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 $k\in ⟨H,gH{g}^{-1}⟩$ such that $kH{k}^{-1}=gH{g}^{-1}$.

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 $gA{g}^{-1}=B$, there exists $k\in N_G(H)$ such that $kA{k}^{-1}=B$.

Related facts

Applications

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

Proof

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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

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

Proof: Steps (5) and (6) together give the proof.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$H\le N_G(A)$ and $H\le N_G(B)$ (where $N_G(B)$ denotes the normalizer of the subset $B$)		$A,B$ are normal subsets of $H$.		Direct from the given
2	$gH{g}^{-1}\le N_G(B)$			Step (1)	$H\le N_G(A)$ by Step (1). Since conjugation by $g$ is an automorphism, it preserves normalizers, so $gH{g}^{-1}\le N_G(gA{g}^{-1})=N_G(B)$.
3	$⟨H,gH{g}^{-1}⟩\le N_G(B)$			Steps (1), (2)	This follows by combining the preceding steps.
4	There exists $x\in N_G(B)$ such that $xH{x}^{-1}=gH{g}^{-1}$		$H$ is pronormal	Step (3)	By pronormality of $H$, there exists $x\in ⟨H,gH{g}^{-1}⟩$ such that $xH{x}^{-1}=gH{g}^{-1}$. Step (3) yields that $x\in N_G(B)$.
5	The element $k=x^{-1}g$ is an element of $N_G(H)$.			Step (4)	Since $xH{x}^{-1}=gH{g}^{-1}$, we have $x^{-1}(gH{g}^{-1})x=x^{-1}(xH{x}^{-1})x=H$.
6	The element $k=x^{-1}g$ satisfies $kA{k}^{-1}=B$.		$gA{g}^{-1}=B$	Step (4)	$gA{g}^{-1}=B$, and $x^{-1}Bx=B$ (since $x\in N_G(B)$ by Step (4)), so $kA{k}^{-1}=x^{-1}(gA{g}^{-1})x=x^{-1}Bx=B$.