Here is a lattice of subgroups D4.
The lattice isn't modular iff there is a "pentagon" as a sublattice.
As we can see $\left \{ \rho_{0} \right\} - \left \{ \rho_{0}, \mu_1 \right\} - \left \{ \rho_{0}, \rho_{2},\mu_1, \mu_2 \right\} - D_4 - \left \{ p_{0}, \delta _2 \right\} - \left \{ p_{0} \right\}$ is a pentagon. Is it enough?
I also know the Modular law, here we have: Let $a = \left \{ \rho_{0}, \mu_1 \right\} , b = \left \{ \rho_{0}, \rho_{2},\mu_1, \mu_2 \right\}, x = \left \{ \rho_{0}, \delta_{2} \right\}$, than $a ∨ (x ∧ b) = \left \{ \rho_{0}, \mu_1 \right\} ∨ (\left \{ \rho_{0}, \delta_{2} \right\} ∧ \left \{ \rho_{0}, \rho_{2},\mu_1, \mu_2 \right\}) =\left \{ \rho_{0}, \mu_1 \right\} ∨ \left \{ \rho_{0}, \delta_{2} \right\}$
And I don't understand what will be a supremum of this, how does the operations ∨ and ∧ work in this case? Thank you for any clarifications.