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Math Help - prove H_T less than or equal to S_n

  1. #1
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    prove H_T less than or equal to S_n

    Let T=[(a_1, a_2,...,a_t) C {1,2,...n}] and H_T={beta is a member of S_n|beta(a_i)=a_i, for all a_i which is a member of T}. H_t is called the stabilizer of T in S_n, stab(T). Prove that H_T < S_n. What is the order of H_T
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    do you understand what H_T does?
    Quote Originally Posted by mandy123 View Post
    Let T=[(a_1, a_2,...,a_t) C {1,2,...n}] and H_T={beta is a member of S_n|beta(a_i)=a_i, for all a_i which is a member of T}. H_t is called the stabilizer of T in S_n, stab(T). Prove that H_T < S_n.
    it suffices to show that H_T is a subgroup of S_n. or do you know that and are wondering how to do it?

    you must show that H_T is nonempty. (consider the identity function)

    then show that it is closed under the composition of functions

    then show that each element has an inverse

    which of these do you have problems with?

    What is the order of H_T
    H_T leaves the elements in T fixed and permutes the other elements. how many ways can we permute the remaining elements?
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  3. #3
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    Unhappy

    ok to say it is nonempty:
    beta(a_i)=a_i, so a_i^2=a_i So H is nonempty?

    to show closure:
    x,y is a member of H_t
    beta(x)=a_i and beta(y)=a_i

    Inverse: Dont know how to do this part

    Did I get the other 2 right?
    I dont understand this stuff.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by mandy123 View Post
    ok to say it is nonempty:
    beta(a_i)=a_i, so a_i^2=a_i So H is nonempty?

    to show closure:
    x,y is a member of H_t
    beta(x)=a_i and beta(y)=a_i

    Inverse: Dont know how to do this part

    Did I get the other 2 right?
    I dont understand this stuff.
    note that the identity function is in H_T so that it is not empty


    to show closure, we need to show that if \alpha (a_i) ,~\beta (a_i) \in H_T then (without loss of generality) \beta \circ \alpha \in H_T

    so assume \alpha , ~\beta \in H_T. then \alpha (a_i) = a_i and \beta (a_i) = a_i

    thus ( \beta \circ \alpha ) (a_i) = \beta ( \alpha (a_i)) = \beta (a_i) = a_i

    so that \beta \circ \alpha \in H_T since ( \beta \circ \alpha )(a_i) = a_i


    For inverses, note that if \alpha \in H_T then \alpha ^{-1} \in S_n works as the inverse of \alpha, and it is also in H_T, since if we assume \alpha \in H_T, then

    \alpha (a_i) = a_i

    \Rightarrow \alpha ^{-1} (\alpha (a_i)) = \alpha ^{-1} (a_i)

    \Rightarrow a_i = \alpha ^{-1}(a _ i)

    so that \alpha ^{-1} \in H_T

    thus, H_T \le S_n
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