Local powering-invariant subgroup containing the center is intermediately local powering-invariant in nilpotent group: Difference between revisions

Latest revision as of 01:43, 3 July 2026

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 $K$ .

Related facts

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

Facts used

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

Proof

Given: A nilpotent group $G$ , a subgroup $H$ of $G$ that is local powering-invariant and such that $Z (G) \leq H$ 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	$Z (G) \leq K$ .		$Z (G) \leq H, H \leq K$		given-direct
2	$K$ is nilpotent.	Fact (3)	$G$ is nilpotent, $K \leq G$		Given-fact direct
3	$K$ is $p$ -torsion-free.	Fact (2)	$h \in K$ has a unique $p^{t h}$ 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 $p^{t h}$ 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.

@@ Line 34: / Line 34: @@
 | 5 || The map <math>t \mapsto t^p</math> is injective in <math>G</math>. || Fact (2) ||<math>G</math> 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 <math>x \in K</math>is the unique <math>p^{th}</math> root of <math>h</math> in all of <math>G</math>. || || <math>x \in K</matH> satisfies <math>x^p = h</math>. || Step (5) || given-step direct
+| 6 || The element <math>x \in K</math>is the unique <math>p^{th}</math> root of <math>h</math> in all of <math>G</math>. || || <math>x \in K</matH> satisfies <math>x^p = h</math> || Step (5) || given-step direct
 |-
-| 7 || The element <math>x</math> of Step (6) is in <math>H</math>. || || <math>H</math> is local powering-invariant in <math>G</math>. || Step (6) || Step-given combination direct.
+| 7 || The element <math>x</math> of Step (6) is in <math>H</math>. || || <math>H</math> is local powering-invariant in <math>G</math> || Step (6) || Step-given combination direct.
 |}