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 $G$.

Related facts

Opposite facts

• Local powering-invariant and normal not implies quotient-local powering-invariant
• Center not is quotient-local powering-invariant

Applications

• Local powering-invariance is quotient-transitive in nilpotent group
• Upper central series members are local powering-invariant in nilpotent group

Facts used

1. Quotient-local powering-invariant implies local powering-invariant
2. Equivalence of definitions of nilpotent group that is torsion-free for a set of primes
3. Nilpotency is quotient-closed

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.