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

  1. #1
    Senior Member abhishekkgp's Avatar
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    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.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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?
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    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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...
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Drexel28 View Post
    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.
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    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.
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  6. #6
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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?
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  7. #7
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    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.
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  8. #8
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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.
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  9. #9
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    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.
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  10. #10
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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...
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  11. #11
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    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
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  12. #12
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    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\}.
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    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by Swlabr View Post
    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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