Powering-invariant not implies local powering-invariant

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties. That is, it states that every subgroup satisfying the first subgroup property (i.e., powering-invariant subgroup) need not satisfy the second subgroup property (i.e., local powering-invariant)
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Statement

It is possible to have a group G and a powering-invariant subgroup H of G that is not local powering-invariant. In other words, the following are true:

  1. H is powering-invariant in G: For any prime number p, if G is powered over p, H is also powered over p.
  2. H is not local powering-invariant in G: There exists a natural number n and an element g \in H such that there is a unique x \in G satisfying x^n = g, but x \notin H.

Proof

Set G = \mathbb{Z} and H as the subgroup 2\mathbb{Z}.

  • H is powering-invariant in G: G is not powered over any prime, so H is is powering-invariant in G for vacuous reasons.
  • H is not local powering-invariant in G: The element 2 \in H has a unique square root (corresponding to n = 2) in G, but this square root is not in H.