Characteristic central factor of WNSCDIN implies WNSCDIN

This article describes a computation relating the result of the Composition operator (?) on two known subgroup properties (i.e., Characteristic central factor (?) and WNSCDIN-subgroup (?)), to another known subgroup property (i.e., WNSCDIN-subgroup (?))
View a complete list of composition computations

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., characteristic central factor) must also satisfy the second subgroup property (i.e., left-transitively WNSCDIN-subgroup)
View all subgroup property implications | View all subgroup property non-implications
Get more facts about characteristic central factor|Get more facts about left-transitively WNSCDIN-subgroup

Statement

Suppose $H\leq K\leq G$ are groups. Suppose $H$ is a characteristic central factor of $K$: in other words, $H$ is both a Characteristic subgroup (?) and a Central factor (?) of $K$. Suppose $K$ is a WNSCDIN-subgroup of $G$. Then, $H$ is also a WNSCDIN-subgroup of $G$.

In other words, if $H$ is a characteristic central factor of $K$, then $H$ is a left-transitively WNSCDIN-subgroup of $K$.

Facts used

1. Characteristic of normal implies normal

Proof

Given: A group $G$, a WNSCDIN-subgroup $K$ of $G$. A characteristic central factor $H$ of $K$.

To prove: $H$ is a WNSCDIN-subgroup of $G$. In other words, if $A$ and $B$ are normal subsets of $H$ that are conjugate in $G$, then $A$ and $B$ are conjugate in $N_{G}(H)$.

Proof:

1. $A,B$ are normal subsets of $K$: Any inner automorphism of $K$ restricts to an inner automorphism of $H$. In particular, since $A$ is invariant under all inner automorphisms of $H$, it is invariant under all inner automorphisms of $H$. Thus, $A$ (and similarly $B$), are normal subsets of $K$.
2. $A,B$ are conjugate in $N_{G}(K)$: This follows from the previous step, and the fact that $K$ is WNSCDIN in $G$.
3. $N_{G}(K)\leq N_{G}(H)$: $H$ is characteristic in $K$ and $K$ is normal in $N_{G}(K)$. Thus, by fact (1), $H$ is normal in $N_{G}(K)$. Thus, $N_{G}(K)\leq N_{G}(H)$.
4. $A,B$ are conjugate in $N_{G}(H)$: This follows from the previous two steps.