Take the tour ×  Let $F_n$ be a free group on $n$ letters, and
fix some prime $p \geq 2$. Define  $$K_{n,p}=\text{ker}(F_n
\rightarrow H_1(F_n;\mathbb{Z}/p))$$   $$V_{n,p} =
H_1(K_{n,p};\mathbb{Q}).$$  For $x \in K_{n,p}$, let $[x]_{n,p}
\in V_{n,p}$ be the associated element. An element $x \in F_n$
is primitive if there is a free basis for $F_n$ containing $x$.
Observe that for $x \in F_n$, we have $x^p \in K_{n,p}$. Define
$$S_{n,p} = \{\text{$[x^p]_{n,p}$ $|$ $x \in F_n$ is
primitive}\}.$$  Question : Does $S_{n,p}$ span $V_{n,p}$? My
guess is that the answer is no, but I cannot figure out how to
prove it.  It is clear that $S_{n,p}$ would span if the
primitivity condition were dropped. Of course, every element of
$F_n$ is a product of primitive elements; however, these
observations are not as helpful as they might appear since
$$[(xy)^p]_{n,p} \neq [x^p]_{n,p} + [y^p]_{n,p}.$$  The correct
formula is actually  $$[(xy)^p]_{n,p} = [x^p]_{n,p} +
[y^p]_{n,p} + [y^{-p} x^{-p} (xy)^p]_{n,p},$$  where of cours