Maximal conjugate-permutable implies normal

Statement

Suppose $H$ is a proper 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 Normal subgroup (?) of $G$.

Related facts

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. Conjugate-permutability is conjugate-join-closed: If $H$ is a conjugate-permutable subgroup of $G$, and $g\in G$, then $H{H}^g$ is also a conjugate-permutable subgroup of $G$.
2. Product of conjugates is proper: If $H$ is a proper subgroup of $G$, and $g\in G$, then $H{H}^g$ 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 $K=H{H}^g$. 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 $G=H{H}^g$, 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.