1-closed transversal not implies permutably complemented

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 $$G$$ 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.

Example of a non-abelian group of prime exponent
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.