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Math Help - prove by orbit-stabilizer

  1. #1
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    prove by orbit-stabilizer



    i got stuck on part three
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  2. #2
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    Prove C(A) is a subgroup.
    This should be a straightforward calculation.

    Show \rho: G\times G\mapsto G defines an action on itself.
    Again a straightforward calculation, just look at definition of group action.

    |G| = |C(G)|+\sum_{x}|G|/|C(\{ x\})|
    This is just the conjugacy class equation.
    Here C(G) = Z(G), i.e. the group center. While [G:C(\{x\})] = |G|/C(\{ x\}). And this follows because x is a sum which goes through all the non-trivial conjugacy classes, i.e. when the orbit is greater than 1.

    Show |C(G)| > 1
    This result is known as Burnside's Lemma.
    Note, |G| = |C(G)| + \sum_{x} |G|/|C(\{ x\})|.
    Now since |C(\{ x\})| \not =  p^r it means |G|/|C\{ x\}| > 1. Thus, the sum is divisible by p and |G| = p^r is divisible by p. It must mean that |C(G)| is divisible by p.
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  3. #3
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    please give proof

    please give proof of the last two parts...
    i m exactly confused with that...
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  4. #4
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    Quote Originally Posted by szpengchao View Post
    please give proof of the last two parts
    Let G be a finite group. Define a relation x- y if and only if " x is conjugate to y".
    And "conjugate" means there is g\in G so that x=gyg^{-1}.
    It is easy to check that x-x, i.e. " x is conjugate to itself".
    Also x-y implies y-x.
    And finally, if x-y and y-z implies x-z.
    This tells us that - is an equivalence relation on G.

    Let x\in G and define [x] = \{ a\in G | x-a\}, more simply it is the set of all elements conjugate with x.
    This is called the "conjugacy class of x".

    It should be known to you that equivalence relations gives rise to partitions.
    Meaning every element is in some conjugacy class and that distinct conjugacy classes are disjoint.
    Therefore, |G| = \sum_{x}| \text{conj}(x)| where \text{conj}(x) is the conjugacy class of x, and the sum ranges over different conjugacy classes.

    Now let us take out all the single element conjugacy classes.
    Thus. |G| = |A| + \sum_{x}|\text{conj}(x)| where x is a sum over conjugacy classes with more than one element and A is the set of all single element conjugacy classes.

    Let us examine A in more detail. Note that a is conjugate to itself only i.e. its conjugacy class consists only of itself if and only if gag^{-1} = a for all g. Thus A is precisely C(G).

    Thus, far we have, |G| = |C(G)| + \sum_{x} |\text{conj}(x)|.

    We are going to state a theorem: there is a bijection between the element of the conjugacy class and the cosets of C(\{ x\}).

    The proof is not difficult. Define \phi: \text{conj}(x) \mapsto G/C(\{ x\}) as \phi (a) = aC(\{ x\}). Try that, I think that will work.

    Thus, this tells us that |\text{conj}(x)| = |G/C(\{x\})| = |G|/|C(\{x\}).
    The last equality follows by Lagrange's theorem.

    Thus, |G| = |C(G)| + \sum_{x} |G|/|C(\text{x})|.
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