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If a magma which has an identity element is also a semigroup and a quasigroup, it can be shown that this is indeed a group. I'm looking for a counter example: a magma which is a quasigroup (for every pair of elements $a,b \in M$ there are elements $x,y \in M$ such that $a=xb$ and $a=by$) and a semigroup ($(xy)z=x(yz)$ for all $x,y,z \in M$) yet which lacks an identity element and is therefor not a group.

Juan Perrisimo
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  • A lot more on groups,semigroups,quasigroups and the various generalizations that result can be found at this wonderful post by Peter Cameron. https://cameroncounts.wordpress.com/2011/07/31/semigroups-quasigroups/ – Mathemagician1234 Jun 24 '16 at 18:56

2 Answers2

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There is no counter-example.

Suppose $G$ is a semigroup and a quasigroup.

Let $a\in G$. Let $e$ be the unique solution to $ea=a$. Take any $b\in G$. Let $d\in G$ be the unique solution to $ad=b$. Then $eb=e(ad)=(ea)d=ad=b$. So $e$ is a left identity.

Now given any $b\in G$ there is a left inverse by the quasigroup property.

But it is well-known that a left identity and a left inverse is enough. See eg Right identity and Right inverse implies a group

Community
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almagest
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  • A left (right) identity implies the existence of a two sided identity and similarly for a left identity for every element. It's how I first learned group theory and the tedious proof is a really good exercise for a beginner. – Mathemagician1234 Jun 24 '16 at 19:15
  • But I am afraid I chickened out. At 66 I have long since forgotten it and just know I would get irritated in the time it would take me to reconstruct it! – almagest Jun 24 '16 at 19:19
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Yes, any nonempty quasigroup that is also a semigroup is a group. For a proof see here. The assumptions force a nonempty quasigroup to have a neutral element.

Dietrich Burde
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