7
$\begingroup$

The usual axioms I've seen for a group are: associativity; existence of two-sided identity; existence of two-sided inverses for all elements.

$$\forall a,b,c\in G: a\left(bc\right)=\left(ab\right)c$$ $$\exists e\in G, \forall a\in G: ae=a=ea$$ $$\forall a\in G \exists a'\in G: aa'=e=a'a$$

I recently came across a different axiomatisation, and there were no proofs of equivalence. They were: associativity; existence of left-identity; existence of left-inverses.

$$\forall a,b,c\in G: a\left(bc\right)=\left(ab\right)c$$ $$\exists e\in G, \forall a\in G: ea=a$$ $$\forall a\in G, \exists a'\in G: a'a=e$$

Are these equivalent? I kind of doubt it, since we have associative semigroups with left but not right identities, but maybe the left-inverses part changes things.

There was a proof that given these axioms, a left-inverse is a right-inverse, and hence that the original inverses axiom is proven, but what about right-identity?

Proof: Let $g\in G$ then $g$ has a left inverse, call it $g'\in G$ and this too has a left inverse, call it $g''\in G$. Then, $g'g=e$, $g''g'=e$ and so $$gg'=egg'=g''g'gg'=g''g'=e$$ so $g'$ is the right-inverse of $g$ also.

$\endgroup$
9
  • $\begingroup$ All semigroups are associative. $\endgroup$
    – Shaun
    Sep 29, 2019 at 21:04
  • $\begingroup$ Yes, fair point on language. I meant to emphasise that these satisfy 2/3 of the alternative group axioms. $\endgroup$ Sep 29, 2019 at 21:06
  • $\begingroup$ The word "magma" would be more appropriate then. $\endgroup$
    – Shaun
    Sep 29, 2019 at 21:08
  • $\begingroup$ I can edit if you like, but do you have any thoughts on the question? $\endgroup$ Sep 29, 2019 at 21:09
  • 2
    $\begingroup$ A left identity $e$ satisfies, given the existence of inverses, $ae=aa'a=ea=a$, for $a\in G$ $\endgroup$
    – user418131
    Sep 29, 2019 at 21:10

1 Answer 1

7
$\begingroup$

I agree with the proof that the left inverse (wrt the left identity) is also a right inverse.

Now let $e$ be the left inverse and $g \in G$. Then

$$ge = g(g'g) = (gg')g = eg= g$$ where we use that $g'$ is both-sided.

Then $e$ is also a right identity.

$\endgroup$
3
  • $\begingroup$ There we go! Nice, didn't see that myself. Guess I find it a little odd to establish that inverses are two sided before identity is, since what is a right-inverse if the identity isn't two sided? $\endgroup$ Sep 29, 2019 at 21:13
  • $\begingroup$ @JoshuaTilley you work with the inverse wrt the left-identity (the only one we have at the outset) and $g'$ is two-sided for that and that very fact then enables this double identity, i.e. that $e$ is identity from the right too. $\endgroup$ Sep 29, 2019 at 21:16
  • $\begingroup$ I understand the argument. My point is that on its own, $aa'=e$ for an arbitrary left-identity $e$ does not make $a'$ behave as a right-inverse, that was my point. Of course it does given the other requirements, as per your answer. $\endgroup$ Sep 29, 2019 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.