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

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)
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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

• Join of 3-subnormal subgroups need not be subnormal
• 2-subnormality is conjugate-join-closed
• 2-subnormal implies join-transitively subnormal
• 4-subnormal not implies finite-conjugate-join-closed subnormal

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

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