Abstract Algebra/Group Theory/Group/Cancellation

Theorem

Let G be a Group.

1. $\forall \;g,a,b\in G:(g\ast a=g\ast b)\rightarrow (a=b)$

2. $\forall \;g,a,b\in G:(a\ast g=b\ast g)\rightarrow (a=b)$

Proof

0. Choose ${\color {OliveGreen}g},a,b\in G$ such that ${\color {OliveGreen}g}\ast a={\color {OliveGreen}g}\ast b$
1. ${\color {BrickRed}g^{-1}}\in G$	definition of inverse of g in G (usage 1)
2. ${\color {BrickRed}g^{-1}}\ast ({\color {OliveGreen}g}\ast a)={\color {BrickRed}g^{-1}}\ast ({\color {OliveGreen}g}\ast b)$	0.
3. $({\color {BrickRed}g^{-1}}\ast {\color {OliveGreen}g})\ast a=({\color {BrickRed}g^{-1}}\ast {\color {OliveGreen}g})\ast b$	$\ast$ is associative in G
4. $e_{G}\ast a=e_{G}\ast b$	g-1 is inverse of g (usage 3)
5. $a=b\,$	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. $\ast$ has to be the binary operator of the group.
3. G has to be a group.