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Thread: 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 $\displaystyle C(A)$ is a subgroup.
    This should be a straightforward calculation.

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

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

    Show $\displaystyle |C(G)| > 1$
    This result is known as Burnside's Lemma.
    Note, $\displaystyle |G| = |C(G)| + \sum_{x} |G|/|C(\{ x\})|$.
    Now since $\displaystyle |C(\{ x\})| \not = p^r$ it means $\displaystyle |G|/|C\{ x\}| > 1$. Thus, the sum is divisible by $\displaystyle p$ and $\displaystyle |G| = p^r$ is divisible by $\displaystyle p$. It must mean that $\displaystyle |C(G)|$ is divisible by $\displaystyle p$.
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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 $\displaystyle G$ be a finite group. Define a relation $\displaystyle x- y$ if and only if "$\displaystyle x$ is conjugate to $\displaystyle y$".
    And "conjugate" means there is $\displaystyle g\in G$ so that $\displaystyle x=gyg^{-1}$.
    It is easy to check that $\displaystyle x-x$, i.e. "$\displaystyle x$ is conjugate to itself".
    Also $\displaystyle x-y$ implies $\displaystyle y-x$.
    And finally, if $\displaystyle x-y$ and $\displaystyle y-z$ implies $\displaystyle x-z$.
    This tells us that $\displaystyle -$ is an equivalence relation on $\displaystyle G$.

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

    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, $\displaystyle |G| = \sum_{x}| \text{conj}(x)|$ where $\displaystyle \text{conj}(x)$ is the conjugacy class of $\displaystyle x$, and the sum ranges over different conjugacy classes.

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

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

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

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

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

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

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