# 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 $K$.

## 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 $Z(G) \le H$ 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 $Z(G) \le K$. $Z(G) \le H, H \le K$. 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), $h \in H \le K$ 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.