Normal subgroup contained in the hypercenter satisfies the subgroup-to-quotient powering-invariance implication

Original formulation
Suppose $$G$$ is a group and $$H$$ is a normal subgroup contained in the hypercenter of $$G$$. Suppose $$p$$ is a prime number such that $$G$$ and $$H$$ are both powered over the prime $$p$$. Then, the quotient group $$G/H$$ is also powered over $$p$$.

Corollary formulation
Suppose $$G$$ is a group and $$H$$ is a normal subgroup contained in the hypercenter of $$G$$ that is also a powering-invariant subgroup of $$G$$. Then, $$H$$ is a quotient-powering-invariant subgroup of $$G$$.

Facts used

 * 1) Upper central series members are normal (in fact, they are strictly characteristic subgroups)
 * 2) uses::Normality is strongly intersection-closed
 * 3) uses::Upper central series members are powering-invariant
 * 4) uses::Central implies normal satisfying the subgroup-to-quotient powering-invariance implication
 * 5) uses::Third isomorphism theorem

Case of containment in a member of the finite upper central series
Note that this case suffices for finite groups and nilpotent groups.

Given: A group $$G$$ with upper central series $$Z^0(G) \le Z^1(G) \le \dots$$. A normal subgroup $$H$$ of $$G$$ contained in $$Z^n(G)$$ for some positive integer $$n$$. A prime number $$p$$ such that both $$G$$ and $$H$$ are powered over $$p$$.

To prove: $$G/H$$ is powered over $$p$$.

Proof:

Modification for transfinite upper central series case
We need to modify the above to do transfinite induction. The main change is that the $$i$$ will everywhere be interpreted as an ordinal (possibly infinite). In addition, Step (7) will now need an accompanying step that handles the limit ordinal situation. The handling of the limit ordinal situation is similar to the proof for quotient-powering-invariance is union-closed, but done in slightly greater generality.