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As noted in this old question it's easy to see that no single identity is equivalent to the conjunction of the commutative and associative laws.

Question. In the language of a binary operation, is there a single nontrivial identity which implies both the commutative law and the associative law?

By "nontrivial" I mean that it does not imply that the operation is constant.

I suppose the answer is well known to experts on universal algebra. If possible, please give the explanation "for dummies", as any technicalities are likely to go over my head.

bof
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    There's a "near-miss" in $(xy)z=y(zx)$, see https://arxiv.org/abs/math/0701713 for various interesting results. – Chris Culter Jun 17 '20 at 08:54
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    In the presence of an identity element (which you seem not to assume), the interchange rule $(ab)(cd)=(ac)(bd)$ implies both commutativity and associativity. – Berci Jun 17 '20 at 09:02
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    @ChrisCulter Thamks for the interesting reference. I don't believe that even the identity $(xy)z=u(vw)$ implies commutativity. – bof Jun 17 '20 at 09:34
  • @Berci That is nice and interesting. As you noticed, my question does not assume an identity element. – bof Jun 17 '20 at 09:37

1 Answers1

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Yes.

According to Theorem $1$ of

Minimal Identities for Boolean Groups
N. S. MENDELSOHN AND R. PADMANABHAN
JOURNAL OF ALGEBRA 34, 451-457 (1975)

a structure $\langle A, \star\rangle$ satisfies the single identity $(x\star y)\star (x\star (z\star y)) \approx z$ if and only if the structure is an abelian group of exponent $2$.

[The fact that such an identity exists follows from work of Higman and Neumann from 1952, but the paper I cite gives short identities.]

Keith Kearnes
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