Permutability is not finite-intersection-closed
This article gives the statement, and possibly proof, of a subgroup property (i.e., permutable subgroup) not satisfying a subgroup metaproperty (i.e., finite-intersection-closed 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 permutable subgroup|Get more facts about finite-intersection-closed subgroup property|
The intersection of two permutable subgroups of a group need not be permutable.
It is possible to find a group and subgroups and of such that and are both permutable subgroups (viz quasinormal subgroups) but is not.
Related facts that don't hold for permutable subgroups
Related facts that do hold for permutable subgroups
- Permutability is strongly join-closed
- Permutability satisfies image condition
- Permutability satisfies inverse image condition
- Permutability satisfies intermediate subgroup condition
- Permutability satisfies transfer condition
Construction of the counterexample
Setup: Let be an odd prime.
- is a semidirect product of cyclic group of prime-square order and cyclic group of prime order. More specifically it is a group generated by two elements subject to the relations and . Alternatively is the semidirect product of the additive group modulo by the multiplicative group of order in the multiplicative group of automorphisms. Note that is a non-abelian group of order .
- is a cyclic group of prime-square order: It is a cyclic group of order , generated by an element .
- , and .
We claim that and are both permutable in , but their intersection is not permutable.
- is permutable: is a direct factor of so it is clearly a normal subgroup and hence a permutable subgroup.
- is permutable: Since permutability satisfies the inverse image condition, we see that if is permutable in , then is permutable in . Thus, it suffices to show that is permutable as a subgroup of . This can easily be checked by verifying that commutes with all the cyclic subgroups of . (a proof of this is provided in an example for permutable not implies normal).
- is not permutable in : Consider the cyclic subgroup generated by . The claim is that . To prove this notice that . This is clearly not in .
Further fact shown by the example
This example shows some further facts:
- The intersection of a permutable subgroup with a direct factor need not be a permutable subgroup. In this example, for instance, is a direct factor, but its intersection with is still not a permutable subgroup.
- A permutable subgroup of a direct factor need not be a permutable subgroup. In this case is a permutable subgroup inside , which itself is a direct factor.
- Permutability is not a direct product-closed subgroup property