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Math Help - Subsets and subgroups

  1. #1
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    Subsets and subgroups

    Really struggling on where to start with this question:

    a. Let G be a group, let H be a subgroup of G and let g be a fixed element of G. Show that the subset gHg-1 = {ghg-1: h ϵ H} is also a subgroup of G.

    b. Suppose g, x ϵ G. Show that (gxg-1 )n =gxng-1 for all n ϵ Z, and hence (or otherwise) that o(gxg-1) = o(x)

    Any help would be really appreciated thanks
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  2. #2
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    Re: Subsets and subgroups

    Quote Originally Posted by NathanBUK View Post
    a. Let G be a group, let H be a subgroup of G and let g be a fixed element of G. Show that the subset gHg-1 = {ghg-1: h ϵ H} is also a subgroup of G.
    Here is a useful theorem on subgroups.
    If G is a group and H\subseteq H then H is a subgroup if annd only if \{x,y\}\subseteq H impies xy^{-1}\in H.

    Take x=gh_1g^{-1}~\&~h=gh_2g^{-1}.
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  3. #3
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    Re: Subsets and subgroups

    for part b): if n > 0

    (gxg^{-1})^n = (gxg^{-1})(gxg^{-1})\cdots(gxg^{-1}) (n times)

     = gx(g^{-1}g)x(g^{-1}g)\cdots xg^{-1} = gxexe\cdots xg^{-1} = gx^ng^{-1}

    if n = 0:

    (gxg^{-1})^0 = e = gg^{-1} = geg^{-1} = gx^0g^{-1}.

    if n < 0, so n = -k, for k > 0:

    (gxg^{-1})^n = (gxg^{-1})^{-k} = ((gxg^{-1})^{-1})^k = ((g^{-1})^{-1}x^{-1}g^{-1})^k

     = (gx^{-1}g^{-1})^k = g(x^{-1})^kg^{-1} = gx^{-k}g^{-1} = gx^ng^{-1}

    now if |x| = n, then xn = e, so (gxg-1)n = gxng-1 = geg-1 = gg-1 = e.

    this means that |gxg-1| is at most n.

    suppose that (gxg-1)k = e, for some 0 < k < n.

    then gxk = eg = g, whence xk = g-1g = e, contradicting the fact that the order of x is n.

    thus n is the smallest positive integer for which (gxg-1)n = e, so is the order of gxg-1.
    Last edited by Deveno; February 7th 2013 at 09:58 AM.
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