# Theorem

Let G be any Group.

Let ${\displaystyle Ga=\lbrace g\ast a\;|\;g\in G\rbrace }$

${\displaystyle \forall \;a\in G:Ga=G}$

# Proof

Part A. ${\displaystyle \color {RawSienna}Ga\subseteq G}$

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

Part B. ${\displaystyle \color {RawSienna}G\subseteq Ga}$

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

Part C. ${\displaystyle \color {RawSienna}Ga=G}$

 ${\displaystyle Ga=G\,}$ ${\displaystyle G\subseteq Ga}$ and ${\displaystyle Ga\subseteq G}$,