If V is a congruence-permutable variety... If V is a congruence-permutable variety such that every subdirectly irreducible algebra
is simple, show that every finite algebra in V is isomorphic to a direct product of simple
algebras.
I'm not sure how to get started on this. Any help is appreciated.
 A: I'll give you a sketch of the proof and hopefully you can fill in the details. I put hints under spoiler tags if you need more help. To begin, let $\mathbf{A}$ be a minimal counterexample (so $\mathbf{A}$ is a finite algebra in $\mathcal{V}$ that is not a direct product of simple algebras).


*

*If $\mathbf{A}$ is a finite algebra in $\mathcal{V}$, then $\mathbf{A}$ is a subdirect product of finitely many simple algebras $\mathbf{B}_1,\mathbf{B}_2,\dots, \mathbf{B}_n$. Hint: 

use the Birkhoff's subdirect representation theorem and the fact that every SI is simple


*Let $\eta_i$ be the kernel of the projection from $\mathbf{A}$ onto $\mathbf{B}_i$. Then $\eta_i$ is a maximal proper congruence of $\mathbf{A}$. Hint:

 use the 1st isomorphism theorem and the fact that each $\mathbf{B}_i$ is simple.


*Let $J\subseteq\{1,2,\dots,n\}$ be minimal with respect to the property that $\bigcap_{j\in J}\eta_j=0_A$. Let $\theta_j= \bigcap_{k\in J-\{j\}}\eta_k$. Then $\eta_j\circ\theta_j=1_A$. Hint:

 use maximality of $\eta_j$ and the fact that CP implies $\alpha\vee\beta=\alpha\circ\beta$


*Conclude that $\mathbf{A}\cong \mathbf{B}_j\times\mathbf{A}/\theta_j$. Hint:

 show $\eta_j\cap\theta_j=0_A$ and use the fact that $\alpha\cap\beta=0_C$ and $\alpha\circ\beta=1_C$ imply that $\mathbf{C}\cong\mathbf{C}/\alpha\times\mathbf{C}/\beta$

Since $\vert A/\theta_j\vert <\vert A\vert$, we get that $\mathbf{A}/\theta_j$ is a product of simple algebras. So then $\mathbf{A}$ is too. A contradiction! Hence every finite algebra is a product of simple algebras. (You can probably prove this directly, but oh well.)
