3-subnormal implies finite-conjugate-join-closed subnormal

From Groupprops
Jump to: navigation, search
This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., 3-subnormal subgroup) must also satisfy the second subgroup property (i.e., finite-conjugate-join-closed subnormal subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about 3-subnormal subgroup|Get more facts about finite-conjugate-join-closed subnormal subgroup

Statement

Any 3-subnormal subgroup of a group is finite-conjugate-join-closed subnormal: a join of finitely many conjugates of such a subgroup is a subnormal subgroup. Further, the join of r such conjugates has subnormal depth at most 2^r + 1.

Related facts

Facts used

  1. 2-subnormal implies join-transitively subnormal
  2. Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal

Proof

Proof idea

The proof follows directly from facts (1) and (2), and the fact that a 2-subnormal subgroup of a normal subgroup is 3-subnormal.

Proof details

PLACEHOLDER FOR INFORMATION TO BE FILLED IN: [SHOW MORE]