Abstract Algebra/Group Theory/Group/Ga = G

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,

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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. $y\in G{\color {OliveGreen}a}$

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

$Ga=G\,$

$G\subseteq Ga$ and $Ga\subseteq G$,