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Math Help - Group actions

  1. #1
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    Group actions

    Let A be a nonempty set and k\in\mathbb{N} with k\leq |A|. The symmetric group S_A acts on B consisting of all subsets of cardinality k by \sigma \{a_1,\cdots ,a_k\}=\{\sigma(a_1),\cdots , \sigma(a_k)\}.

    Prove that this is a group action.

    \sigma_{g_1g_2}(a)=(g_1g_2)(a)=g_1(g_2(a))=\sigma_  {g_1}(\sigma_{g_2}(a))

    Is this what it is asking for?
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  2. #2
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    Re: Group actions

    um,... what do you mean by "a"?
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    Re: Group actions

    Quote Originally Posted by Deveno View Post
    um,... what do you mean by "a"?
    a I would assume is an element.
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    Re: Group actions

    Quote Originally Posted by dwsmith View Post
    a I would assume is an element.
    An element of what?
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    Re: Group actions

    Quote Originally Posted by emakarov View Post
    An element of what?
    Probably of A.
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  6. #6
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    Re: Group actions

    when discussing math, it never hurts to define your terms, so as to be perfectly clear what you mean. at worst, you will take up more space than necessary, saying things that are already understood. but you avoid vagueness, and lessen the risk of misunderstanding.

    i'm not in your course, i'm not reading your notes, or privy to the notational shorthand used therein. and contrary to public belief, there is no one "standard way" of expressing mathematical concepts (although certain conventions are wide-spread).

    let's look at an example of the construction, here.

    suppose A = {a1,a2,a3,a4}, then Sym(A) is a group isomorphic to S4. let's say that k = 2, so:

    B = {{a1,a2},{a1,a3},{a1,a4},{a2,a3},{a2,a4},{a3,a4}} = {b1,b2,b3,b4,b5,b6}

    so let's say σ is the element (a1 a2) in Sym(A). what is the associated mapping b-->σb in B?

    well, {σ(a1),σ(a2)} = {a2,a1} = {a1,a2}. so we have b1-->b1 that is:

    σb1 = b1. similarly {σ(a1), σ(a3)} = {a2,a3} = b4, so σb2 = b4. so σ inducees the permutation on B:

    b1-->b1
    b2-->b4
    b3-->b5
    b4-->b2
    b5-->b3
    b6-->b6, which is the permutation (b2 b4)(b3 b5).

    do you see what is happening here?
    Last edited by Deveno; October 8th 2011 at 10:39 AM.
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  7. #7
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    Re: Group actions

    Quote Originally Posted by Deveno View Post
    when discussing math, it never hurts to define your terms, so as to be perfectly clear what you mean. at worst, you will take up more space than necessary, saying things that are already understood. but you avoid vagueness, and lessen the risk of misunderstanding.

    i'm not in your course, i'm not reading your notes, or privy to the notational shorthand used therein. and contrary to public belief, there is no one "standard way" of expressing mathematical concepts (although certain conventions are wide-spread).
    This is word for word from Abstract Algebra by Dummit and Foote. I give you verbatim the text. I am not told where a is from either. But my guess is a would be in A.
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  8. #8
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    Re: Group actions

    suppose |A| = n. then we start with a group isomorphic to Sn. if |B| = k, we end up with an element in Sm, where m = n!/(k!(n-k)!

    you can do one of 2 things: show the subgroup of Sm we end up with is a quotient group (homomorphic image) of Sn (this is the "abstract approach")

    or show that the "G-set multiplication" is associative, with identity id_A (the "concrete approach"). what's your poison?
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  9. #9
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    Re: Group actions

    Quote Originally Posted by Deveno View Post
    suppose |A| = n. then we start with a group isomorphic to Sn. if |B| = k, we end up with an element in Sm, where m = n!/(k!(n-k)!
    By assumption, B consists of all subsets of A of cardinality k, so |B| = {n\choose k}=n!/(k!(n-k)!=m, and an element of A produces (through action) a permutation of B, i.e., an element of S_m.

    Quote Originally Posted by Deveno View Post
    you can do one of 2 things: show the subgroup of Sm we end up with is a quotient group (homomorphic image) of Sn (this is the "abstract approach")

    or show that the "G-set multiplication" is associative, with identity id_A (the "concrete approach").
    I think we are moving too fast here. The question so far is whether S_A (or \mbox{Sym}_A) acts on A or B, i.e., whether a\in A or a\in B in the equation (\sigma_1\sigma_2)(a)=\sigma_1(\sigma_2(a)).

    The problem is clear that S_A acts on B. Therefore, it asks to prove that

    (\sigma_1\sigma_2)(b)=\sigma_1(\sigma_2(b)) for all b\in B

    and

    e(b) = b for all b\in B.

    Here \sigma_1,\sigma_2\in S_A, \sigma_1\sigma_2 is the result of the group operation on \sigma_1 and \sigma_2 and \sigma_2(b) is the result of the group action of \sigma_2 on b. And, of course, e is the unit of the group S_A.

    Quote Originally Posted by Deveno View Post
    what's your poison?
    Oh, no...
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  10. #10
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    Re: Group actions

    i've never read Dummit and Foote, some people love it, some people hate it. from what i gather, it takes the "abstract approach" rather hard-core, which is probably the source of dwsmith's bewilderment. i'm not trying to "move too fast", the way i would prefer to handle group actions (especially on a "first-look") is by seeing it as a way to introduce group multiplication into a set, which appears to be at odds with the way that his text is dealing with it.

    which is what emarkov is doing in his post, good job
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