# Abnormal implies WNSCC

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., abnormal subgroup) must also satisfy the second subgroup property (i.e., WNSCC-subgroup)
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## Statement

### Statement with symbols

Suppose $H$ is an abnormal subgroup of a group $G$. Then, $H$ is a WNSCC-subgroup of $G$: for any two normal subsets $A,B$ of $H$ with $gAg^{-1} = B$ for some $g \in G$, we have $A = B$.

## Definitions used

(These definitions use the left action convention, but using the right action convention does not change any of the proofs).

### Abnormal subgroup

Further information: Abnormal subgroup

A subgroup $H$ of a group $G$ is termed abnormal in $G$ if, for any $g \in G$, we have $g \in \langle H, gHg^{-1} \rangle$.

### WNSCC-subgroup

Further information: WNSCC-subgroup

A subgroup $H$ of a group $G$ is termed WNSCC in $G$ if, for any two normal subsets $A,B$ of $H$ such that there exists $g \in G$ with $gAg^{-1} = B$, we have $A = B$.

## Proof

Given: A group $G$, an abnormal subgroup $H$. Two normal subsets $A,B$ of $H$ such that there exists $g \in G$ with $gAg^{-1} = B$.

To prove: $A = B$.

Proof:

1. $H \le N_G(A)$ and $H \le N_G(B)$ (here $N_G(B)$ denotes the normalizer of the subset $B$ in $G$): This is a direct consequence of the fact that $A, B$ are normal subsets of $H$.
2. $gHg^{-1} \le N_G(B)$: Since $H \le N_G(A)$, and conjugation by $g$ is an automorphism, we get $gHg^{-1} \le N_G(gAg^{-1})$, yielding $gHg^{-1} \le B$.
3. $\langle H, gHg^{-1} \in N_G(B)$: This follows from the previous two steps.
4. $g \in N_G(B)$: Since $H$ is abnormal, $g \in \langle H, gHg^{-1}$. Combining this with the previous step yields $g \in N_G(B)$.
5. $A = B$: Since $g \in N_G(B)$, we have $g^{-1} \in N_G(B)$, so $g^{-1}Bg = B$. Also, by assumption, $g^{-1}Bg = A$. Combining these, we get $A = B$.