# Normal not implies potentially verbal

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

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## Contents

## Definition

It is possible to have a group and a normal subgroup of that is *not* a potentially verbal subgroup, i.e., there is no group containing for which is a verbal subgroup of .

## Related facts

- NPC theorem: Normal equals
*potentially characteristic*. - Potentially normal-subhomomorph-containing equals normal
- Normal not implies potentially fully invariant

## Facts used

- Verbal implies fully invariant, fully invariant implies normal
- Equivalence of definitions of complete direct factor

## Proof

### Example involving complete group

Let be a nontrivial complete group. Define and . Clearly, is a normal subgroup of .

Suppose is a group containing , such that is verbal in . In particular, is normal in . Since is complete, it is a direct factor, so there exists a group that is a complement to , so as an internal direct product. Further, since is a subgroup of , has a subgroup, say , isomorphic to .

Then, consider the endomorphism of that sends to the trivial subgroup and isomorphically to the subgroup does *not* send to within itself. Thus, is not fully invariant in , and hence, is not verbal in .

### Example involving infinite simple non-abelian group

Suppose is an infinite simple non-abelian group that satisfies no nontrivial identity. An example is the finitary alternating group on an infinite set (see finitary alternating group on infinite set implies no nontrivial identity, alternating groups are simple). Define and is the first direct factor of . Let be the second direct factor.

- is a direct factor of : This is direct from the definition.
- is not a potentially verbal subgroup of :
- For any nontrivial word , the verbal subgroup generated by that with letters from is the whole of : the verbal subgroup corresponding to that word, with letters coming from , is a verbal subgroup of . Hence, it is normal in , and so is either trivial or the whole group. It is not trivial, because satisfies no nontrivial identity, so it is the whole group.
- For any nontrivial word , the verbal subgroup generated by that with letters from is the whole of : This follows for the same reason as the previous step.
- For any nontrivial word , the verbal subgroup generated by that with letters from is the whole of : This follows from the previous step.
- If is a subgroup of a group , any nontrivial verbal subgroup of contains . In particular, it cannot be equal to : This is because any nontrivial word gives rise to a subgroup containing , because even restricting the letters to within , we obtain the whole of .