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Let $\mathcal{N}_5$ refer to the Pentagon lattice, or the lattice generated by the set $\{0, a, b, c, 1\}$ subject to $1 > a$, $1 > c$, $a > b$, $b > 0$ and $c > 0$.

My aim is to find the congruences on $\mathcal{N}_5$. I know that in any lattice, a congruence class under a congruence relation is a convex sublattice. Thus, one approach I could take is enumerating all convex sublattices of $\mathcal{N}_5$ and trying to see which of those correspond to congruence classes under various congruences on the lattice.

However, this seems like a lot of enumeration, and I feel like there must be a more elegant way to approach this problem (consider, by contrast, the ease of characterizing congruences on $n$-element chains $C_n$). So do I truly just have to bite the bullet and start enumerating, or is there something else I could do?

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I think it's easier to just enumerate the congruences by hand.

  • The only congruence with $1\sim 0$ is the trivial congruence.

  • If $b\sim c$ then $a\sim b\sim c\sim 1$ and so we only have one nontrivial congruence identifying $b$ and $c$.

  • We can collapse $a$ and $b$ without collapsing anything else; the result is then the "diamond" lattice, whose congruence lattice you can compute separately.

  • There are two congruences satisfying $a\sim c$ and $b\not\sim c$, depending on whether or not $b$ and $0$ are identified.

Through this sort of analysis (really the above is most of the argument), you can determine the entire congruence lattice of $\mathcal{N}_5$ without too much trouble.


If you want to be more methodical, you can also think about building the congruence lattice "layer by layer" - e.g. for the first layer, identify for each pair of distinct elements $(x,y)$ what is the smallest congruence containing that pair. There aren't too many of these to work through.

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  • $\begingroup$ I think you might have consider a different labelling than the OP, since $b\sim c$ or $a\sim c$ implies $0\sim 1$ and everything is related. The lattice of congruences of $N_5$ can be found, for example in this question. $\endgroup$
    – amrsa
    Commented Jun 13 at 10:51

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