We start from the following results, which are consequences
of theorems by Freese, Hobby, and McKenzie.
\begin{quote}
Let $\mathbf{A}$ be a finite algebra in a congruence modular
variety such
that $\mathrm{Con} (\mathbf{A})$ has a $(0,1)$-sublattice $\mathbb{L}$
that is simple, complemented, and has at least three elements.
Then $\mathbf{A}$ is abelian.
\end{quote}
Similarly if $\mathbb{L}$
has no $2$-element homomorphic image, then
$\mathbb{A}$ is solvable.
We derive a similar condition for nilpotency and investigate
what could be converses to these results.