Fully invariant not implies abelian-potentially verbal in abelian group

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This article gives the statement and possibly, proof, of a non-implication relation between two subgroup properties, when the big group is a abelian group. That is, it states that in a abelian group, every subgroup satisfying the first subgroup property (i.e., fully invariant subgroup) need not satisfy the second subgroup property (i.e., abelian-potentially verbal subgroup)
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Statement

It is possible to have an abelian group A and a fully invariant subgroup B of A such that A is not an abelian-potentially verbal subgroup of B, i.e., there is no abelian group C containing B such that A is a verbal subgroup of C.

Facts used

  1. Divisible abelian subgroup of abelian group contains no proper nontrivial verbal subgroup

Proof

Let B be the p-quasicyclic group, i.e., the group of all (p^k)^{th} roots of unity for all k, under multiplication. Let A be the subgroup of B comprising the group of p^{th} roots of unity. Thus, A is a fully invariant subgroup of B that is neither trivial nor the whole group.

Since B is a divisible abelian group, fact (1) tells us that for any abelian group C containing B, any verbal subgroup of C contained in B is equal to either the trivial subgroup or B. In particular, A is not a verbal subgroup of C.