I'm sorry this is a stupid question.

 

Obviously if ag = bg then ag(g inverse) = bg(g inverse) => a = b

 

I shouldn't post questions late at night.

 

 

 

 

 

 

In the definitionof a group action say of a group G on a set S,

there are two conditions:

 

1 for a in S and e in G where e is the identity, ae = a.

 

2 for a in S and g and h in G, (ag)h = a(gh)

 

 

Where does it say that if a and b in S and g in G, that ag is not equal to bg?

 

They always say that for ag = z then obiously there is an inverse g- that zg- = a.

 

One could add another condition but nobody ever does.

 

 

 

Thanks

Edited August 12, 201312 yr by jerryb

 

Where does it say that if a and b in S and g in G, that ag is not equal to bg?

 

I don't quite understand what you mean?

 

Why should ag = bg if a and b are different points of S?

 

What does the condition (group axiom) that every member of G has an inverse imply about a and b and the equality of ag and bg?

 

 

 

I find it helps to discuss such statements in relation to a particular example, did you have one in mind?

Edited August 12, 201312 yr by studiot

See the revised question.

Where does it say that if a and b in S and g in G, that ag is not equal to bg?

 

They always say that for ag = z then obiously there is an inverse g- that zg- = a.

 

One could add another condition but nobody ever does.

 

The statement [latex]a\ne b\Rightarrow ag\ne bg[/latex] is the contrapositive of [latex]ag=bg\Rightarrow a=b[/latex] (which is what you proved in the opening statement of your post). There is no need to state it as an extra condition.