# 1-closed transversal not implies permutably complemented

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., subgroup having a 1-closed transversal) neednotsatisfy the second subgroup property (i.e., permutably complemented subgroup)

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## Contents

## Statement

### Property-theoretic statement

A subgroup having a 1-closed transversal need not be a permutably complemented subgroup.

### Verbal statement

It is possible to have a group and a subgroup of such that has a left transversal that is a 1-closed subset of but does not have *any* permutable complement.

## Proof

### Example of a non-abelian group of prime exponent

`Further information: prime-cube order group:U(3,p)`

Let be a prime number. Let be a non-abelian group of order and exponent (hence is isomorphic to prime-cube order group:U(3,p)). Let be the center of . is a normal subgroup of and it has no permutable complement because nilpotent and non-abelian implies center is not complemented (which in turn follows from the fact that nilpotent implies center is normality-large). On the other hand, does have a 1-closed transversal.

More generally, any subgroup of a group of prime exponent has a 1-closed transversal but is not necessarily permutably complemented.