I'm trying to come up with a good explanation for my students of why the [finite lattice representation problem](https://en.wikipedia.org/wiki/Finite_lattice_representation_problem) is difficult. I've already shown that the "greedy approach" to representing the lattice drawn in [this post](https://mathoverflow.net/questions/85724/given-a-lattice-l-with-n-elements-are-there-finite-groups-h-g-such-that-l-c) as the congruence lattice on a finite algebra - namely, realize the lattice as a sublattice of a lattice of partitions on some large enough finite set $X$, and then look at the algebra on $X$ given by all functions respecting each partition in that sublattice - breaks down.


I'd now like to explain why this *doesn't* lead to an easy *negative* solution. Towards that end:



> 
> Is there a not-too-hard-to-verify example of a bounded lattice $L$ such that
> 
> 
> * $L$ is isomorphic to the congruence lattice of some finite algebra, but
> * the smallest $n$ such that $L$ embeds into the bounded lattice of partitions of $\{1,...,n\}$ is strictly less than the smallest size of an algebra with congruence lattice $\cong L$?
> 
> 
> 


Essentially, a concrete example of this would give a convincing argument that the failure of the greedy strategy doesn't actually tell us much.



Example of trickiness of finite lattice representation problem?
$M\_4$, the modular lattice of height two with four atoms is an example.
$M\_4$ arises as the subgroup lattice of the symmetric group on $3$ letters, hence it arises as the congruence lattice of a regular $S\_3$-set. This is an algebra with $6$ elements. There is no representation of $M\_4$ as the congruence lattice of a finite algebra $\mathbf{A}$ where $\mathbf{A}$ has fewer than $6$ elements. This is proved in 
[A minimal congruence lattice representation for $M\_{p+1}$](https://www.cambridge.org/core/journals/journal-of-the-australian-mathematical-society/article/abs/minimal-congruence-lattice-representation-for-mathbbmp1/F1031B09770B7060128554C0A2E86EB5)  

Roger Bunn, David Grow, Matt Insall, Philip Thiem  

J. Aust. Math. Soc. 108 (2020), 332-340
Their main result is that if $p$ is an odd prime, then a minimal congruence lattice representation of $M\_{p+1}$ has size $2p$, so a minimal congruence lattice representation of $M\_4 = M\_{3+1}$ has size $2\cdot 3 = 6$.
A few years ago I posted [a shorter proof](https://arxiv.org/abs/2008.04784) of the Bunn-Grow-Insall-Thiem theorem to the arxiv. 
On the other hand, it is possible to embed $M\_4$ into the lattice of partitions of a $5$-element set, and $5<6$. The following four partitions of $\{1,2,3,4,5\}$ generate such an $M\_4$:
12/34/5  

13/25/4  

14/35/2  

15/24/3












