jerryb Posted August 12, 2013 Share Posted August 12, 2013 (edited) 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, 2013 by jerryb Link to comment Share on other sites More sharing options...
studiot Posted August 12, 2013 Share Posted August 12, 2013 (edited) 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, 2013 by studiot Link to comment Share on other sites More sharing options...
jerryb Posted August 12, 2013 Author Share Posted August 12, 2013 See the revised question. Link to comment Share on other sites More sharing options...
Olinguito Posted August 21, 2013 Share Posted August 21, 2013 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. Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now