# No subgroup property between normal Sylow and subnormal or between Sylow retract and retract is conditionally lattice-determined

## Statement

It is possible to have a group $G$, a lattice automorphism $\varphi: L(G) \to L(G)$ of the lattice of subgroups, and subgroups $H_1, H_2$ of $G$ with $\varphi(H_1) = H_2$, such that:

### Property-theoretic implication

No subgroup property of any of the following kinds is a conditionally lattice-determined subgroup property:

## Proof

### General example

Choose primes $p,q$ such that $q$ divides $p - 1$. Let $G$ be the semidirect product of the group of order $p$ by the subgroup of order $q$ in its automorphism group. $G$ is a group of order $pq$. Its lattice has size $p + 3$, including the trivial subgroup, whole group, and $p+1$ intermediate mutually incomparable subgroups, one of order $p$ and $p$ of order $q$.

Let $H_1$ be the subgroup of order $p$ and $H_2$ be any subgroup of order $q$. The map from $L(G)$ to itself interchanging $H_1$ and $H_2$ is a lattice automorphism, and it interchanges the two subgroups. Also, $H_1$ and $H_2$ satisfy the stated conditions, completing the proof.

### Smallest example

The smallest example is obtained by setting $p = 3, q = 2$, giving:

• $G$ equals symmetric group:S3
• $H_1$ equals A3 in S3
• $H_2$ equals S2 in S3

For more on the subgroup structure, see subgroup structure of symmetric group:S3.

The lattice of subgroups is also shown below: