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:

• $H_1$ is a normal Sylow subgroup but not a Sylow direct factor (and hence, not a retract).
• $H_2$ is a Sylow retract but not a normal Sylow subgroup.

Property-theoretic implication

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

• Any property equal to or weaker than normal Sylow subgroup but stronger than or equal to subnormal subgroup. This includes the properties: fully invariant subgroup, characteristic subgroup, normal subgroup, normal Hall subgroup, complemented normal subgroup.
• Any property equal to or weaker than Sylow retract but stronger than or equal to retract. This includes the properties Sylow retract, Hall retract, and retract.

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This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about subnormal subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about normal subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., characteristic subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about characteristic subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., fully invariant subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about fully invariant subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal Hall subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about normal Hall subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., normal Sylow subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about normal Sylow subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., complemented normal subgroup) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about complemented normal subgroup|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., Sylow retract) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about Sylow retract|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., Hall retract) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about Hall retract|Get more facts about conditionally lattice-determined subgroup property|

This article gives the statement, and possibly proof, of a subgroup property (i.e., retract) not satisfying a subgroup metaproperty (i.e., conditionally lattice-determined subgroup property).
View all subgroup metaproperty dissatisfactions | View all subgroup metaproperty satisfactions|Get help on looking up metaproperty (dis)satisfactions for subgroup properties
Get more facts about retract|Get more facts about conditionally lattice-determined subgroup property|

Related facts

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

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: