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) \le 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) \le K. Z(G) \le H, H \le K. given-direct
2 K is nilpotent. Fact (3) G is nilpotent, K \le G Given-fact direct
3 K is p-torsion-free. Fact (2) h \in K has a unique p^{th} 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 Kis the unique p^{th} 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.