# Fully invariant not implies abelian-potentially verbal in abelian group

From Groupprops

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) neednotsatisfy the second subgroup property (i.e., abelian-potentially verbal subgroup)

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## Statement

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

## Facts used

## Proof

Let be the -quasicyclic group, i.e., the group of all roots of unity for all , under multiplication. Let be the subgroup of comprising the group of roots of unity. Thus, is a fully invariant subgroup of that is neither trivial nor the whole group.

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