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Proof of abelian-ness if every element is also its own inverse

  1. Oct 15, 2010 #1
    Salutations! I just want to make sure I am on the right track...

    1. The problem statement, all variables and given/known data

    Let G be a group. Prove that if every element in a group is equal to its own inverse, then G is abelian

    2. The attempt at a solution

    Pick two elements [itex] a, b \in \thinspace G[/itex]. By assumption, [itex]a = a^{-1}[/itex] and [itex] b = b^{-1}[/itex]. Our goal is to show that [itex] ab = ba[/itex]. By our assumption, the following holds
    ab = a^{-1}b^{-1}
    Since G is a group, it is closed under inverses, therefore we can take the inverse of both sides
    (ab)^{-1} = \bigl (a^{-1} b^{-1} \bigl )^{-1} = \bigl (b^{-1} \bigl )^{-1} \bigl (a^{-1} \bigl )^{-1} = ba

    Last edited: Oct 15, 2010
  2. jcsd
  3. Oct 15, 2010 #2
    ab is an element of G, what does that mean (ab)^-1 is by assumption?
  4. Oct 15, 2010 #3
    Oh! By assumption [itex] ab = (ab)^{-1} [\latex] as well. That definitely makes this a simple one-liner when I apply the inverse and initial assumpition to the right side. Thanks!
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