Conjugate-join-closed subnormal implies join-transitively subnormal

Statement
Suppose $$H$$ is a conjugate-join-closed subnormal subgroup of a group $$G$$: for any subset $$S$$ of $$G$$, the subgroup $$H^S = \langle H^s \mid s \in S \rangle$$ is a subnormal subgroup of $$G$$.

Then, $$H$$ is a join-transitively subnormal subgroup of $$G$$: for any subnormal subgroup $$K$$ of $$G$$, the [[join of subgroups $$\langle H, K \rangle$$ is also a subnormal subgroup of $$G$$.

Facts used

 * 1) uses::Join of subnormal subgroups is subnormal iff their commutator is subnormal: Suppose $$H, K \le G$$ are subnormal subgroups. Then $$[H,K]$$ is subnormal if and only if $$H^K$$ is subnormal, if and only if $$\langle H, K \rangle$$ is subnormal.

Related facts

 * Join-transitively subnormal implies finite-automorph-join-closed subnormal
 * Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal
 * 2-subnormal implies join-transitively subnormal
 * 2-subnormality is conjugate-join-closed

Proof
Given: A conjugate-join-closed subnormal subgroup $$H$$ of a group $$G$$, a subnormal subgroup $$K$$ of $$G$$.

To prove: $$\langle H, K \rangle$$ is subnormal.

Proof:


 * 1) (Given data used: $$H$$ is conjugate-join-closed subnormal in $$G$$): Since $$H$$ is conjugate-join-closed subnormal in $$G$$, $$H^K$$ is subnormal in $$G$$.
 * 2) (Given data used: $$H, K$$ are subnormal in $$G$$; Fact used: fact (1)): By fact (1), we get that $$\langle H, K \rangle$$ is subnormal in $$G$$.