Abstract Algebra/Group Theory/Group/Double Inverse

= Theorem = Let G be any group with operation $$\ast$$.


 * $$\forall \; g \in G: [g^{-1}]^{-1} = g $$


 * In Group G, inverse of inverse of any element g is g.

= Proof =
 * {| style="width: 60%"

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 * 0. Choose $$ {\color{OliveGreen} g} \in G $$
 * 1. $$ \exist \; {\color{BrickRed} g^{-1}} \in G: {\color{OliveGreen} g} \ast {\color{BrickRed} g^{-1}} = {\color{BrickRed} g^{-1}} \ast {\color{OliveGreen} g} = e_{G} $$
 * definition of inverse of g in G (usage 1,3)
 * 2. $$ {\color{OliveGreen} g} \ast {\color{BrickRed} a} = {\color{BrickRed} a} \ast {\color{OliveGreen} g} = e_{G} $$
 * let a = g−1
 * 3. $$ {\color{BrickRed} a} \ast {\color{OliveGreen} g} = {\color{OliveGreen} g} \ast {\color{BrickRed} a} = e_{G} $$
 * 4. $$ [{\color{BrickRed} a}]^{-1} = {\color{OliveGreen} g} $$
 * definition of inverse of a in G (usage 2)
 * 5. $$ [{\color{BrickRed} g^{-1}}]^{-1} = {\color{OliveGreen} g} $$
 * as a = g−1
 * }

= Diagrams =