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

Statement

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. Normality is strongly intersection-closed
3. Upper central series members are powering-invariant
4. Central implies normal satisfying the subgroup-to-quotient powering-invariance implication
5. Third isomorphism theorem

Proof

Case of containment in a member of the finite upper central series

Given: A group $G$ with upper central series $Z^{0}(G)\leq Z^{1}(G)\leq \dots$. 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:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	Consider the subgroup series $1=H\cap Z^{0}(G)\leq H\cap Z^{1}(G)\leq H\cap Z^{2}(G)\leq \dots \leq H\cap Z^{n}(G)=H$.
2	Each subgroup of the form $H\cap Z^{i}(G),0\leq i\leq n$, is a normal subgroup of $G$.	Facts (1), (2)	$H$ is normal in $G$.		Fact-given direct.
3	For any $i$ with $0\leq i\leq n-1$, $(H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))$ is central in $G/(H\cap Z^{i}(G))$.			Step (2) (for making sense of quotients)	We have $[G,H\cap Z^{i+1}(G)]\leq [G,H]\cap [G,Z^{i+1}(G)]\leq H\cap Z^{i}(G)$, and the claim follows.
4	For any $i$ with $0\leq i\leq n-1$, we have that if $G/(H\cap Z^{i}(G))$ is $p$-powered and $(H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))$ is $p$-divisible (here, $p$-divisible means every element has a not necessarily unique $p^{th}$ root in the group), then $G/(H\cap Z^{i+1}(G))$ is $p$-powered.	Facts (4), (5)		Step (3)	First note that if $G/(H\cap Z^{i}(G))$ is $p$-powered, then any $p$-divisible subgroup must be $p$-powered because of global uniqueness of $p^{th}$ roots. Thus, $(H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))$ is $p$-powered. Step (3) tells us that $(H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))$ is central in $G/(H\cap Z^{i}(G))$, hence by Fact (4) and the preceding sentences, we get that the quotient group ${\frac {G/(H\cap Z^{i}(G))}{H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))}}$ is $p$-powered. By Fact (5), this is isomorphic to $G/(H\cap Z^{i+1}(G))$, completing the proof.
5	Each of the subgroups $H\cap Z^{i}(G)$ is $p$-powered and hence $p</amth>-divisible.||Fact(3)||<math>G$ and $H$ are both $p$-powered.		By Fact (3), each $Z^{i}(G)$ is $p$-powered. Combine with the fact that $H$ is $p$-powered to get that the intersection is.
6	Each quotient group $(H\cap Z^{i+1}(G))/(H\cap Z^{i}(G))$ is $p$-divisible for $0\leq i\leq n-1$ (here, $p$-divisible means every element has a not necessarily unique $p^{th}$ root in the group).			Step (5)	This follows directly from Step (5), and the observation that any any quotient of a $p$-divisible group by a normal subgroup is $p$-divisible.
7	For any $i$ with $0\leq i\leq n-1$, we have that if $G/(H\cap Z^{i}(G))$ is $p$-powered, so is $G/(H\cap Z^{i+1}(G))$.			Steps (4), (6)	Step-combination direct.
8	$G/H$ is $p$-powered.		$G$ is $p$-powered.	Step (7)	Step (7) and use the principle of mathematical induction, starting from $i=0$ (i.e., $G$ is $p$-powered), and inducting all the way till we reach that $G/(H\cap Z^{n}(G))=G/H$ is $p$-powered.