Join-transitively subnormal of normal implies finite-conjugate-join-closed subnormal

Statement with symbols
Suppose $$H \le K \le G$$ are subgroups such that $$K$$ is a normal subgroup of $$G$$ and $$H$$ is a join-transitively subnormal subgroup of $$K$$ (in other words, the join of $$H$$ with any subnormal subgroup of $$G$$ is subnormal). Then, $$H$$ is a finite-conjugate-join-closed subnormal subgroup of $$G$$: a join of finitely many conjugate subgroups of $$H$$ in $$G$$ is again subnormal.

Related facts

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

Facts used

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

Hands-on proof
Given: $$H \le K \le G$$, such that $$K$$ is normal in $$G$$ and $$H$$ is join-transitively subnormal in $$K$$.

To prove: For any finite set $$S \subseteq G$$, the subgroup $$H^S$$, defined as the join of subgroups $$H^g, g \in S$$, is subnormal in $$G$$.

Proof:


 * 1) Each $$H^g$$ is join-transitively subnormal in $$K$$: Conjugation by $$g$$ defines an automorphism of $$K$$, since $$K$$ is normal in $$G$$. Thus, each $$H^g$$ is the image of $$H$$ under an automorphism of $$K$$. Since automorphism preserve subgroup properties, each $$H^g$$ is join-transitively subnormal in $$K$$.
 * 2) $$H^S$$ is subnormal in $$K$$: Since $$S$$ is a finite set, $$H^S$$ is the join of finitely many subgroups $$H^g$$, each of which is join-transitively subnormal in $$K$$. By induction, we see that $$H^S$$ is also join-transitively subnormal in $$K$$; in particular, it is subnormal in $$K$$.
 * 3) $$H^S$$ is subnormal in $$G$$: Since $$H^S$$ is subnormal in $$K$$ and $$K$$ is normal in $$G$$, we obtain that $$H^S$$ is subnormal in $$G$$.

Proof using given facts
The proof follows directly by piecing together facts (1) and (2).