Subnormality is permuting upper join-closed in finite

History
The result was proved both by Maier and by Wielandt.

When the whole group is finite
Suppose $$G$$ is a finite group and $$H$$ is a subgroup of $$G$$. Suppose $$K_1, K_2$$ are intermediate subgroups of $$G$$ such that $$K_1K_2 = K_2K_1$$ (i.e., they are fact about::permuting subgroups) and $$H$$ is a subnormal subgroup in both $$K_1$$ and $$K_2$$. Then, $$H$$ is also subnormal in the product of subgroups $$K_1K_2$$.

Equivalent formulation when the whole group is not finite
Suppose $$G$$ is a group and $$H$$ is a finite subgroup of $$G$$. Suppose $$K_1, K_2$$ are intermediate finite subgroups of $$G$$ such that $$K_1K_2 = K_2K_1$$ (i.e., they are fact about::permuting subgroups) and $$H$$ is a subnormal subgroup in both $$K_1$$ and $$K_2$$. Then, $$H$$ is also subnormal in the product of subgroups $$K_1K_2$$.

Note that these two formulations are equivalent because even if $$G$$ is not finite, the product $$K_1K_2$$ is still finite since both $$K_1$$ and $$K_2$$ are finite.

Related facts

 * Subnormality is not permuting upper join-closed
 * Subnormality is not finite-upper join-closed (we can get counterexamples even in finite groups)