# Local powering-invariant subgroup containing the center is intermediately local 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 local powering-invariant subgroup of $G$. Explicitly, suppose $K$ is a subgroup of $G$ containing $H$. Then, $H$ is a local powering-invariant subgroup of $K$.

## 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
3. Nilpotency is subgroup-closed

## Proof

Given: A nilpotent group $G$, a subgroup $H$ of $G$ that is local powering-invariant and such that $Z(G) \le H$ where $Z(G)$ is the center of $G$. A subgroup $K$ of $G$ containing $H$. A prime number $p$ and an element $h \in H$ such that there is a unique element $x \in K$ satisfying $x^p = h$.

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 $Z(G) \le K$. $Z(G) \le H, H \le K$. given-direct
2 $K$ is nilpotent. Fact (3) $G$ is nilpotent, $K \le G$ Given-fact direct
3 $K$ is $p$-torsion-free. Fact (2) $h \in K$ has a unique $p^{th}$ root in $K$. We use the equivalence (3) implies (1) within the multi-part equivalence of Fact (2).
4 $Z(G)$ is $p$-torsion-free. Fact (1) Steps (1), (3) Step-fact combination direct
5 The map $t \mapsto t^p$ is injective in $G$. Fact (2) $G$ is nilpotent Step (4) Step-fact combination direct (specifically, we want to use the implication from (4) to (1) in the multi-part equivalence of Fact (2))
6 The element $x \in K$is the unique $p^{th}$ root of $h$ in all of $G$. $x \in K$ satisfies $x^p = h$. Step (5) given-step direct
7 The element $x$ of Step (6) is in $H$. $H$ is local powering-invariant in $G$. Step (6) Step-given combination direct.