# 1-closed transversal not implies permutably complemented

## Statement

### Verbal statement

It is possible to have a group $G$ and a subgroup $H$ of $G$ such that $H$ has a left transversal that is a 1-closed subset of $G$ but $H$ 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 $p \ne 2$ be a prime number. Let $G$ be a non-abelian group of order $p^3$ and exponent $p$ (hence $G$ is isomorphic to prime-cube order group:U(3,p)). Let $H$ be the center of $G$. $H$ is a normal subgroup of $G$ 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, $H$ 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.