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

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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:

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Related facts

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:

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

The lattice of subgroups is also shown below:

S3latticeofsubgroups.png