# Alternative axioms for groups.

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.

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

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.
• @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. Sep 29, 2019 at 21:16
• 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. Sep 29, 2019 at 21:18