4),(1\:2)\rangle$acts on$\displaystyle \{1, 2, 3, 4, 5, 6\}$and your blocks will be$\displaystyle \{1, 2, 3, 4\}, \{5\}, \{6\}$. 13. Originally Posted by Swlabr Oh, I think I understand what you mean now. So just take your blocks to be the individual orbits. These partition your set in the desired fashion. (e.g.$\displaystyle S_4 = \langle (1\:2\:3\:4),(1\:2)\rangle$acts on$\displaystyle \{1, 2, 3, 4, 5, 6\}$and your blocks will be$\displaystyle \{1, 2, 3, 4\}, \{5\}, \{6\}$.$\displaystyle B=\{1,2,3,4\}$is a block--agreed. but$\displaystyle \sigma_1 \cdot \{1,2,3,4\}, \sigma_2 \cdot \{1,2,3,4\},\ldots,\sigma_{24} \cdot \{1,2,3,4\}$do not form a partition of$\displaystyle \{1,2,3,4,5,6\}$. here$\displaystyle \sigma_i, 1 \leq i \leq 24$are elements of$\displaystyle S_4\$

More over the trivial blocks in this case are {1},{2},{3},{4},{5},{6},{1,2,3,4,5,6}.