Pronormality is not finite-upper join-closed
This article gives the statement, and possibly proof, of a subgroup property (i.e., pronormal subgroup) not satisfying a subgroup metaproperty (i.e., finite-upper join-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Upper join-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 pronormal subgroup|Get more facts about finite-upper join-closed subgroup propertyGet more facts about upper join-closed subgroup property|
We can have a subgroup such that is a pronormal subgroup in two intermediate subgroups and , but is not pronormal in the join .
Related facts about pronormality
- Pronormality satisfies intermediate subgroup condition: If and is pronormal in , then is also pronormal in .
Related facts about upper join-closedness
- 2-subnormality is not upper join-closed
- Subnormality is not upper join-closed
- Normality is upper join-closed
- Characteristicity is not upper join-closed
Let be the symmetric group on the set , be the subgroup comprising permutations on , be the subgroup comprising permutations on , and be the two-element subgroup generated by . Observe that:
- is a pronormal subgroup in both and . In fact, it is a maximal subgroup in both, and maximal implies pronormal.
- is not pronormal in . That's because the subgroup generated by is a conjugate of , and these two subgroups are not conjugate in the subgroup they generate.