1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centralizer proof

  1. Mar 31, 2010 #1
    1. The problem statement, all variables and given/known data
    If a is an element of a group and |a|=n, prove that C(a)=C(a^k) when k is relatively prime to n.

    2. Relevant equations
    If n and k are relatively prime, then there exists integers s and t such that ns+kt=1.
    The centralizer a in G, C(a) is the set of all elements in a group G that commute with a. C(a)={g an element of G|ga-ag}

    3. The attempt at a solution
    I tried proving it by double containment, but I couldn't show that C(a^K) is contained in C(a). Should I try a proof by contradiction? I will be grateful for any help.
  2. jcsd
  3. Mar 31, 2010 #2
    I think double containment is the easiest method (I have never heard the term double containment, but I'm assuming it refers to proving [itex]C(a) \subseteq C(a^K)[/itex] and [itex]C(a^K) \subseteq C(a)[/itex]).

    You want to show that if b is in C(a^K), then b is in C(a). In other words assume:
    [tex]ba^Kb^{-1} = a^K[/itex]
    i.e. a^K and b commute. You then wish to show [itex]bab^{-1}=a[/itex]. Here I have just stated the goal so we're clear about that.

    To say that K and n are relatively prime is equivalent to saying that there exists integers x, y such that xK+yn=1. Now you have:
    [tex](bab^{-1})^1 = (bab^{-1})^{xK+yn}[/tex]
    so it suffices to show:
    [tex](bab^{-1})^{xK+yn} = a[/tex]
    See if you can do this. Remember that you have:
    [tex]ba^K b^{-1} = a^K \qquad a^n = 1[/tex]
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook