Pronormal subgroup is normal subset-conjugacy-determined in normalizer

History

This theorem is attributed to Burnside.

Statement

Suppose $H$ is a Pronormal subgroup (?) of a 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: $x^{g} = g^{- 1} x g$ .

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 $H \leq N_{G} (A)$ by assumption, we get $H^{g} \leq N_{G} (A)^{g} = N_{G} (B)$ . Thus, both $H$ and $H^{g}$ are subgroups of $N_{G} (B)$ .

By the assumption that $H$ is pronormal in $G$ , we can find $x \in ⟨ H, H^{g} ⟩$ such that $H^{x} = H^{g}$ . In particular, $x \in N_{G} (B)$ .

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

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

References

Textbook references

Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 240, Theorem 1.1, Section 7.1 (Local fusion)