Local powering-invariant subgroup containing the center is intermediately powering-invariant in nilpotent group

Statement

Suppose $G$ is a nilpotent group and $H$ is a subgroup containing the center of $G$ that is also a local powering-invariant subgroup of $G$. Then, $H$ is an intermediately powering-invariant subgroup of $G$. Explicitly, suppose $K$ is a subgroup of $G$ containing $H$. Then, $H$ is a powering-invariant subgroup of $K$.

Related facts

• Local powering-invariant subgroup containing the center is intermediately local powering-invariant in nilpotent group

Facts used

1. Torsion-freeness for a prime is subgroup-closed
2. Equivalence of definitions of nilpotent group that is torsion-free for a set of primes

Proof

Given: A nilpotent group $G$, a subgroup $H$ of $G$ that is local powering-invariant and such that $Z(G)\leq H$ where $Z(G)$ is the center of $G$. A subgroup $K$ of $G$ containing $H$. A prime number $p$ such that $K$ is $p$-powered. An element $h\in H$.

To prove: There exists a unique element $x\in H$ such that $x^{p}=h$.

Proof:

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$Z(G)\leq K$.		$Z(G)\leq H,H\leq K$.		given-direct
2	$K$ is $p$-torsion-free.		$K$ is $p$-powered (we are using the powering-injectivity here)		given-direct
3	$Z(G)$ is $p$-torsion-free.	Fact (1)		Steps (1), (2)	Step-fact combination direct
4	The map $t\mapsto t^{p}$ is injective in $G$.	Fact (2)	$G$ is nilpotent	Step (3)	Step-fact combination direct (specifically, we want to use the implication from (4) to (1) in the multi-part equivalence of Fact (2))
5	There exists an element $x\in K$ satisfying $x^{p}=h$.		$K$ is $p$-powered (we are using the $p$-divisibility here), $h\in H\leq K$		given-direct
6	The element $x\in K$ of the preceding step is the unique $p^{th}$ root of $h$ in all of $G$.			Steps (4), (5)	Step-combination direct
7	The element $x$ of Step (5) is in $H$.		$H$ is local powering-invariant in $G$.	Step (6)	Step-given combination direct.