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

• Jan 31st 2011, 08:55 AM
abhishekkgp
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)$are all distinct images of B under the elements of G, then these form a partition of A. Can the same be proved if G is not transitive on the finite set A. • Jan 31st 2011, 12:17 PM Drexel28 Quote: Originally Posted by abhishekkgp Let G be a transitive permutation group on the finite set A. a 'block' is a non-empty subset B of A such that for all$\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?
• Feb 1st 2011, 12:07 AM
Swlabr
Quote:

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$...does that make sense?...Basically, every element of G takes everything in A and sticks it somewhere within the orbit. So the orbit is partitioned nicely (because G acts transitively on the orbit), and we can just take everything else' to be another partition... • Feb 1st 2011, 05:29 AM abhishekkgp Quote: Originally Posted by Drexel28 Your terminology is unfamiliar to me. Are you saying that$\displaystyle G$acts on$\displaystyle A$transitively? Is that what your first sentence means? yes, i mean the same. I have taken the question from the text on abstract algebra by dummit and foote. • Feb 1st 2011, 05:51 AM abhishekkgp Quote: Originally Posted by Swlabr Yes, if I understand your definition of a block' correctly. I mean, let$\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.
• Feb 1st 2011, 09:00 AM
Swlabr
Quote:

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$. is this correct too?? If these two are correct then i will be able to understand your post. THANKS IN ADVANCE. Yes, these are correct. Can you see why they are? • Feb 1st 2011, 08:16 PM abhishekkgp Quote: Originally Posted by Swlabr Yes, these are correct. Can you see why they are? yes swlabr. i can see why they are correct. I infer that your solution uses the fact that G acts transitively on A. does this mean that the distinct images of the block B$\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.
• Feb 1st 2011, 11:56 PM
Swlabr
Quote:

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$. So your blocks are the partitions of G_x along with A_0. • Feb 2nd 2011, 02:54 AM abhishekkgp Quote: Originally Posted by Swlabr Yes, if I understand your definition of a block' correctly. I mean, let$\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.(Doh)(Happy)
• Feb 2nd 2011, 04:38 AM
Swlabr
Quote:

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 $. May be i am just being a 'block'head here but i am only a newbie at group theory so please help this poor guy.(Doh)(Happy) What do you mean by$\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$, as required... • Feb 2nd 2011, 09:23 AM abhishekkgp Quote: Originally Posted by Swlabr What do you mean by$\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$• Feb 2nd 2011, 11:55 PM Swlabr Quote: Originally Posted by abhishekkgp no no.. when i write$\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\}$. • Feb 3rd 2011, 12:39 AM abhishekkgp Quote: Originally Posted by Swlabr Oh, I think I understand what you mean now. So just take your blocks to be the individual orbits. These partition your set in the desired fashion. (e.g.$\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}.