0. Choose ${\color {OliveGreen}g}\in G$ . Then, inverse g_{1}^{−1} of g is also in G.


1. Assume g has a different inverse g_{2}^{−1} in G


 2. $({\color {BrickRed}g_{1}^{1}}\ast {\color {OliveGreen}g})\ast {\color {Purple}g_{2}^{1}}={\color {BrickRed}g_{1}^{1}}\ast ({\color {OliveGreen}g}\ast {\color {Purple}g_{2}^{1}})$

$\ast$ is associative on G

 3. $e_{G}\ast {\color {Purple}g_{2}^{1}}={\color {BrickRed}g_{1}^{1}}\ast e_{G}$

g_{1}^{1} and g_{2}^{1} are inverses of g on G (usage 3)

 4. ${\color {Purple}g_{2}^{1}}={\color {BrickRed}g_{1}^{1}}$ , contradicting 1.

e_{G} is identity of G (usage 3)
