Find all pairs of lattices (L, K ) (up to isomorphism)

Question

The question is that Find all pairs of lattices $(L, K )$ (up to isomorphism)

a) $L \times K$ contains $20$ elements;

(b) $L$ is non-distributive;

(c) $K$ has at least $3$ elements;

(d) the greatest element of $L \times K$ covers precisely $4$ elements.

I got one is $L$ is $5$ elements pentagon lattice and $K$ has $4$ elements like Diamond. I want to know what else pairs of lattice also work.

Answer 1

Since $L$ is non-distributive, it follows that $|L|\geq 5$.
As $|K|\geq3$ and $|L\times K|=|L|\times|K|=20$, we obtain $|L|=5$ and $|K|=4$.
So $L$ is either the diamond $\mathbf M_3$ or the pentagon $\mathbf N_5$.

Now here's an easy lemma for you to prove:

Lemma. Let $P$ and $Q$ be partially ordered sets. If $a,b \in P$ and $c,d \in Q$, then $(a,c) \prec (b,d)$ in $P\times Q$ iff $a=b$ and $c \prec d$ in $Q$, or $a\prec b$ in $P$ and $c=d$.

(Note about notation: $x\prec y$ means that $x$ is covered by $y$.)

Using the Lemma, we see that the number of elements covered by the top element of $L\times K$ is the sum of the number of elements (in $L$) covered by the top of $L$ with the number of elements (in $K$) covered by the top of $K$.
As the top element of $\mathbf M_3$ covers $3$ elements, the top element of the corresponding $K$ covers only one element, whence $K=\mathbf 4$, the four-element chain.
The top element of $\mathbf N_5$ covers $2$ elements, whence the corresponding $K$ is such that its top covers $2$ elements too, so $K=\mathbf 2^2$, the four-element lattice which is not a chain.

Summing up, either $L=\mathbf M_3$ and $K=\mathbf 4$ or $L=\mathbf N_5$ and $K=\mathbf 2^2$.