Conjugate-permutable implies subnormal in finite

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties, when the big group is a finite group. That is, it states that in a Finite group (?), every subgroup satisfying the first subgroup property (i.e., Conjugate-permutable subgroup (?)) must also satisfy the second subgroup property (i.e., Subnormal subgroup (?)). In other words, every conjugate-permutable subgroup of finite group is a subnormal subgroup of finite group.
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Statement

Any conjugate-permutable subgroup of a finite group is subnormal.

Related facts

• Permutable implies subnormal in finite
• 2-subnormal implies conjugate-permutable
• Subnormal not implies conjugate-permutable
• Conjugate-permutable implies descendant in slender

Facts used

1. Conjugate-permutability satisfies intermediate subgroup condition
2. Maximal conjugate-permutable implies normal

Proof

Given: A finite group $G$, a conjugate-permutable subgroup $H$ of $G$.

To prove: $H$ is a subnormal subgroup of $G$.

Proof:

1. (Fact used: fact (1)): Define a descending chain as follows: $K_{0}=G$, and if $K_{i}$ properly contains $H$, $K_{i+1}$ is a maximal element among the proper conjugate-permutable subgroups of $K_{i}$ that contain $H$.
   1. (Well-definedness): Note that by fact (1), $H$ is conjugate-permutable in $K_{i}$, so the collection of proper conjugate-permutable subgroups of $K_{i}$ containing $H$ is nonempty. Since $G$ is finite, it has a maximal element.
   2. (Terminates at $H$ in finitely many steps): The chain $K_{i}$ is a strictly descending chain of subgroups until it reaches $H$. Since $G$ is finite, it terminates in finitely many steps at $H$. Thus, there exists $n$ such that $K_{n}=H$.
   3. (Fact used: fact (2)): By definition, $K_{i+1}$ is a maximal conjugate-permutable subgroup of $K_{i}$, so fact (2) tells us that $K_{i+1}$ is normal in $K_{i}$.
   4. (Conclusion): The $K_{i}$s thus form a subnormal series for $H$ in $G$, making $H$ a subnormal subgroup of $G$.