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

[Juanriq]

Salutations! I just want to make sure I am on the right track...

Homework Statement

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 $a, b \in \thinspace G$. By assumption, $a = a^{-1}$ and $b = b^{-1}$. Our goal is to show that $ab = ba$. 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
$(ab)^{-1} = \bigl (a^{-1} b^{-1} \bigl )^{-1} = \bigl (b^{-1} \bigl )^{-1} \bigl (a^{-1} \bigl )^{-1} = ba$

Thanks!

 

[deluks917]

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

 

[Juanriq]

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!