Powering-invariant not implies divisibility-closed

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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., divisibility-closed subgroup)
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Statement

It is possible to have a group G and a subgroup H such that:

  1. H is a powering-invariant subgroup of G: If n is a natural number such that every element of G has a unique n^{th} root, then every element of H has a unique n^{th} root within H.
  2. H is not a divisibility-closed subgroup of G: There exists a natural number n such that every element of G has a n^{th} root (not necessarily unique) but not every element of H has a n^{th} root within H.

Related facts

Proof

Proof idea

The key fact is that any finite subgroup of a group must be powering-invariant, but it need not be divisibility-closed. We will construct an example where the subgroup is finite.

Proof details

For any prime number p:

  • Let G be the p-quasicyclic group.
  • Let H be the subgroup comprising the elements of order 1 or p.

Clearly:

  • H, being finite, is powering-invariant (in fact, both G and H are powered over precisely the set of primes other than p).
  • However, H is not divisibility-closed: G is p-divisible, but H is not.