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Show that given conditions, element is in center of group G

  1. Oct 13, 2018 #1
    1. The problem statement, all variables and given/known data
    Let ##G## be a finite group and ##m## a positive integer which is relatively prime to ##|G|##. If ##b\in G## and ##a^mb=ba^m## for all ##a\in G##, show that ##b## is in the center of ##G##.

    2. Relevant equations


    3. The attempt at a solution
    Let ##|G| = n## and ##b\in G##. Note that by Bézout 's identity ##nx + my = 1## for some ##x,y\in\mathbb{Z}##. Also, note that ##a^m = ba^mb^{-1}##. So

    $$
    \begin{align*}
    a^m &= ba^mb^{-1}\\
    (a^m)^y&= (ba^mb^{-1})^y\\
    a^{my} &= ba^{my}b^{-1}\\
    a^{1-nx} &= ba^{1-nx}b^{-1}\\
    a(a^n)^{-x}&= ba(a^n)^{-x}b^{-1}\\
    a(e)^{-x}&= ba(e)^{-x}b^{-1}\\
    a&= bab^{-1}\\
    \end{align*}
    $$

    Since ##a## is an arbitrary element of ##G##, we see that ##b\in Z(G)## ☐

    Is this the correct argument?
     
  2. jcsd
  3. Oct 13, 2018 #2

    fresh_42

    User Avatar
    2018 Award

    Staff: Mentor

    Yes. Because ##m## and ##n## are coprime, ##a## generates the same elements as ##a^m## so they can be interchanged.
     
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