- #1

Juanriq

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

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

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!

## 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 [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!

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