Abstract Algebra/Group Theory/Group/Cancellation

Theorem

Let G be a Group.

1. $\mathrm{\forall }g,a,b\in G:(g\ast a=g\ast b)\to (a=b)$

2. $\mathrm{\forall }g,a,b\in G:(a\ast g=b\ast g)\to (a=b)$

Proof

0. Choose $g,a,b\in G$ such that $g\ast a=g\ast b$
1. $g^{-1}\in G$	[[../Definition of a Group/Definition of Inverse#Usage1|definition of inverse of g in G (usage 1)]]
2. $g^{-1}\ast (g\ast a)=g^{-1}\ast (g\ast b)$	0.
3. $(g^{-1}\ast g)\ast a=(g^{-1}\ast g)\ast b$	[[../Definition of a Group/Definition of Associativity#Usage1|$\ast$ is associative in G]]
4. $e_G\ast a=e_G\ast b$	[[../Definition of a Group/Definition of Inverse#Usage3|g-1 is inverse of g (usage 3)]]
5. $a=b$	[[../Definition of a Group/Definition of Identity#Usage3|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. Template:Anchorif 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.