2 added 565 characters in body

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.

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 course we have $x^{-p} x^{-p} (xy)^p \in [F_n,F_n] \subset K_{n,p}$. Alas, however, in general we will not have $[x^{-p} x^{-p} (xy)^p]_{n,p} = 0$.

1

# Generating set for abelianization of "mod $p$ commutator subgroup" of a free group

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.