Join of normal and subnormal implies subnormal of same depth

Statement
Let $$k \ge 1$$ be a natural number. Then, the join of any normal subgroup of a group with any $$k$$-subnormal subgroup is $$k$$-subnormal (Note: When we say a subgroup is $$k$$-subnormal, we mean that its fact about::subnormal depth is at most equal to $$k$$).

In symbols, if $$K$$ is a $$k$$-subnormal subgroup of $$G$$ and $$H$$ is a normal subgroup of $$G$$, then the join of subgroups $$\langle H, K \rangle$$ (which in this case is also the product of subgroups $$HK$$) is also a $$k$$-subnormal subgroups.

Facts used

 * 1) uses::Join lemma for normal subgroup of subgroup with normal subgroup of whole group: This states that if $$A \underline{\triangleleft} B \le G$$ and $$C \underline{\triangleleft} G$$, then $$\langle A, C \rangle \underline{\triangleleft} \langle B, C \rangle$$.

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

To prove: $$\langle H, K \rangle$$ is $$k$$-subnormal in $$G$$.

Proof: