Abnormal implies WNSCC

Statement with symbols
Suppose $$H$$ is an abnormal subgroup of a group $$G$$. Then, $$H$$ is a WNSCC-subgroup of $$G$$: for any two normal subsets $$A,B$$ of $$H$$ with $$gAg^{-1} = B$$ for some $$g \in G$$, we have $$A = B$$.

Definitions used
(These definitions use the left action convention, but using the right action convention does not change any of the proofs).

Abnormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed abnormal in $$G$$ if, for any $$g \in G$$, we have $$g \in \langle H, gHg^{-1} \rangle$$.

WNSCC-subgroup
A subgroup $$H$$ of a group $$G$$ is termed WNSCC in $$G$$ if, for any two normal subsets $$A,B$$ of $$H$$ such that there exists $$g \in G$$ with $$gAg^{-1} = B$$, we have $$A = B$$.

Stronger facts

 * Pronormal implies WNSCDIN

Proof
Given: A group $$G$$, an abnormal subgroup $$H$$. Two normal subsets $$A,B$$ of $$H$$ such that there exists $$g \in G$$ with $$gAg^{-1} = B$$.

To prove: $$A = B$$.

Proof:


 * 1) $$H \le N_G(A)$$ and $$H \le N_G(B)$$ (here $$N_G(B)$$ denotes the normalizer of the subset $$B$$ in $$G$$): This is a direct consequence of the fact that $$A, B$$ are normal subsets of $$H$$.
 * 2) $$gHg^{-1} \le N_G(B)$$: Since $$H \le N_G(A)$$, and conjugation by $$g$$ is an automorphism, we get $$gHg^{-1} \le N_G(gAg^{-1})$$, yielding $$gHg^{-1} \le B$$.
 * 3) $$\langle H, gHg^{-1} \in N_G(B)$$: This follows from the previous two steps.
 * 4) $$g \in N_G(B)$$: Since $$H$$ is abnormal, $$g \in \langle H, gHg^{-1}$$. Combining this with the previous step yields $$g \in N_G(B)$$.
 * 5) $$A = B$$: Since $$g \in N_G(B)$$, we have $$g^{-1} \in N_G(B)$$, so $$g^{-1}Bg = B$$. Also, by assumption, $$g^{-1}Bg = A$$. Combining these, we get $$A = B$$.