Pronormal implies WNSCDIN

This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., pronormal subgroup) must also satisfy the second subgroup property (i.e., WNSCDIN-subgroup)
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Statement

Verbal statement

Any pronormal subgroup of a group is a WNSCDIN-subgroup: it is a weak normal subset-conjugacy-determined subgroup inside its normalizer relative to the whole group.

Definitions used

(These definitions use the left action convention. The proof using the right action convention is the same).

Pronormal subgroup

Further information: Pronormal subgroup

A subgroup $H$ of a group $G$ is termed a pronormal subgroup if, for any $g\in G$, there exists $k\in \langle H,gHg^{-1}\rangle$ such that $kHk^{-1}=gHg^{-1}$.

WNSCDIN-subgroup

Further information: WNSCDIN-subgroup

A subgroup $H$ of a group $G$ is termed a WNSCDIN-subgroup if, for any normal subsets $A,B$ of $H$, and any $g\in G$ such that $gAg^{-1}=B$, there exists $k\in N_{G}(H)$ such that $kAk^{-1}=B$.

Related facts

Applications

• Sylow implies WNSCDIN
• Center of pronormal subgroup is subset-conjugacy-determined in normalizer
• Abnormal implies WNSCC

Proof

Given: A group $G$, a pronormal subgroup $H$ of $G$. Normal subsets $A,B$ of $H$. An element $g\in G$ such that $gAg^{-1}=B$.

To prove: There exists $k\in N_{G}(H)$ such that $kAk^{-1}=B$.

Proof:

1. $H\leq N_{G}(A)$ and $H\leq N_{G}(B)$ (where $N_{G}(B)$ denotes the normalizer of the subset $B$): This follows from the fact that $A,B$ are normal subsets of $H$.
2. $gHg^{-1}\leq N_{G}(B)$: We have $H\leq N_{G}(A)$. Since conjugation by $g$ is an automorphism, it preserves normalizers, so $gHg^{-1}\leq N_{G}(gAg^{-1})=N_{G}(B)$.
3. $\langle H,gHg^{-1}\rangle \leq N_{G}(B)$: This follows from the previous two steps.
4. There exists $x\in N_{G}(B)$ such that $xHx^{-1}=gHg^{-1}$: This follows from the previous step, and the fact that $H$ is pronormal.
5. The element $k=x^{-1}g$ is an element of $N_{G}(H)$ such that $kAk^{-1}=B$:
   1. $k\in N_{G}(H)$: Since $xHx^{-1}=gHg^{-1}$, we have $x^{-1}(gHg^{-1})x=x^{-1}(xHx^{-1})x=H$.
   2. $kAk^{-1}=B$: We have $k=x^{-1}g$, with $gAg^{-1}=B$, and $x^{-1}Bx=B$ (since $x\in N_{G}(B)$), so $kAk^{-1}=x^{-1}(gAg^{-1})x=x^{-1}Bx=B$.