Pronormal subgroup is normal subset-conjugacy-determined in normalizer

History
This theorem is attributed to Burnside.

Statement
Suppose $$H$$ is a fact about::pronormal subgroup of a fact about::group $$G$$ and $$N_G(H)$$ be its normalizer in $$G$$. Suppose $$A, B$$ are two normal subsets of $$H$$: they are both invariant under the action of $$H$$ by conjugation. Then, if $$A$$ and $$B$$ are conjugate in $$G$$, they are also conjugate in $$N_G(H)$$.

Note that the element of $$N_G(H)$$ may not act on $$A$$ in precisely the same way as the element of $$G$$. For the sharper result that the two elements act in the same way, we need to take both $$A$$ and $$B$$ to be inside the center of $$H$$.

Related facts

 * Center of pronormal subgroup is subset-conjugacy-determined in normalizer: The proof is the same, except that we use centralizers instead of normalizers, and obtain a sharper result stating that the element of the normalizer acts in exactly the same way as the element of the group.
 * Abelian pronormal subgroup is subset-conjugacy-determined in normalizer
 * Abelian and abnormal implies subset-conjugacy-closed

Proof
This proof uses the right action convention: $$x^g = g^{-1}xg$$.

Given: A pronormal subgroup $$H$$ of a group $$G$$ with normalizer $$N_G(H)$$. $$A$$ and $$B$$ are subsets of $$H$$ invariant under the action of $$H$$ by conjugation. There exists $$g \in G$$ such that $$A^g = B$$.

To prove: There exists $$h \in N_G(H)$$ such that $$A^h = B$$. Note that the action of $$h$$ by conjugation on $$A$$ may not be precisely the same as the action of $$g$$ on $$A$$.

Proof: Let $$N_G(A)$$ denote the set of all $$y \in G$$ such that $$A^y = A$$. Define $$N_G(B)$$ similarly. Then, $$N_G(A^g) = N_G(A)^g = N_G(B)$$. Since $$H \le N_G(A)$$ by assumption, we get $$H^g \le N_G(A)^g = N_G(B)$$. Thus, both $$H$$ and $$H^g$$ are subgroups of $$N_G(B)$$.

By the assumption that $$H$$ is pronormal in $$G$$, we can find $$x \in \langle H, H^g \rangle$$ such that $$H^x = H^g$$. In particular, $$x \in N_G(B)$$.

Define $$h = gx^{-1}$$. $$H^h = (H^g)^{x^{-1}} = (H^x)^{x^{-1}} = H^{xx^{-1}} = H$$. Thus, $$h \in N_G(H)$$. Further, $$A^h = A^{gx^{-1}} = (A^g)^{x^{-1}} = B^{x^{-1}} = B$$, where the last step follows from the fact that $$x \in N_G(B)$$.

Thus, $$h$$ satisfies the required conditions and we are done.

Textbook references

 * , Page 240, Theorem 1.1, Section 7.1 (Local fusion)