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

**abhishekkgp** · January 31st 2011, 08:55 AM

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.

**Drexel28** · January 31st 2011, 12:17 PM

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?

**Swlabr** · February 1st 2011, 12:07 AM

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...

**abhishekkgp** · February 1st 2011, 05:29 AM

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.

**abhishekkgp** · February 1st 2011, 05:51 AM

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.
THANKS IN ADVANCE.

**Swlabr** · February 1st 2011, 09:00 AM

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.
THANKS IN ADVANCE.

Yes, these are correct. Can you see why they are?

**abhishekkgp** · February 1st 2011, 08:16 PM

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.

**Swlabr** · February 1st 2011, 11:56 PM

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.

**abhishekkgp** · February 2nd 2011, 02:54 AM

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.

**Swlabr** · February 2nd 2011, 04:38 AM

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.

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...

**abhishekkgp** · February 2nd 2011, 09:23 AM

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 $\sigma(2)=\sigma \cdot 2=3$

**Swlabr** · February 2nd 2011, 11:55 PM

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 $\{1, 2, 3, 4\}, \{5\}, \{6\}$ .

**abhishekkgp** · February 3rd 2011, 12:39 AM

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}.

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