Let $F_n$ be a free group on $n$ letters, and fix some prime $p \geq 2$.  Define

$$K_{n,p}=\text{ker}(F_n \rightarrow H_1(F_n;\mathbb{Z}/p))$$

and

$$V_{n,p} = H_1(K_{n,p};\mathbb{Q}).$$

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.