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

• Jan 31st 2011, 09:55 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 $\sigma \text \in G$ either $\displaystyle \sigma (B)=B$ or $\sigma(B) \cap B=\phi$. (here $\sigma(B)$ is the set $\left\{\sigma(b)|b \in B \right\}$).

It can be easily shown that if B is a block and ${\sigma}_1 (B),\text{} {\sigma}_2 (B), \ldots, \text{} {\sigma}_n (B)$ 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, 01:17 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 $\sigma \text \in G$ either $\displaystyle \sigma (B)=B$ or $\sigma(B) \cap B=\phi$. (here $\sigma(B)$ is the set $\left\{\sigma(b)|b \in B \right\}$).

It can be easily shown that if B is a block and ${\sigma}_1 (B),\text{} {\sigma}_2 (B), \ldots, \text{} {\sigma}_n (B)$ 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 $G$ acts on $A$ transitively? Is that what your first sentence means?
• Feb 1st 2011, 01:07 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 $\sigma \text \in G$ either $\displaystyle \sigma (B)=B$ or $\sigma(B) \cap B=\phi$. (here $\sigma(B)$ is the set $\left\{\sigma(b)|b \in B \right\}$).

It can be easily shown that if B is a block and ${\sigma}_1 (B),\text{} {\sigma}_2 (B), \ldots, \text{} {\sigma}_n (B)$ 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 $A_0 = A\setminus Gx$, where $Gx$ is the orbit of $G$. Then clearly $\sigma(A_0)\cap A_0 = \emptyset$, because $A_0 \cap Gx=\emptyset$...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, 06:29 AM
abhishekkgp
Quote:

Originally Posted by Drexel28
Your terminology is unfamiliar to me. Are you saying that $G$ acts on $A$ 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, 06:51 AM
abhishekkgp
Quote:

Originally Posted by Swlabr
Yes, if I understand your definition of a block' correctly. I mean, let $A_0 = A\setminus Gx$, where $Gx$ is the orbit of $G$. Then clearly $\sigma(A_0)\cap A_0 = \emptyset$, because $A_0 \cap Gx=\emptyset$...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 $A_0$ is non-empty. is this correct??
Then $\sigma (A_0) \cap A_0 \neq \phi$. is this correct too??

If these two are correct then i will be able to understand your post.
• Feb 1st 2011, 10:00 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 $A_0$ is non-empty. is this correct??
Then $\sigma (A_0) \cap A_0 \neq \phi$. is this correct too??

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

Yes, these are correct. Can you see why they are?
• Feb 1st 2011, 09:16 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 ${\sigma}_1 (B),\text{} {\sigma}_2 (B), \ldots, \text{} {\sigma}_n (B)$ under the elements of G do not necessarily form a partition of A if G does not act transitively on A.
• Feb 2nd 2011, 12:56 AM
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 ${\sigma}_1 (B),\text{} {\sigma}_2 (B), \ldots, \text{} {\sigma}_n (B)$ 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 $G_x = A\setminus A_0$. So your blocks are the partitions of G_x along with A_0.
• Feb 2nd 2011, 03:54 AM
abhishekkgp
Quote:

Originally Posted by Swlabr
Yes, if I understand your definition of a block' correctly. I mean, let $A_0 = A\setminus Gx$, where $Gx$ is the orbit of $G$. Then clearly $\sigma(A_0)\cap A_0 = \emptyset$, because $A_0 \cap Gx=\emptyset$...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 $A_0 = A \setminus Gx$.
I agree with the fact that $A_0 \cap Gx=\emptyset$, because this follows from the definition of $A_0$.
But i don't agree with $\sigma(A_0) \cap A_0=\emptyset$. 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 $A_0$ is non-empty( as in this case $Gx \subset A$), and there must be some element $a \in A_0$ such that $\sigma(a) \notin Gx \text{ } \forall \text{ } \sigma \in G$.

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, 05:38 AM
Swlabr
Quote:

Originally Posted by abhishekkgp
I have quoted the solution you have given.
you have defined $A_0 = A \setminus Gx$.
I agree with the fact that $A_0 \cap Gx=\emptyset$, because this follows from the definition of $A_0$.
But i don't agree with $\sigma(A_0) \cap A_0=\emptyset$. 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 $A_0$ is non-empty( as in this case $Gx \subset A$), and there must be some element $a \in A_0$ such that $\sigma(a) \notin Gx \text{ } \forall \text{ } \sigma \in G$.

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 $\sigma(a)$? Do you mean $\sigma(a)=\{a\cdot g: g \in G\}$. As in, $\sigma(a)$ is the orbit of $a$. This is, I believe, what is happening.

Then by definition, $\sigma(a) \in Gx$ for all $a \in A$, and so $\sigma(a) \not\in A_0$ for all $a \in A$, as required...
• Feb 2nd 2011, 10:23 AM
abhishekkgp
Quote:

Originally Posted by Swlabr
What do you mean by $\sigma(a)$? Do you mean $\sigma(a)=\{a\cdot g: g \in G\}$. As in, $\sigma(a)$ is the orbit of $a$. This is, I believe, what is happening.

Then by definition, $\sigma(a) \in Gx$ for all $a \in A$, and so $\sigma(a) \not\in A_0$ for all $a \in A$, as required...

no no.. when i write $\sigma(a)$ i mean $\sigma \cdot a$. the orbit of a will be represented by $Ga$.

I wrote $\sigma \cdot a$ as $\sigma(a)$ 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 $S_4$ act on {1,2,3,4}. then for $\sigma=\left\{2,3,1,4\right\} \in S_4$ we write $\sigma(2)=\sigma \cdot 2=3$
• Feb 3rd 2011, 12:55 AM
Swlabr
Quote:

Originally Posted by abhishekkgp
no no.. when i write $\sigma(a)$ i mean $\sigma \cdot a$. the orbit of a will be represented by $Ga$.

I wrote $\sigma \cdot a$ as $\sigma(a)$ 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 $S_4$ act on {1,2,3,4}. then for $\sigma=\left\{2,3,1,4\right\} \in S_4$ we write $\sigma(2)=\sigma \cdot 2=3$

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. $S_4 = \langle (1\:2\:3\:4),(1\:2)\rangle$ acts on $\{1, 2, 3, 4, 5, 6\}$ and your blocks will be $\{1, 2, 3, 4\}, \{5\}, \{6\}$.
• Feb 3rd 2011, 01:39 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. $S_4 = \langle (1\:2\:3\:4),(1\:2)\rangle$ acts on $\{1, 2, 3, 4, 5, 6\}$ and your blocks will be $\{1, 2, 3, 4\}, \{5\}, \{6\}$.

$B=\{1,2,3,4\}$ is a block--agreed. but $\sigma_1 \cdot \{1,2,3,4\}, \sigma_2 \cdot \{1,2,3,4\},\ldots,\sigma_{24} \cdot \{1,2,3,4\}$ do not form a partition of $\{1,2,3,4,5,6\}$. here $\sigma_i, 1 \leq i \leq 24$ are elements of $S_4$

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