Suppose $\mathfrak{U}$ is a group variety. Now let’s define $L(\mathfrak{U})$ as the class of all such groups $G$, such that $\forall g \in G$ $\langle \langle g \rangle \rangle \in \mathfrak{U}$ (here $\langle \langle ... \rangle \rangle$ stands for normal closure). It is not hard to see, that $L(\mathfrak{U})$ is closed under subgroups, quotients and direct products, and thus is a variety too. So $L$ is an operator on group varieties. If I am not mistaken, it is usually called Levi operator.
It is always true, that $\mathfrak{U} \subset L(\mathfrak{U})$, however the converse is not true in general: for example, Levi operator maps the variety of abelian groups to the variety of $2$-Engel groups. The variety of all groups and the Burnside varieties are nevertheless preserved by it.
Now, let’s call a variety of groups a one-word variety if it can be defined by one word (for example the varieties of groups of exponent dividing $m$ are defined by $x^m$, the variety of abelian groups is defined by $[x, y]$, etc.).
My question is:
Does $\mathfrak{U}$ being a one word variety imply that $L(\mathfrak{U})$ is one?
For Burnside varieties and the variety of abelian groups this is true. However, I do not know how to prove this statement in general. Or is there a counterexample...