# WNSCDIN implies every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed

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., WNSCDIN-subgroup) must also satisfy the second subgroup property (i.e., subgroup in which every normalizer-relatively normal conjugation-invariantly relatively normal subgroup is weakly closed)
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## Statement

If $K$ is a WNSCDIN-subgroup of $G$, and $H$ is a normalizer-relatively normal conjugation-invariantly relatively normal subgroup of $K$ with respect to $G$, then $H$ is a weakly closed subgroup of $K$ with respect to $G$.

## Definitions used

### WNSCDIN-subgroup

Further information: WNSCDIN-subgroup

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

### Normalizer-relatively normal subgroup

Further information: Normalizer-relatively normal subgroup

Suppose $H \le K \le G$. $H$ is normalizer-relatively normal in $K$ with respect to $G$ if $H$ is normal in $N_G(K)$. In other words, $N_G(K) \le N_G(H)$.

### Conjugation-invariantly relatively normal subgroup

Further information: Conjugation-invariantly relatively normal subgroup

Suppose $H \le K \le G$. $H$ is conjugation-invariantly relatively normal in $K$ with respect to $G$ if $H$ is normal in every conjugate of $K$ in $G$ that contains $H$.

### Weakly closed subgroup

Further information: Weakly closed subgroup

Suppose $H \le K \le G$. $H$ is weakly closed in $K$ relative to $G$ if, for any $x \in G$ such that $xHx^{-1} \le K$, we have $xHx^{-1} \le H$.

## Proof

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Given: Groups $H \le K \le G$ such that $K$ is a WNSCDIN-subgroup of $G$. $H$ is normal in $N_G(K)$, and $H$ is normal in $gKg^{-1}$ for all $g \in G$ such that $H \le gKg^{-1}$. We have $x \in G$ such that $xHx^{-1} \le K$.

To prove: $xHx^{-1} \le H$ (in fact, we'll prove equality).

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 $H$ is a normal subgroup of $x^{-1}Kx$ $xHx^{-1} \le K$ Since $xHx^{-1} \le K$, conjugating both sides by $x^{-1}$ yields $H \le x^{-1}Kx$. By the assumption that $H$ is normal in every conjugate of $K$ containing it, $H$ is a normal subgroup of $x^{-1}Kx$.
2 $xHx^{-1}$ is a normal subgroup of $K$. Step (1) This follows from the previous step, and conjugating by $K$.
3 $H$ is a normal subgroup of $K$. normality satisfies intermediate subgroup condition $H \le K$, $H$ is normal in $N_G(K)$ Step-fact direct
4 $H$ and $xHx^{-1}$ are conjugate in $N_G(K)$ $K$ is a WNSCDIN-subgroup of $G$ Steps (2), (3) Given-step direct
5 $H = xHx^{-1}$ $H$ is normal in $N_G(K)$ Step (4) Given-step direct