Maximal conjugate-permutable implies normal

Statement
Suppose $$H$$ is a proper fact about::conjugate-permutable subgroup of a group $$G$$ such that $$H$$ is not properly contained in any proper conjugate-permutable subgroup of $$G$$. Then, $$H$$ is a fact about::normal subgroup of $$G$$.

Similar facts

 * Pronormal implies self-conjugate-permutable
 * Maximal implies self-conjugate-permutable
 * Conjugate-permutable and self-conjugate-permutable implies normal

Applications

 * Conjugate-permutable implies subnormal in finite

Facts used

 * 1) uses::Conjugate-permutability is conjugate-join-closed: If $$H$$ is a conjugate-permutable subgroup of $$G$$, and $$g \in G$$, then $$HH^g$$ is also a conjugate-permutable subgroup of $$G$$.
 * 2) uses::Product of conjugates is proper: If $$H$$ is a proper subgroup of $$G$$, and $$g \in G$$, then $$HH^g$$ is a proper subset of $$G$$.

Proof
Given: A group $$G$$, a proper conjugate-permutable subgroup $$H$$ of $$G$$ such that $$H$$ is not contained in any proper conjugate-permutable subgroup of $$G$$.

To prove: $$H$$ is normal in $$G$$: for any $$g \in G$$, $$H^g \le H$$.

Proof:


 * 1) (Fact used: fact (1), conjugate-permutability is conjugate-join-closed): Suppose $$K = HH^g$$. Then, since $$H$$ is conjugate-permutable in $$G$$, fact (1) tells us that $$K$$ is also conjugate-permutable in $$G$$.
 * 2) (Given data used: $$H$$ is maximal conjugate-permutable): By our assumption, either $$H = K$$ or $$K = G$$.
 * 3) (Fact used: fact (2), product of conjugates is proper): If $$K = G$$, we have $$G = HH^g$$, which yields a contradiction by fact (2).
 * 4) Combining steps (2) and (3), we see that $$K = H$$, forcing $$H^g \le H$$, as required.