Jump to content

jerryb

Members
  • Posts

    2
  • Joined

  • Last visited

Everything posted by jerryb

  1. See the revised question.
  2. 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
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.