Nilpotent-quotient implies subgroup-to-quotient powering-invariance implication

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., nilpotent-quotient subgroup) must also satisfy the second subgroup property (i.e., normal subgroup satisfying the subgroup-to-quotient powering-invariance implication)
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Statement

Suppose $G$ is a group and $H$ is a normal subgroup of $G$ that is a nilpotent-quotient subgroup, i.e., the quotient group $G/H$ is a nilpotent group (note that $G$ and $H$ themselves may or may not be nilpotent). Then, $H$ is a normal subgroup satisfying the subgroup-to-quotient powering-invariance implication in $G$. Explicitly, this means that if $p$ is a prime number such that both $G$ and $H$ are $p$-powered, then so is $G/H$.

Facts used

1. Divisibility is inherited by quotient groups
2. Equivalence of definitions of nilpotent group that is torsion-free for a set of primes: We use the equivalence of (1) and (2) within the multi-part equivalence. This says that a nilpotent group has no non-identity elements of order $p$ if and only if its $p^{th}$ power map is injective.

Proof

Given: Group $G$, normal subgroup $H$ such that both $G$ and $H$ are $p$-powered for some prime number $p$.

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

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$G/H$ is $p$-divisible, i.e., any element in it has a $p^{th}$ root. In other words, the $p^{th}$ power map in it is surjective.	Fact (1)	$G$ is $p$-powered, hence $p$-divisible, $H$ is normal in $G$.
2	$G/H$ is $p$-torsion-free.		$G$ and $H$ are both $p$-powered.		Any element of $H$ has a unique $p^{th}$ root in $H$, and also a unique $p^{th}$ root in $G$, so those unique roots coincide, which means it has no $p^{th}$ root outside $H$. Thus, $G/H$ has no non-identity element of order $p$.
3	The $p^{th}$ power map in $G/H$ is injective.	Fact (2)	$G/H$ is nilpotent.	Step (2)	Fact-given-step combination direct.
4	$G/H$ is $p$-powered, i.e., its $p^{th}$ power map is bijective.			Steps (1), (3)	Step-combination direct.