Equivalence of definitions of weakly abnormal subgroup

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 fact about::self-normalizing subgroup of $$G$$.
 * 3) If $$H \le L \le G$$, then $$H$$ is a fact about::contranormal subgroup of $$L$$.

(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 $$K = N_G(K)$$.