# Local powering-invariant and normal iff quotient-local powering-invariant in nilpotent group

## Statement

Suppose $G$ is a nilpotent group and $H$ is a subgroup of $G$. The following are equivalent:

1. $H$ is a local powering-invariant normal subgroup of $G$, i.e., $H$ is both a local powering-invariant subgroup of $G$ and a normal subgroup of $G$.
2. $H$ is a quotient-local powering-invariant subgroup of $G$.

## Proof

### (2) implies (1)

This follows from Fact (1) (note that normality is definitional).

### (1) implies (2)

Given: A nilpotent group $G$, a subgroup $H$ of $G$. $H$ is normal and local powering-invariant in $G$. An element $g \in G$ and a prime number $p$ such that the equation $u^p = g$ has a unique solution $u \in G$. $\varphi:G \to G/H$ is the quotient map.

To prove: $\varphi(u)$ is the unique $p^{th}$ root of $\varphi(g)$ in the quotient group $G/H$.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The map $t \mapsto t^p$ is injective from $G$ to itself. Fact (2) (implication (3) implies (1) in the equivalence on the fact page) $G$ is nilpotent and $g \in G$ has a unique $p^{th}$ root. Step-fact combination direct.
2 For any $x \in G$ satisfying $x^p \in H$, we have $x \in H$. $H$ is local powering-invariant in $G$. Step (1) Let $h = x^p$. By Step (1), the $p$-power map is injective in $G$, hence $x$ is the unique $p^{th}$ root of $h$ in $G$. Since $H$ is local powering-invariant, this forces $x \in H$.
3 $G/H$ is $p$-torsion-free. Step (2) If an element of $G/H$ has order $p$, any inverse image of it in $G$ is an element outside $H$ whose $p^{th}$ power is in $H$, contradicting Step (2).
4 $G/H$ is nilpotent. Fact (3) $G$ is nilpotent Given-fact direct
5 The map $t \mapsto t^p$ is injective from $G/H$ to itself. Fact (2) Steps (3), (4) Direct by combining the steps and the fact.
6 $\varphi(u)$ is the unique $p^{th}$ root of $\varphi(g)$ in $G/H$. $u^p = g$ Step (5) Applying $\varphi$ to $u^p = g$ gives that $(\varphi(u))^p = \varphi(g)$. The injectivity of the map by Step (5) now gives uniqueness.