2-subnormal implies join-transitively subnormal

Statement with symbols
Suppose $$H, K \le G$$ are subgroups such that $$H$$ is a 2-subnormal subgroup of $$G$$ and $$K$$ is a subnormal subgroup of $$G$$. Then, the join $$\langle H, K \rangle$$ is a subnormal subgroup of $$G$$, and its subnormal depth in $$G$$ is at most twice the subnormal depth of $$K$$.

Related facts

 * Weaker than in assumption and conclusion::Join of normal and subnormal implies subnormal of same depth
 * Permutable and subnormal implies join-transitively subnormal
 * Subnormality is normalizing join-closed
 * Subnormality is permuting join-closed
 * 2-subnormality is conjugate-join-closed

Facts used

 * 1) uses::2-subnormality is conjugate-join-closed: A join of any collection of 2-subnormal subgroups that are conjugate to each other is again 2-subnormal.
 * 2) uses::Join of subnormal subgroups is subnormal iff their commutator is subnormal: Suppose $$H, K$$ are subnormal subgroups of a group $$G$$. Then, consider the subgroups $$[H,K]$$ (the commutator) of $$H$$ and $$K$$), the subgroup $$H^K$$ (the join of $$H^k$$ for all $$k \in K$$) and the subgroup $$\langle H, K \rangle$$. If any one of these is subnormal, so are the other two. Further, if $$s_1, s_2, s_3$$ denote respectively the subnormal depths of $$[H,K], H^K, \langle H, K \rangle$$, we have $$s_3 \le s_2k$$.

Proof
Given: A group $$G$$, a $$2$$-subnormal subgroup $$H$$, a $$k$$-subnormal subgroup $$K$$.

To prove: $$\langle H, K \rangle$$ is a $$2k$$-subnormal subgroup of $$G$$.

Proof: