Nilpotent-quotient implies subgroup-to-quotient powering-invariance implication

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This article gives the statement and possibly, proof, of an implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., nilpotent-quotient subgroup) must also satisfy the second subgroup property (i.e., normal subgroup satisfying the subgroup-to-quotient powering-invariance implication)
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Statement

Suppose G is a group and H is a normal subgroup of G that is a nilpotent-quotient subgroup, i.e., the quotient group G/H is a nilpotent group (note that G and H themselves may or may not be nilpotent). Then, H is a normal subgroup satisfying the subgroup-to-quotient powering-invariance implication in G. Explicitly, this means that if p is a prime number such that both G and H are p-powered, then so is G/H.

Facts used

  1. Divisibility is inherited by quotient groups
  2. Equivalence of definitions of nilpotent group that is torsion-free for a set of primes: We use the equivalence of (1) and (2) within the multi-part equivalence. This says that a nilpotent group has no non-identity elements of order p if and only if its p^{th} power map is injective.

Proof

Given: Group G, normal subgroup H such that both G and H are p-powered for some prime number p.

To prove: G/H is p-powered.

Proof:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 G/H is p-divisible, i.e., any element in it has a p^{th} root. In other words, the p^{th} power map in it is surjective. Fact (1) G is p-powered, hence p-divisible, H is normal in G.
2 G/H is p-torsion-free. G and H are both p-powered. Any element of H has a unique p^{th} root in H, and also a unique p^{th} root in G, so those unique roots coincide, which means it has no p^{th} root outside H. Thus, G/H has no non-identity element of order p.
3 The p^{th} power map in G/H is injective. Fact (2) G/H is nilpotent. Step (2) Fact-given-step combination direct.
4 G/H is p-powered, i.e., its p^{th} power map is bijective. Steps (1), (3) Step-combination direct.