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 $H\leq K\leq G$ 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.

Related facts

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 $G$.

Proof by construction of a subnormal series

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