# Theorem

Let G be a Group.
1. ${\displaystyle \forall \;g,a,b\in G:(g\ast a=g\ast b)\rightarrow (a=b)}$
2. ${\displaystyle \forall \;g,a,b\in G:(a\ast g=b\ast g)\rightarrow (a=b)}$

# Proof

 0. Choose ${\displaystyle {\color {OliveGreen}g},a,b\in G}$  such that ${\displaystyle {\color {OliveGreen}g}\ast a={\color {OliveGreen}g}\ast b}$ 1. ${\displaystyle {\color {BrickRed}g^{-1}}\in G}$ definition of inverse of g in G (usage 1) 2. ${\displaystyle {\color {BrickRed}g^{-1}}\ast ({\color {OliveGreen}g}\ast a)={\color {BrickRed}g^{-1}}\ast ({\color {OliveGreen}g}\ast b)}$ 0. 3. ${\displaystyle ({\color {BrickRed}g^{-1}}\ast {\color {OliveGreen}g})\ast a=({\color {BrickRed}g^{-1}}\ast {\color {OliveGreen}g})\ast b}$ ${\displaystyle \ast }$  is associative in G 4. ${\displaystyle e_{G}\ast a=e_{G}\ast b}$ g-1 is inverse of g (usage 3) 5. ${\displaystyle a=b\,}$ eG is identity of G(usage 3)

# Diagrams

 if a*g = b*g... a = a*g*g-1 b*g*g-1 = b then a = b.

# Usage

1. if a, b, x are in the same group, and x*a = x*b, then a = b

# Notice

1. a, b, and g have to be all in the same group.
2. ${\displaystyle \ast }$  has to be the binary operator of the group.
3. G has to be a group.