Theorem[edit | edit source]

In a group, each element only has one inverse.

Proof[edit | edit source]

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