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

by Juanriq
Tags: abelianness, element, inverse, proof
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Oct15-10, 10:15 AM
P: 42
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

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Oct15-10, 10:39 AM
P: 367
ab is an element of G, what does that mean (ab)^-1 is by assumption?
Oct15-10, 12:31 PM
P: 42
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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