Local powering-invariant subgroup containing the center is intermediately 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 powering-invariant subgroup of G. Explicitly, suppose K is a subgroup of G containing H. Then, H is a 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

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 such that K is p-powered. An element h \in 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 p-torsion-free. K is p-powered (we are using the powering-injectivity here) given-direct
3 Z(G) is p-torsion-free. Fact (1) Steps (1), (2) Step-fact combination direct
4 The map t \mapsto t^p is injective in G. Fact (2) G is nilpotent Step (3) Step-fact combination direct (specifically, we want to use the implication from (4) to (1) in the multi-part equivalence of Fact (2))
5 There exists an element x \in K satisfying x^p = h. K is p-powered (we are using the p-divisibility here), h \in H \le K given-direct
6 The element x \in K of the preceding step is the unique p^{th} root of h in all of G. Steps (4), (5) Step-combination direct
7 The element x of Step (5) is in H. H is local powering-invariant in G. Step (6) Step-given combination direct.