Conjugate-permutable implies subnormal in finite

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) uses::Conjugate-permutability satisfies intermediate subgroup condition
 * 2) uses::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 $$H$$.
 * 2) (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.
 * 3) (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 $$K_n = H$$.
 * 4) (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 $$K_i$$.
 * 5) (Conclusion): The $$K_i$$s thus form a subnormal series for $$H$$ in $$G$$, making $$H$$ a subnormal subgroup of $$G$$.