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: $x^{g}=g^{-1}xg$.

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 $H\leq N_{G}(A)$ 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 $x\in \langle H,H^{g}\rangle$ 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.

References

Textbook references

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