Center of pronormal implies SCDIN

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., center of pronormal subgroup) must also satisfy the second subgroup property (i.e., SCDIN-subgroup)
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Statement

Statement with symbols

Suppose $G$ is a group and $H$ is a pronormal subgroup of $G$. Then, $Z(H)$, the center of $H$, is a SCDIN-subgroup of $G$.

Facts used

1. Pronormal implies MWNSCDIN
2. Characteristic central factor of MWNSCDIN implies MWNSCDIN
3. Abelian and MWNSCDIN implies SCDIN
4. Center of pronormal subgroup is subset-conjugacy-determined in normalizer
5. Characteristic of normal implies normal
6. Center is characteristic

Proof

Proof using facts (1)-(3)

Given: Pronormal subgroup $H$ of a group $G$.

To prove: $Z(H)$ is a SCDIN-subgroup of $G$.

Proof:

1. $H$ is MWNSCDIN in $G$: This follows from fact (1).
2. $Z(H)$ is a characteristic central factor of $H$: Both characteristicity and being a central factor are direct from the definitions (See fact (6)) for a proof of characteristicity).
3. $Z(H)$ is MWNSCDIN in $G$: This follows from the previous two steps and fact (2).
4. $Z(H)$ is SCDIN in $G$: This follows from the previous step and fact (3), along with the fact that by definition, $Z(H)$ is abelian.

Proof using fact (4)

Fact (4) yields that any two subsets $A,B$ of $Z(H)$ that are conjugate by $g\in G$ are conjugate by $h\in N_G(H)$ such that $g,h$ has the same element-wise action on $A$. To complete the proof, we only need to show that $N_G(H)\le N_G(Z(H))$. This follows from fact (5): $Z(H)$ is characteristic in $H$ (fact (6)), which is normal in $N_G(H)$, so $Z(H)$ is normal in $N_G(H)$. Thus, $N_G(H)\le N_G(Z(H))$.