Center of pronormal subgroup is subset-conjugacy-determined in normalizer

Statement
Suppose $$G$$ is a group and $$H$$ is a fact about::pronormal subgroup. Suppose $$A,B$$ are subsets of $$Z(H)$$ (the fact about::center of $$H$$) and $$g \in G$$ is such that $$A^g = B$$. Then, there exists $$h \in N_G(H)$$ such that, for all $$a \in A$$, $$a^h = a^g$$.

(Here, $$a^g = g^{-1}ag$$ denotes conjugation by $$g$$, following the right-action convention. The statement and proof remain the same for the left-action convention).

Pronormal subgroup
A subgroup $$H$$ of a group $$G$$ is termed a pronormal subgroup if for any $$g \in G$$, there exists $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g$$.

Stronger facts

 * Pronormal subgroup is normal subset-conjugacy-determined in normalizer

Corollaries

 * Center of Sylow subgroup is subset-conjugacy-determined in normalizer
 * Abelian pronormal subgroup is subset-conjugacy-determined in normalizer
 * Abelian and abnormal implies subset-conjugacy-closed

Proof
Given: $$H \le G$$ is a pronormal subgroup, $$A^g = B$$ for $$A, B \subseteq Z(H)$$ and $$g \in G$$.

To prove: There exists $$h \in N_G(H)$$ such that $$a^g = a^h$$ for all $$a \in A$$.

Proof: Since $$A^g = B$$, we have $$C_G(A)^g = C_G(B)$$. Now, since $$A \subseteq Z(H)$$, we have $$H \le C_G(A)$$, so $$H^g \le C_G(B)$$. Also, $$B \subseteq Z(H)$$, so $$H \le C_G(B)$$.

Thus, $$\langle H, H^g \rangle \le C_G(B)$$. By pronormality, there exists $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g$$. In particular, there exists $$x \in C_G(B)$$ such that $$H^x = H^g$$.

Now consider $$h = gx^{-1}$$. We check the two required conditions:


 * $$h \in N_G(H)$$: Clearly, $$H^h = H^{gx^{-1}} = (H^g)^{x^{-1}} = (H^x)^{x^{-1}} = H^{xx^{-1}} = H$$. So, $$H^h = H$$, and hence $$h \in N_G(H)$$.
 * $$a^g = a^h$$ for all $$a \in A$$: Pick $$a \in A$$. We have $$a^h = a^{gx^{-1}} = (a^g)^{x^{-1}}$$. Now, $$a^g \in B$$ and $$x \in C_G(B)$$, so $$x^{-1}$$ fixes $$a^g$$. Thus, $$a^h = a^g$$.