What does it mean for a group to be free in a variety of groups I'm reading the paper "Finitely generated cyclic extensions
of free groups are residually finite" by Baumslag. One of the hypotheses of the main proposition in the paper is that for a group $G$ and $N < G$ a subgroup, $N$ is free in a nilpotent variety $\mathcal{V}$ of prime exponent $p$.
I understand that a variety of groups, roughly speaking, is a collection $\mathcal{V}$  of groups defined by some equation. It is nilpotent if every group in $\mathcal{V}$ is nilpotent, and has exponent $p$ if every group in $\mathcal{V}$ has exponent $p$.
What I don't understand is what Baumslag means by the fact that $N$ is free in $\mathcal{V}$? The source of my confusion lies in what comes next, mainly that the group $N$ is assumed to be a free group. Surely if $N$ is free, then it can never be contained in a variety of exponent $p$?
 A: A group $N$ is free in the variety $\mathcal{V}$ if and only if there exists a subset $X$ of $N$ such that:

*

*$N\in\mathcal{V}$; and

*$\langle X\rangle = N$; and

*For every $G\in\mathcal{V}$ and every set-theoretic map $f\colon X\to G$, there exists a unique group homomorphism $\phi\colon N\to G$ such that $\phi|_X = f$.

In other words, $N$ is in $\mathcal{V}$ and satisfies the universal property of "free group on $X$", but relative only to groups in $\mathcal{V}$.  Sometimes we say that $N$ is "relatively free" in $\mathcal{V}$ (and the usual free groups are said to be "absolutely free").
It is not hard to show  that  $N$ is free in $\mathcal{V}$ if and only if there exists an absolutely free group $F$ such that $N\cong F/\mathcal{V}(F)$, where $\mathcal{V}(F)$ is the verbal subgroup of $F$ corresponding to $\mathcal{V}$: the smallest normal subgroup of $F$ such that the quotient lies in $\mathcal{V}$.
The canonical place to learn about varieties is Hanna Neumann's Varieties of Groups, Springer-Verlag, 1967, MR 0215899.
