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Homework Help: Induced binary operator

  1. Feb 6, 2010 #1
    1. The problem statement, all variables and given/known data
    suppose that * is an associative binary operation on a set S. Let H = {a elementof S | a*x = x*a forall x elementof S}. Show that H is closed under *.


    2. Relevant equations



    3. The attempt at a solution
    i don't really know where to begin. I know i need to show forall a,b elementof H, a*b elementof H. i know * is associative, and H is the subset of all commutive elements of S. (the book says, "we think of H as consisting of all elements of S that commute with every element is S." same thing?) where do i go from here? what does * being associative have to do with commutive elements?
     
  2. jcsd
  3. Feb 6, 2010 #2

    vela

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    Let [itex]a, b \in H[/itex]. To show closure, you want to prove that [itex]a*b \in H[/itex]. In other words, you want to show that for all [itex]x \in S[/itex], [itex](a*b)*x = x*(a*b)[/itex]. Make sense?
     
  4. Feb 6, 2010 #3
    so could i say:
    since * is associative, then for all a, b, x in S, (a*b)*x = a*(b*x), and since H is the set of all commutative elements, then (a*b)*x = a*(b*x) = x*(b*a) = x*(a*b), and H is closed under S. does that make sense, or did i just reiterate what you said? ( i guess i want to reiterate what you said with more detail.)
     
  5. Feb 6, 2010 #4

    vela

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    Yeah, essentially, but you need to pay more attention to the details. The statement only holds for a, b in H, not in S generally, but it needs to hold for all x in S. Also, just do one step at a time so it's clear that each step is justified. When you wrote a*(b*x)=x*(b*a), how do you know x and a can switch places like that? You are using neither commutativity nor associativity alone, so it's not immediately apparent to the reader that the step is correct.
     
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