Subnormality is normalizing join-closed

Statement

Suppose $H,K\le G$ are Subnormal subgroup (?)s, with the property that $K\le N_G(H)$: in other words, $K$ normalizes $H$. Then the join of subgroups $⟨H,K⟩$ is also subnormal. Moreover, the Subnormal depth (?) of $⟨H,K⟩$ is bounded from above by the products of subnormal depths of $H$ and $K$.

Related facts

• Join of normal and subnormal implies subnormal of same depth
• 2-subnormality is conjugate-join-closed

Facts used

1. Join of normal and subnormal implies subnormal of same depth

References

Textbook references

• A Course in the Theory of Groups by Derek J. S. Robinson, ISBN 0387944613, ^{More info}, Page 387, Section 13.1 (Joins and intersections of subnormal subgroups)