$\mathbf{N}_5$ as a congruence lattice 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?
 A: 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.
