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Math Help - |x|=|gxg^{-1}|

  1. #1
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    |x|=|gxg^{-1}|

    Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by x\mapsto gxg^{-1}. For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and gxg^{-1} have the same order for all x in G and that for any subset A of G, |A|=|gAg^{-1}|.

    I have shown this an isomorphism. I am not sure how to show x and gxg^{-1} have the same order.
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    MHF Contributor Drexel28's Avatar
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    Re: |x|=|gxg^{-1}|

    Quote Originally Posted by dwsmith View Post
    Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by x\mapsto gxg^{-1}. For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and gxg^{-1} have the same order for all x in G and that for any subset A of G, |A|=|gAg^{-1}|.

    I have shown this an isomorphism. I am not sure how to show x and gxg^{-1} have the same order.
    If \phi is an isomorphism then \phi(x)^{|x|}=\phi(x^{|x|})=\phi(e)=e and e=\phi^{-1}(e)=\phi^{-1}(\phi(x)^{|\phi(x)|})=x^{|\phi(x)|} imply |\phi(x)|\mid |x| and |x|\mid |\phi(x)| respectively....so.
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  3. #3
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    Re: |x|=|gxg^{-1}|

    there are different ways to do this:

    1). suppose that x has order n. then x^n = e, so (gxg^-1)^n = (gxg^-1)(gxg^-1).....(gxg^-1) (n times)

    = gx(g^-1g)x(g^-1g).....xg^-1 = (gx)(x)(x)....(x)g^-1 = g(x^n)g^-1 = geg^-1 = gg^-1 = e.

    (ok, technically using induction on n would be better, but you get the idea). thus |gxg^-1| divides n.

    now suppose that |gxg^-1| = k < n. then (gxg^-1)^k = g(x^k)g^-1 = e, so x^k = g^-1g = e, a contradiction.

    hence |gxg^-1| = |x|.

    2). since x-->gxg^-1 is an isomorphism, the order of gxg^-1 must be the order of x. why? suppose not.

    case 2a) |x| < |gxg^-1|. in this case we can find a k with x^k = e, but (gxg^-1)^k ≠ e.

    using our isomorphism e = geg^-1 = g(x^k)g^-1 = (gxg^-1)^k, contradicting our choice of k.

    case 2b) |gxg^-1| < |x|. then we have (gxg^-1)^k = e, but x^k ≠ e, for some k.

    since e = (gxg^-1)^k = g(x^k)g^-1, we have that x^k is in the kernel of our isomorphism.

    but since an isomorphism is injective, the kernel is {e}, contradiction.

    thus |gxg^-1| = |x|.

    now, since x-->gxg^-1 is an isomorphism, it is bijective, so |A| and |gAg^-1| have to be equal.
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    Re: |x|=|gxg^{-1}|

    Quote Originally Posted by Drexel28 View Post
    \phi^{-1}(\phi(x)^{|\phi(x)|})
    Why is this phi of x to the order of phi of x?
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