In the book The Structure of Finite Algebras, by Hobby and McKenzie, page 23, it says, in a paragraph before the exercises:
The first substantial result proved as an application of tame congruence theory was that $\mathbf{Sub\,A}_{4}$ (and many other tight lattices, in particular $\mathbf{M}_{n}$ if $n \geq 3$) cannot be isomorphic to the congruence lattice of any finite algebra with just one basic operation, such as a semigroup. (This result will not be proved in this book. The proof can be found in [22].)
[22] refers to a paper of McKenzie: Finite Forbidden Latices, in Universal Algebra and Lattice Theory, number 1004, Springer Lecture Notes.
Could this be a typo? and it's mean to be "$\mathbf{M}_n$ if $n > 3$"?
I'm asking because $\mathbf{M}_3$ is, up to isomorphism, the partition lattice on a $3$-element set, and I think I know (and it seems obvious to me) that every partition lattice is a congruence lattice of an algebra with the same underlying set.
For example the left-zero semigroup on that set (or the right-zero semigroup, or the null semigroup...).
The reason is that in these semigroups all equivalences are congruences (are they not?).
I tried to find some errata, but didn't succeed...
Is there some subtlety that I'm missing here?
Even if you change $n\geq 3$ to $n\gt 3$, the statement is still incorrect. $M_n$ is the congruence lattice of a finite algebra with one basic operation if and only if $n$ is a prime power plus $1$. One can rule out other values of $n$ with the results of the Finite Forbidden Lattices paper. But when $n=q+1$, where $q$ is a prime power, one can represent $M_n$ as the congruence lattice of a 2-dimensional affine vector space over the $q$-element field. This algebra can be presented with one basic operation.