Subnormality is permuting upper join-closed in finite

This article gives the statement, and possibly proof, of a subgroup property (i.e., subnormal subgroup of finite group) satisfying a subgroup metaproperty (i.e., permuting upper join-closed subgroup property)
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History

The result was proved both by Maier and by Wielandt.

Statement

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 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 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)