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

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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

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)H where Z(G) is the center of G. A subgroup K of G containing H. A prime number p and an element hH such that there is a unique element xK satisfying xp=h.

To prove: There exists a unique element xH such that xp=h.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 Z(G)K. Z(G)H,HK given-direct
2 K is nilpotent. Fact (3) G is nilpotent, KG Given-fact direct
3 K is p-torsion-free. Fact (2) hK has a unique pth 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 ttp 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 xKis the unique pth root of h in all of G. xK satisfies xp=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.