# There exists an abelian group of prime power order that is lattice-isomorphic to a non-abelian group not of prime power order

## Statement

It is possible to have two groups $G_1, G_2$ such that:

• The lattice of subgroups of $G_1$ is isomorphic to the lattice of subgroups of $G_2$
• $G_1$ is a finite abelian group, finite nilpotent group, and group of prime power order
• $G_2$ is not a finite abelian group, finite nilpotent group, or group of prime power order.

## Proof

Choose primes $p,q$ such that $q$ divides $p - 1$. Choose the following:

• $G_1$ is the elementary abelian group of order $p^2$. It is clearly finite abelian, finite nilpotent, and of prime power order..
• $G_2$ is the external semidirect product of the cyclic group of order $p$ by the cyclic subgroup of order $q$ in its automorphism group. It is clearly non-abelian and not of prime power order (its order is $pq$).

Both $G_1$ and $G_2$ have a lattice of size $p+3$, including the trivial subgroup, whole group, and $p + 1$ intermediate, mutually incomparable subgroups. These lattices are clearly isomorphic.

The smallest example is $p = 3, q = 2$, giving $G_1$ as elementary abelian group:E9 and $G_2$ as symmetric group:S3.