Pronormality is not finite-intersection-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-intersection-closed subgroup property).This also implies that it does not satisfy the subgroup metaproperty/metaproperties: Intersection-closed subgroup property (?), .
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An intersection of two pronormal subgroups of a group need not be pronormal.
Further information: pronormal subgroup
A subgroup of a group is termed pronormal in if, given any , and are conjugate in the subgroup they generate.
The example of the symmetric group
Further information: symmetric group:S4
Let be the symmetric group on the set . Let be the subgroup comprising the identity and double transpositions. Let be the subgroup generated by two disjoint single transpositions. Then, is a two-element subgroup.
- is pronormal: In fact, is a normal subgroup of .
- is pronormal: Any conjugate of is either equal to or intersects trivially, in which case they generate the whole group (in other words, is a subgroup whose join with any distinct conjugate is the whole group). Thus, is pronormal in (See fact (1)).
- is not pronormal: Indeed, this subgroup and a conjugate of it generate the subgroup , within which they are not conjugate.