# Maximal conjugate-permutable implies normal

This article gives the proof of a maximality equivalence. In other words, there are two subgroup properties: a stronger one (Normal subgroup (?)) and a weaker one (Conjugate-permutable subgroup (?)). However, any subgroup maximal among the proper subgroups with the weaker property also has the stronger property, and is thus also maximal among proper subgroups with the stronger property.
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## 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 $G$.

## Facts used

1. 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. 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.