Abstract Algebra/Group Theory/Group/Inverse is Unique

= Theorem =


 * In a group, each element only has one inverse.

= Proof =

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 * 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. $$ ({\color{BrickRed}g^{-1}_1} \ast {\color{OliveGreen}g}) \ast {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1} \ast ({\color{OliveGreen}g} \ast {\color{Purple}g^{-1}_2})$$
 * 2. $$ ({\color{BrickRed}g^{-1}_1} \ast {\color{OliveGreen}g}) \ast {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1} \ast ({\color{OliveGreen}g} \ast {\color{Purple}g^{-1}_2})$$

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 * $\ast$ is associative on G
 * 3. $$ e_{G} \ast {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1} \ast e_G$$
 * 3. $$ e_{G} \ast {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1} \ast e_G$$

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 * g1-1 and g2-1 are inverses of g on G (usage 3)
 * 4. $$ {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1}$$, contradicting 1.
 * 4. $$ {\color{Purple}g^{-1}_2} = {\color{BrickRed}g^{-1}_1}$$, contradicting 1.


 * eG is identity of G (usage 3)
 * }