# Inverse of Group Inverse

## Theorem

Let $\struct {G, \circ}$ be a group.

Let $g \in G$, with inverse $g^{-1}$.

Then:

$\paren {g^{-1} }^{-1} = g$

## Proof 1

Let $g \in G$.

Then:

 $\displaystyle g$ $\in$ $\displaystyle G$ $\displaystyle \leadsto \ \$ $\displaystyle e$ $=$ $\displaystyle g^{-1} \circ g$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle \paren {g^{-1} }^{-1} \circ e$ $=$ $\displaystyle \paren {g^{-1} }^{-1} \circ \paren {g^{-1} \circ g}$ $\displaystyle \leadsto \ \$ $\displaystyle \paren {g^{-1} }^{-1} \circ e$ $=$ $\displaystyle \paren {\paren {g^{-1} }^{-1} \circ g^{-1} } \circ g$ Definition of Associative Operation $\displaystyle \leadsto \ \$ $\displaystyle \paren {g^{-1} }^{-1} \circ e$ $=$ $\displaystyle e \circ g$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle \paren {g^{-1} }^{-1}$ $=$ $\displaystyle g$ Definition of Identity Element

$\blacksquare$

## Proof 2

Let $g \in G$.

Then:

 $\displaystyle g g^{-1}$ $=$ $\displaystyle e$ Definition of Inverse Element $\displaystyle \leadsto \ \$ $\displaystyle g$ $=$ $\displaystyle \paren {g^{-1} }^{-1}$ Group Product Identity therefore Inverses

$\blacksquare$