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Here, I (basically) stated the group axioms as follows.

  1. $(xy)z=x(yz)$
  2. $xe=x, ex=x$
  3. $xx^{-1}=e$

In that post, answerers Martin and Ittay were critical of the above list for not including $x^{-1}x=e$, even though it follows from the above three. Pece's answer also included $x^{-1}x=e$ without comment.

How can I tell whether a system of axioms of 'complete'? Is this even a rigorous concept?

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    $\begingroup$ I don't see anyone lambasting you in any of the answers. Martin says "By the way, Definition 2' is conceptually not complete yet", and Pece merely includes the left inverse without criticizing that you didn't include it. Regarding Martin's statement, did you see his last comment? I'm wondering whether there's a misunderstanding here, because he seems to be specifically criticizing Definition 2', not, as you write, your choice of group axioms in general. $\endgroup$ – joriki May 23 '13 at 14:23
  • $\begingroup$ @joriki, Ittay criticizes it as well. $\endgroup$ – goblin May 23 '13 at 14:24
  • $\begingroup$ Where does Ittay criticize it? $\endgroup$ – joriki May 23 '13 at 14:36
  • $\begingroup$ Sentence 0: "First, a group is a monoid where every element has a two-sided inverse, not just a one-sided inverse." $\endgroup$ – goblin May 23 '13 at 14:37
  • $\begingroup$ @joriki, the above was in reply to your comment. $\endgroup$ – goblin May 23 '13 at 14:46
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even if we delete ex=x, it is still a group. which follows like this:

$$ x^{-1}x=x^{-1}xe=x^{-1}x(x^{-1}(x^{-1})^{-1})=x^{-1}(xx^{-1})(x^{-1})^{-1}=x^{-1}(e)(x^{-1})^{-1}=x^{-1}(x^{-1})^{-1}=e $$

and

$$ ex=xx^{-1}x=x(x^{-1}x)=xe $$

so it is a group.

I am not sure if there is a general way to tell a group, but the basic rule of a group are those four: closure, associativity, identity and inverse.

As long as we can conclude the four rules from what we have given, then it is complete. So I think concentrate on the basic idea of the system is important, all others come from the basics.

As for group, we often consider subgroup, which will simplify the judging process, we just consider $xy^{-1}$ belongs to the subset or not.

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    $\begingroup$ x's right inverse $\endgroup$ – delta May 23 '13 at 15:03

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