Max-equivalent to normal and intermediate subgroup condition implies subnormal in finite

Statement
Suppose $$p$$ is a subgroup property satisfying the following two conditions:


 * If $$H$$ is maximal among the proper subgroups of a group $$G$$ satisfying $$p$$, then $$H$$ is a normal subgroup.
 * $$p$$ satisfies the intermediate subgroup condition: If $$H \le K \le G$$ and $$H$$ has property $$p$$ in $$G$$, then $$H$$ has property $$p$$ in $$K$$.

Then, every subgroup of a finite group satisfying property $$p$$ is a subnormal subgroup.

Generalizations

 * Max-equivalent to normal implies descendant in slender

Applications

 * Permutable implies subnormal in finite
 * Conjugate-permutable implies subnormal in finite

Proof
These two proofs are essentially the same; one is presented in an inductive style, and the other is presented in terms of construction of the subnormal series.

Inductive proof
Given: A subgroup $$H$$ of a group $$G$$ satisfying property $$p$$. $$p$$ is max-equivalent to normality and satisfies the intermediate subgroup condition.

To prove: $$H$$ is subnormal in $$G$$.

Proof:


 * 1) Consider the collection $$\mathcal{C}$$ of proper subgroups of $$G$$ containing $$H$$ and satisfying property $$p$$. Note that $$\mathcal{C}$$ is nonempty (it contains $$H$$) and finite, so it has maximal elements under inclusion. Let $$K$$ be such an element.
 * 2) Since $$p$$ is max-equivalent to normality, $$K$$ is normal in $$G$$.
 * 3) Since $$p$$ satisfies the intermediate subgroup condition, $$H$$ satisfies $$p$$ in $$K$$. Thus, by induction on the order, $$H$$ is subnormal in $$K$$.
 * 4) Combining the previous two steps, we see that $$H$$ is subnormal in $$G$$.