Local powering-invariant and normal iff quotient-local powering-invariant in nilpotent group

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Suppose G is a nilpotent group and H is a subgroup of G. The following are equivalent:

  1. H is a local powering-invariant normal subgroup of G, i.e., H is both a local powering-invariant subgroup of G and a normal subgroup of G.
  2. H is a quotient-local powering-invariant subgroup of G.

Related facts

Opposite facts


Facts used

  1. Quotient-local powering-invariant implies local powering-invariant
  2. Equivalence of definitions of nilpotent group that is torsion-free for a set of primes
  3. Nilpotency is quotient-closed


(2) implies (1)

This follows from Fact (1) (note that normality is definitional).

(1) implies (2)

Given: A nilpotent group G, a subgroup H of G. H is normal and local powering-invariant in G. An element g \in G and a prime number p such that the equation u^p = g has a unique solution u \in G. \varphi:G \to G/H is the quotient map.

To prove: \varphi(u) is the unique p^{th} root of \varphi(g) in the quotient group G/H.


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The map t \mapsto t^p is injective from G to itself. Fact (2) (implication (3) implies (1) in the equivalence on the fact page) G is nilpotent and g \in G has a unique p^{th} root. Step-fact combination direct.
2 For any x \in G satisfying x^p \in H, we have x \in H. H is local powering-invariant in G. Step (1) Let h = x^p. By Step (1), the p-power map is injective in G, hence x is the unique p^{th} root of h in G. Since H is local powering-invariant, this forces x \in H.
3 G/H is p-torsion-free. Step (2) If an element of G/H has order p, any inverse image of it in G is an element outside H whose p^{th} power is in H, contradicting Step (2).
4 G/H is nilpotent. Fact (3) G is nilpotent Given-fact direct
5 The map t \mapsto t^p is injective from G/H to itself. Fact (2) Steps (3), (4) Direct by combining the steps and the fact.
6 \varphi(u) is the unique p^{th} root of \varphi(g) in G/H. u^p = g Step (5) Applying \varphi to u^p = g gives that (\varphi(u))^p  = \varphi(g). The injectivity of the map by Step (5) now gives uniqueness.