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

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) uses::2-subnormal implies join-transitively subnormal
 * 2) uses::Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal

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.