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 $H \le K \le G$, then $K$ is a Self-normalizing subgroup (?) of $G$.
3. If $H \le L \le G$, then $H$ is a Contranormal subgroup (?) of $L$.

Proof

(1) implies (2)

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

To prove: $K = N_G(K)$.

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

(2) implies (1)

Given: $H \le G$ 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 $H \le L \le G$, $H$ is contranormal in $L$.

Proof: Let $K$ be the normal closure of $H$ in $L$. Since $K$ is normal in $L$, $L \le N_G(K)$. Since $H \le K$, $N_G(K) = K$, so $L \le K$, forcing $K = L$. Thus, $H$ is contranormal in $L$.

(3) implies (2)

Given: A subgroup $H$ of $G$ with the property that if $H \le L \le G$, $H$ 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 $H \le K \le L \le G$. By assumption, $H$ is contranormal in $L$, and $K$ is normal in $L$, forcing $K = L$. Thus, we get $K = N_G(K)$.