Are there any algebraic structures where the congruence lattice and subalgebra lattice are opposites that do not look like groups with operators?
My motivation for this question is noticing that submodules have many nice properties that subgroups of Abelian groups have and that this seems to boil down to the fact that every submodule is both [ideal/normal-subgroup/congruence]-like and a subalgebra of the larger module.
Then I noticed that $R$-modules are Abelian groups with operators. That made me wonder whether there are other algebraic structures out there with a similar nice property to this one.
- In an Abelian group, every subgroup is normal.
- Additionally, for an $R$-module called $M$ where $R$ is a commutative ring, there is 1:1 order-reversing correspondence between submodules of $M$ and concrete quotients of $M$.
A concrete quotient is a nonstandard notion that I'm going to define:
Let a concrete quotient of an algebraic structure $A$ be $(A, \equiv)$ where $\equiv$ is a congruence relation on $A$. This means, for example, that $\mathbb{R}[x, y]/(x)$ as a concrete quotient of modules and $\mathbb{R}[x, y]/(y)$ as a concrete quotient of modules are distinct. They are, however, isomorphic and we can't determine that they are distinct unless we are allowed to peer inside. I am explicitly allowing us to peer inside.
Suppose we have a concrete module quotient $(M, \equiv)$. We can extract the set of elements in the equivalence class of $0$ and verify that it is a module; thus it is a subalgebra of $M$.
Additionally, if we have $M$ and a designated submodule $N$, we can associate two elements $a, b \in M$ precisely when $a - b$ is in $N$. Since we're working in a group, this association is unique.
Thus we have a bijective correspondence between subalgebras of a module $M$ and concrete quotients of $M$.
Concrete quotients are in a bijective correspondence with congruences, so I'll mostly talk about congruences.
This doesn't hold for algebraic structures in general though, even ones based on Abelian groups. If we take something like the integers $\mathbb{Z}$, then $\mathbb{Z}$ has exactly two subalgebras: the trivial ring and $\mathbb{Z}$ itself, but has a non-isomorphic ideal for each element element of $\mathbb{Z}$ up to associates, ordered by divisibility.
A normal subgroup, ideal of a ring, or a submodule of a module, are all convenient representatives for a congruence of an algebraic structure. Since they're subsets of the domain of a larger algebra, they give us an obvious way to compare subalgebras and ideals/congruences/[concrete-quotients].
However, we can temporarily forget how congruences are encoded and just look at the lattice of congruences (ordered by inclusion) and the lattice of subalgebras (ordered by inclusion). In groups and $R$-modules, these lattices are opposites. As a quick check that the lattices are opposites instead of isomorphic, the identity congruence $=$ is the smallest congruence by inclusion, but when modding out an Abelian group $A$ by $=$, you get back $=$ which is the largest subalgebra by inclusion.
So, now we can compare the congruence lattice and the (opposite of) the subalgebra lattice on any algebraic structure.
This brings me to my question: are there any algebraic structures where the congruence lattice and subalgebra lattice are opposites that do not look like groups with operators?
My question is a little bit fuzzy by necessity. Technically, an Abelian group $A$ equipped with a $47$-place function $z$ that always returns $0$ is not a group with operators. But the $z$ is not adding anything and doesn't change the essential nature of the algebra as an Abelian group. I also don't want to commit to a specific technical notion of "looking like an Abelian groups with operators", such as being an extension by definitions, because I don't know enough to formulate that technical condition perfectly on the first try.
Also, groups where every subgroup is normal that aren't Abelian, such as the quaternion group are interesting but aren't the thrust of the question.
I'm also not insisting that the mapping back and forth be as simple as it is in the Abelian group case, where we can extract the kernel from a congruence, or easily extend a chosen kernel to an entire congruence.
In ... -modules, these lattices are opposites.
No, they are isomorphic. If $N$ is a submodule of $M$, then the corresponding congruence is $\theta_N=\{(x,y)\in M^2\;|\;x-y\in N\}$. If $\theta$ is a congruence on $M$, then the corresponding submodule is $N_{\theta}=0/\theta$ ( = the $\theta$-class of $0$). The inverse bijections $\textrm{Sub}(M)\to \textrm{Con}(M):N\mapsto \theta_N$ and $\textrm{Con}(M)\to \textrm{Sub}(M):\theta\mapsto 0/\theta$ are order-preserving, not order-reversing.
Nevertheless, one may still ask each of the questions:
A standard result on this topic is Theorem 2 of
W. A. Lampe
The independence of certain related structures of a universal algebra, III
The subalgebra lattice and congruence lattice are independent
algebra universalis volume 2, pages 286-295 (1972)
Lampe observes that subalgebra lattices and congruence lattices are algebraic lattices, and then shows conversely that if $L$ and $K$ are arbitrary algebraic lattices there exists an algebraic structure $A$ such that $\textrm{Sub}(A)\cong L$ and $\textrm{Con}(A)\cong K$. Taking $L=K$ one gets that for any algebraic lattice there is an algebra $A$ such that $\textrm{Sub}(A)\cong L \cong \textrm{Con}(A)$. If $L$ is both algebraic and dually algebraic, then Lampe's construction also provides an algebra $A$ such that $\textrm{Sub}(A)\cong L \cong \textrm{Con}(A)^{\partial}$. Lampe's algebras are not groups with operators. In fact, groups with operators have modular congruence lattices, and Lampe's construction works even when $\textrm{Con}(A)$ is not modular.
The remarks I have made so far concern single algebras only. You don't need Lampe's construction to answer the two questions above for single algebras. It is easy to construct algebras of any size greater than $1$ where $\textrm{Sub}(A)$ and $\textrm{Con}(A)$ are both $2$-element chains. (E.g., for $|A|=2$ take $\langle \{0,1\}; *, 1\rangle$ as a multiplicative submonoid of $\langle {\mathbb R}; *, 1\rangle$.) I believe that the questions were intended to be about classes of similar algebras where $\textrm{Sub}(A)\cong \textrm{Con}(A)$ holds. Let me say something about that.
Let $C$ be a category with a zero object and assume that kernels and cokernels of all morphisms exist in $C$. Call a morphism normal if it is the kernel of its cokernel and conormal if it is the cokernel of its kernel. Now consider these definitions for prevarieties of similar algebras. (A prevariety is a class of similar algebras closed under the formation of isomorphisms, subalgebras, and products). Call a prevariety normal if every monomorphism is normal and conormal if every epimorphism is conormal. This is a setting where there is a 'natural' correspondence between $\textrm{Sub}(A)$ and $\textrm{Con}(A)$ generalizing the correspondence found for any variety of $R$-modules, i.e. any subobject is the $0$-class of a uniquely determined congruence. It turns out that this 'natural' correspondence between $\textrm{Sub}(A)$ and $\textrm{Con}(A)$ cannot be generalized beyond modules. Theorem 2.11 of
Keith A. Kearnes and Frank Vogt
Bialgebraic contexts from dualities.
Journal of the Australian Mathematical Society 60 (1996), 389-404.
shows that a normal and conormal prevariety is definitionally equivalent to a variety of $R$-modules for some ring $R$.