Let $(X,\le)$ a (finite) modular lattice. Is there a (finite) group $G$ such that the lattice of all normal subgroups of $G$ is isomorphic to $(X,\le)$?
There are many counterexamples. Let $M_n$ be the lattice of height two with $n$ atoms. Say the lattice of normal subgroups of the group $G$ is isomorphic with $M_n$. Let $A,B$ be two atoms in this lattice. Then $AB=G$ and $A \cap B=1$. So, $G=A \times B$. As each normal subgroup of $A$ is normal in $A \times B$, we see that $A$ and $B$ are simple. Unless $A,B$ have the same prime order $p$, the only normal subgroups of $G$ are $A$ and $B$, and $n=2$. If $A,B$ both have order $p$ then $n=p+1$.