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

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! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
! Step no. !! Assertion/construction !! Facts used !! Given data used !! Previous steps used !! Explanation
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| 1 || <math>Z(G) \le K</math>. || || <math>Z(G) \le H, H \le K</math>. || || given-direct
| 1 || <math>Z(G) \le K</math>. || || <math>Z(G) \le H, H \le K</math> || || given-direct
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| 2 || <math>K</math> is nilpotent. || Fact (3) || <math>G</math> is nilpotent, <math>K \le G</math> || || Given-fact direct
| 2 || <math>K</math> is nilpotent. || Fact (3) || <math>G</math> is nilpotent, <math>K \le G</math> || || Given-fact direct
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| 3 || <math>K</math> is <math>p</math>-torsion-free. || Fact (2) || <math>h \in K</math> has a unique <math>p^{th}</math> root in <math>K</math>. || || We use the equivalence (3) implies (1) within the multi-part equivalence of Fact (2).
| 3 || <math>K</math> is <math>p</math>-torsion-free. || Fact (2) || <math>h \in K</math> has a unique <math>p^{th}</math> root in <math>K</math> || || We use the equivalence (3) implies (1) within the multi-part equivalence of Fact (2).
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| 4 || <math>Z(G)</math> is <math>p</math>-torsion-free. || Fact (1) || || Steps (1), (3)  || Step-fact combination direct
| 4 || <math>Z(G)</math> is <math>p</math>-torsion-free. || Fact (1) || || Steps (1), (3)  || Step-fact combination direct

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

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 h∈H such that there is a unique element x∈K satisfying xp=h.

To prove: There exists a unique element x∈H 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,H≤K given-direct
2 K is nilpotent. Fact (3) G is nilpotent, K≤G Given-fact direct
3 K is p-torsion-free. Fact (2) h∈K 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 t↦tp 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∈Kis the unique pth root of h in all of G. x∈K 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.