Register to reply

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

by Juanriq
Tags: abelianness, element, inverse, proof
Share this thread:
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

Phys.Org News Partner Science news on
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
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!

Register to reply

Related Discussions
Stuck, finding inverse in element in ring Z Calculus & Beyond Homework 5
Can I calculate the (multiplicative) inverse of any element in a cyclic group? Calculus & Beyond Homework 3
In a finite group G, the inverse of each element is a power of itself. Calculus & Beyond Homework 2
Abelian group proof Calculus & Beyond Homework 19
Free abelian group proof help Linear & Abstract Algebra 4