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SUBSET K of elements in a group with finite distinct distinct conjugates

  1. Nov 30, 2011 #1
    WTS, is that such set is a subgroup.

    I need to show closure under group operation and inverse.

    I can do the inverse which is usually the hardest part, but I'm stuck on the grp op.

    So let a in K and b in K, both have finite distinct conjugates. Their conjugates are in the group too. WTS is that ab in K too.

    if a, b in K then xax-1 = c in K and yay-1

    consider xy(ab)(xy)-1 note since xax-1 and yay-1 are finite and distinct then x and y are finite and distinct hence xy and its inverse is finite

    hence xy(ab)(xy)-1 in K

    what do you think?
  2. jcsd
  3. Dec 1, 2011 #2


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    Science Advisor

    it looks like you're confusing elements and sets.

    the SET of all conjugates of a {xax-1: x in G} is finite, not the conjugates themselves.

    so you want to show that if the set of conjugates of a is finite, and the set of conjugates of b is finite, so is the set of conjugates of ab.

    it might be helpful to note that if x is ANY element of G:

    x(ab)x-1 = (xax-1)(xbx-1).
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