Abstract Algebra/Group Theory/Group/Double Inverse

Theorem

Let G be any [[../Definition of a Group|group]] with operation $\ast$.

$\mathrm{\forall }g\in G:[g^{-1}{]}^{-1}=g$

In Group G, inverse of inverse of any element g is g.

Proof

0. Choose $g\in G$
1. $\mathrm{\exists }{g}^{-1}\in G:g\ast g^{-1}=g^{-1}\ast g=e_G$	[[../Definition of a Group/Definition of Inverse#Usage1|definition of inverse of g in G (usage 1,3)]]
2. $g\ast a=a\ast g=e_G$	let a = g−1
3. $a\ast g=g\ast a=e_G$
4. $[a{]}^{-1}=g$	[[../Definition of a Group/Definition of Inverse#Usage2|definition of inverse of a in G (usage 2)]]
5. $[g^{-1}{]}^{-1}=g$	as a = g−1

Diagrams

1. inverse of filled circle is empty circle.

2. inverse of empty circle is filled circle, given 1.