Center of pronormal implies SCDIN

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

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, $$N_G(H) \le N_G(Z(H))$$.