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!
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.