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A finite lattice is said to be representable if there exists a finite algebra whose congruence lattice is isomorphic to that lattice.

As I was reading a paper, I came across the line: "The reader can find examples to show that every lattice with fewer than 6 elements is representable."

I was able to easily find examples for seven of the ten lattices by using subgroup lattices of abelian groups.

For the ordinal sum of the trivial lattice with the diamond lattice, I was able to use the congruences of the lattice $\mathbf{N}_5$. For the dual of the above lattice, I eventually found it to be the lattice of normal subgroups of the group $\mathbb{Z}_3 \rtimes \mathbb{Z}_4$.

However, I could not find any algebra with $\mathbf{N}_5$ as its congruence lattice. The main problem is that all the classes of algebras with which I'm familiar are congruence-modular (e.g. groups, rings, modules, and lattices).

Can somebody find an example of a finite algebra that has this property?

Eran
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2 Answers2

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Presenting examples of algebras having a pentagonal congruence lattice is not hard. Showing you how to find such examples is a little harder. If you're interested in how to do this, feel free to post again (or ask the hiring committee in your department to invite me to give a talk about it :).

Example 1. First find a lattice $L \cong N_5$ that is a sublattice of the lattice of equivalence relations on a small set, like $X = \{0, 1, 2, 3\}$, and then find a set $F$ of operations that respect only the equivalence relations in $L$. Then the algebra $\langle X, F \rangle$ has $L$ as its congruence lattice.

For example, let $L = \{0_X, \alpha, \beta, \gamma, 1_X\}$, where $\alpha$, $\beta$, and $\gamma$ are the equivalence relations on $X = \{0,1,2,3\}$ corresponding to the following partitions, respectively:

|0,1|2|3|, |0,1|2,3|, and |0,2|1,3|.

You can see that $L\cong N_5$.

Now define two unary operations $f$ and $g$ on $X$ as follows: $f = (1,0,3,2)$ and $g = (1,0,1,0)$. By this notation I mean $f(0)=1$, $f(1)=0$, $f(2)=3$, etc.

If you enter these two operations into the Universal Algebra Calculator, click the Con tab, and hit Go, you will find that the algebra $\langle X, \{f, g\}\rangle$ has congruence lattice $L$.

(Perhaps you are wondering how I found the operations $f$ and $g$. We have a computer program that does this. I can explain how it works and send it to you if you want.)

Example 2. If you find finite groups $H < G$ such that the interval $[ H, G ] = \{K : H \leq K \leq G\}$ in the subgroup lattice of $G$ is isomorphic to $N_5$, this will give you another example of a pentagonal congruence lattice. This is because of the following standard result: Let $X$ be the set $G/H$ of left cosets of $H$ in $G$. Let the operations $F$ be the set of elements of $G$ acting by left multiplication on the set $X$. Then the algebra $\langle X, F\rangle$ has congruence lattice isomorphic to the interval $[H, G]$.

Perhaps surprisingly, there are no pentagonal intervals in subgroup lattices of very small finite groups. I believe you have to look at groups of order at least 216 to find such intervals (though I am not positive about this). In any case, here is the smallest example I know of:

Let $G$ be (in GAP) SmallGroup(216,153).

(Incidentally, this group has structure description $((C_3 \times C_3) \rtimes Q_8) \rtimes C_3$.)

There is a subgroup $H\cong C_6$ of $G$ (of index 36) such that $[H, G]\cong N_5$.

So, the resulting algebra (i.e., $G$ acting on cosets of $H$) has 36 elements and has congruence lattice isomorphic to $[H, G]$.

I've posted this algebra's UACalc file: Pentagon.ua. If you wanted to you could load this file into UACalc and inspect its congruence lattice, though I'm not sure how illuminating this exercise would be. If you want to know the GAP commands I used to construct the algebra and write it out to Pentagon.ua, let me know.

  • @AlexanderKonovalov Glad to hear it! Of course, as you well know, the congruences (aka "systems of imprimitivity") of this algebra can be seen right in GAP with something like for b in AllBlocks(G) do Print(Orbit(G,b,OnSets), "\n"); od; Still, UACalc is nice if you want to see cool rotating Hasse diagrams! :) – William DeMeo Jan 11 '14 at 04:56
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Here's another example of an algebraic structure that has $N_5$ as its lattice.

I computed this one by hand, but unfortunately the trick it uses, a function forcing $c$ to unify with $0$ if it unifies with anything and a function representing an undirected graph, doesn't generalize to other lattices with unknown representations like this one in William DeMeo's answer to an MO question.

$N_5$ is shown below:

      1
    / |
   /  b
  /   |
 c    a
  \  /
   0

I will now build $A$, an algebra with the desired congruence lattice.

First, I'll define the carrier of $A$ as $\{0, a, b, c\}$.

Next, I define two functions:

$$ D[c](x) = \;\text{$c$ if $x = c$ else $0$} $$ $$ f(x, y) = \text{$x$ if $\{x, y\} = \{a, b\}$ else $y$} $$

$f$ corresponds to the simple graph with edge set $\{\{a, b\}\}$; the idea is to block $a$ and $b$ from being sent to different parts in the partition.

$D[c]$ is well-defined if and only if $c$ is in a part by itself, or $c$ unifies with $0$.

Here is a table checking whether $D[c]$ and $f$ are defined for any given partition. . is true. X is false.

partition    D[c] def?   f def?
0|a|b|c         .          .
0|ab|c          .          .
0|ac|b          X          X
0|bc|a          X          X
0|abc           X          .
0abc            .          .
0a|bc           X          X
0b|ac           X          X
0c|ba           .          .
0ab|c           .          .
0bc|a           .          X
0ac|b           .          X

This gives us the following lattice:


         0abc
       /  |
     /   0ab|c
   /      |
0c|ba    0|ab|c
  \      /
  0|a|b|c

Another moderately useful primitive for constructing small examples is $L[a,b](x) = \text{$b$ if $x=a$ else $0$}$.

$L[a,b]$ is well-defined in a quotient structure if and only if it holds that if $a$ unifies with anything, $b$ unifies with $0$.

Greg Nisbet
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    I just noticed you gave, last year, a new answer to this old question. I also liked this approach (+1), but I disagree with saying that $D_c$ or $f$ are well-defined or not for each partition. At least from the universal algebraic point of view, you defined those two operations very clearly—they are well defined. Then what happens is that some equivalence relations are compatible with none, one or the other, or both operations. But I suppose this as to do with different backgrounds... – amrsa Jun 22 '24 at 21:11
  • @amrsa I don't think it's a difference in background, I'm probably just wrong. I learned a little bit about model theory and universal algebra mostly from using this site and looking at some random textbooks. Is it more idiomatic to describe a function symbol like $f$ and its interpretation as being compatible with an equivalence relation rather than being well-defined on the carrier quotiented by the equivalence relation? – Greg Nisbet Jun 23 '24 at 01:14
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    Yes, it's more idiomatic because they are well-defined, even if they're not compatible with any equivalence relation (except for the trivial ones, with which every function is compatible). There might be other words for it, but they are well-defined because they are so as operations, in this case, maps from $A$ to $A$ and from $A^2$ to $A$. To see that, just apply the usual definition of what is a well-defined map $X\to Y$: to each member of $X$, correspond one, and only one, element of $Y$. In our case, $X$ is $A$ or $A^2$ for $D[c]$ or $f$, respectively, and $Y$ is always $A$. – amrsa Jun 23 '24 at 09:08
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    While I was looking at some notes, I came back to this and realize it doesn't work, sorry. Because 0|ab|c < 0c|ab. Moreover (and this was what trigger this comment, when I was reviewing those notes), a close inspection at the lattice of partitions of a 4-element set shows that the pentagon cannot be obtained including a partition like 0ab|c. Nevertheless, it's quite possible that your reasoning can be changed to a different set of partitions that make a pentagon. But if you don't care to do that, it's ok to me. Best wishes anyway. – amrsa Jun 28 '24 at 21:32
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    @amrsa great find! I will try to come up with another structure or a variation of this one that does give a pentagon and update the answer. – Greg Nisbet Jun 28 '24 at 23:30