# Theorem

In a group, each element only has one inverse.

# Proof

 0. Choose ${\displaystyle {\color {OliveGreen}g}\in G}$ . Then, inverse g1−1 of g is also in G. 1. Assume g has a different inverse g2−1 in G 2. ${\displaystyle ({\color {BrickRed}g_{1}^{-1}}\ast {\color {OliveGreen}g})\ast {\color {Purple}g_{2}^{-1}}={\color {BrickRed}g_{1}^{-1}}\ast ({\color {OliveGreen}g}\ast {\color {Purple}g_{2}^{-1}})}$ ${\displaystyle \ast }$  is associative on G 3. ${\displaystyle e_{G}\ast {\color {Purple}g_{2}^{-1}}={\color {BrickRed}g_{1}^{-1}}\ast e_{G}}$ g1-1 and g2-1 are inverses of g on G (usage 3) 4. ${\displaystyle {\color {Purple}g_{2}^{-1}}={\color {BrickRed}g_{1}^{-1}}}$ , contradicting 1. eG is identity of G (usage 3)