# Theorem

Let G be any Group.

Let $Ga=\lbrace g\ast a\;|\;g\in G\rbrace$ $\forall \;a\in G:Ga=G$ # Proof

Part A. $\color {RawSienna}Ga\subseteq G$ 0. Choose ${\color {OliveGreen}a}\in G$ 1. Choose $x\in G{\color {OliveGreen}a}=\lbrace g\ast {\color {OliveGreen}a}\;|\;g\in G\rbrace$ 2. $\exists \;g\in G:x=g\ast {\color {OliveGreen}a}$ 1. 3. $g\ast {\color {OliveGreen}a}\in G$ closure of G, $g,a\in G$ 4. $x=g\ast {\color {OliveGreen}a}\in G$ 2,

Part B. $\color {RawSienna}G\subseteq Ga$ 5. Choose ${\color {OliveGreen}a}\in G$ 6. $\exists \;{\color {BrickRed}a^{-1}}\in G:{\color {BrickRed}a^{-1}}\ast {\color {OliveGreen}a}=e_{G}$ definition of inverse 7. Choose $y\in G$ 8. $y\ast {\color {BrickRed}a^{-1}}\in G$ closure of G, and, y, a−1 are in G 9. $(y\ast {\color {BrickRed}a^{-1}})\ast {\color {OliveGreen}a}\in G{\color {OliveGreen}a}$ definition of Ga 10. $y\ast ({\color {BrickRed}a^{-1}}\ast {\color {OliveGreen}a})\in G{\color {OliveGreen}a}$ associativity on G (not Ga) 11. $y\ast e_{G}\in G{\color {OliveGreen}a}$ eG is identity of G 12. $y\in G{\color {OliveGreen}a}$ Part C. $\color {RawSienna}Ga=G$ $Ga=G\,$ $G\subseteq Ga$ and $Ga\subseteq G$ ,