2
$\begingroup$

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 $(X,\le)$?

$\endgroup$
0

1 Answer 1

6
$\begingroup$

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 $n=p+1$.

$\endgroup$
1

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.