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A Mal'cev algebra is an algebra $\mathbf{A}$ with a ternary term $t$ such that $\mathbf{A} \models t(x,x,y) \equiv y, t(x,y,y) \equiv x$.

For the remaining of the question, refer to the definitions and results contained in Hobby and McKenzie's book The Structure of Finite Algebras.

I'm trying to prove that every finite Mal'cev algebra $\mathbf{A}$ can only have typ$\{\mathbf{A}\} \subseteq \{\mathbf{2},\mathbf{3}\}$.

My approach is the following: let $(\alpha, \beta)$ be a prime quotient. If $(\alpha, \beta)$ is abelian, $\text{typ}(\alpha, \beta)=\mathbf{2}$ since strong abelianity is incompatible with having a Mal'cev term. But if $(\alpha, \beta)$ is nonabelian, how can I rule out the possibility that $(\alpha, \beta)$ has type $\mathbf{4}$ or $\mathbf{5}$? Any suggestion? (I'd prefer a hint rather than an answer).

Thanks!

Noah Schweber
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Mockingbird
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    I had just answer it when I noticed you prefer a hint. Perhaps an answer with hints to which results to use would be ok? (I mean results from the same book and exercises before that one.) The answer is based ruling out the possibility of the quotient type being 1 or 5, and the with different results for 4 (which also seem to work for 5). – amrsa Feb 16 '24 at 19:44
  • Thank you, that would be fine! (I already know how to rule out the type 1 btw) – Mockingbird Feb 16 '24 at 21:44

1 Answers1

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The references below are to results in the book you refer to.

If $\mathrm{typ}(\alpha,\beta) \in \{\mathbf{4,5}\}$, then by Lemma 5.24 there exist two reflexive and admissible relations $\rho_0$ and $\rho_1$ satisfying $\alpha \neq \rho_i \subseteq \beta$ and $\rho_0 \cap \rho_1 = \alpha$, whence $\alpha \subset \rho_i \subseteq \beta$.

By Lemma 5.22, under these conditions $\rho_i \in \mathbf{Con\,A}$, because $\mathbf A$ is Mal'cev.
Use the conditions above to conclude that $\beta = \alpha$, a contradiction.

amrsa
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