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Proof of abelianness if every element is also its own inverse 
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#1
Oct1510, 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 [itex] ab = a^{1}b^{1} [/itex] Since G is a group, it is closed under inverses, therefore we can take the inverse of both sides [itex] (ab)^{1} = \bigl (a^{1} b^{1} \bigl )^{1} = \bigl (b^{1} \bigl )^{1} \bigl (a^{1} \bigl )^{1} = ba [/itex] Thanks! 


#2
Oct1510, 10:39 AM

P: 367

ab is an element of G, what does that mean (ab)^1 is by assumption?



#3
Oct1510, 12:31 PM

P: 42

Oh! By assumption [itex] ab = (ab)^{1} [\latex] as well. That definitely makes this a simple oneliner when I apply the inverse and initial assumpition to the right side. Thanks!



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