Equivalence of definitions of weakly abnormal subgroup

This article gives a proof/explanation of the equivalence of multiple definitions for the term weakly abnormal subgroup
View a complete list of pages giving proofs of equivalence of definitions

The definitions that we have to prove as equivalent

Here are two equivalent definitions of weakly abnormal for a subgroup $H$ of a group $G$:

1. For any $g\in G$, let $H^{\langle g\rangle }$ be the closure in $G$ of $H$ under the action by conjugation of the cyclic group generated by $g$. Then, $g\in H^{\langle g\rangle }$.
2. If $H\leq K\leq G$, then $K$ is a Self-normalizing subgroup (?) of $G$.
3. If $H\leq L\leq G$, then $H$ is a Contranormal subgroup (?) of $L$.

Proof

(1) implies (2)

Given: $H\leq G$ such that $g\in H^{\langle g\rangle }$ for all $g\in G$. $H\leq K\leq G$.

To prove: $K=N_{G}(K)$.

Proof: Suppose $g\in N_{G}(K)$. Then, $K^{\langle g\rangle }=K$, so $H^{\langle g\rangle }\leq K^{\langle g\rangle }\leq K$. In particular, $g\in H^{\langle g\rangle }\leq K$, so $g\in K$. Thus, $N_{G}(K)\leq K$. $K\leq N_{G}(K)$ is tautological, so $N_{G}(K)=K$.

(2) implies (1)

Given: $H\leq G$ such that, for any intermediate subgroup $K$, $K=N_{G}(K)$.

To prove: $g\in H^{\langle g\rangle }$ for any $g\in G$.

Proof: Let $K=H^{\langle g\rangle }$. Then, by definition of $H^{\langle g\rangle }$, both $g$ and $g^{-1}$ map $K$ to within itself, so conjugation by $g$ gives an automorphism of $K$. Thus, $g\in N_{G}(K)$. By assumption, $K=N_{G}(K)$, so $g\in K$, yielding $g\in H^{\langle g\rangle }$, as desired.

(2) implies (3)

Given: A subgroup $H$ of $G$ with the property that $N_{G}(K)=K$ for every subgroup $K$ of $G$ containing $H$.

To prove: If $H\leq L\leq G$, $H$ is contranormal in $L$.

Proof: Let $K$ be the normal closure of $H$ in $L$. Since $K$ is normal in $L$, $L\leq N_{G}(K)$. Since $H\leq K$, $N_{G}(K)=K$, so $L\leq K$, forcing $K=L$. Thus, $H$ is contranormal in $L$.

(3) implies (2)

Given: A subgroup $H$ of $G$ with the property that if $H\leq L\leq G$, $H$ is contranormal in $L$.

To prove: $N_{G}(K)=K$ for any subgroup $K$ of $G$ containing $H$.

Proof: Let $L=N_{G}(K)$. Then $H\leq K\leq L\leq G$. By assumption, $H$ is contranormal in $L$, and $K$ is normal in $L$, forcing $K=L$. Thus, we get $K=N_{G}(K)$.