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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.

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  • $\begingroup$ 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. $\endgroup$
    – Joshua Tilley
    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? $\endgroup$
    – Joshua Tilley
    Sep 29, 2019 at 21:09
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Joshua Tilley
    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. $\endgroup$
    – Henno Brandsma
    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. $\endgroup$
    – Joshua Tilley
    Sep 29, 2019 at 21:18

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