Abstract Algebra/Group Theory/Group/Ga = G

Theorem

Let G be any Group.

Let $Ga=\{g\ast a|g\in G\}$

$\mathrm{\forall }a\in G:Ga=G$

Proof

Part A. $Ga\subseteq G$

0. Choose $a\in G$
1. Choose $x\in Ga=\{g\ast a|g\in G\}$
2. $\mathrm{\exists }g\in G:x=g\ast a$	1.
3. $g\ast a\in G$	[[../Definition of a Group/Definition of Closure|closure of G]], $g,a\in G$
4. $x=g\ast a\in G$	2,

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Part B. $G\subseteq Ga$

5. Choose $a\in G$
6. $\mathrm{\exists }{a}^{-1}\in G:a^{-1}\ast a=e_G$	[[../Definition of a Group/Definition of Inverse|definition of inverse]]
7. Choose $y\in G$
8. $y\ast a^{-1}\in G$	[[../Definition of a Group/Definition of Closure|closure]] of G, and, y, a−1 are in G
9. $(y\ast a^{-1})\ast a\in Ga$	definition of Ga
10. $y\ast (a^{-1}\ast a)\in Ga$	[[../Definition of a Group/Definition of Associativity|associativity]] on G (not Ga)
11. $y\ast e_G\in Ga$	eG is [[../Definition of a Group/Definition of Identity|identity]] of G
12. $y\in Ga$

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Part C. $Ga=G$

$Ga=G$

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