Let$\newcommand\tf{\text{tf}}\newcommand\tor{\text{tor}}$Let $G$ be a finitely generated group. Let $\gamma_k(G)$ denote the kth$k$th term in the lower central series for $G$, so $\gamma_1(G) = G$ and $\gamma_{k+1}(G) = [\gamma_k(G),G]$ for all $k \geq 1$.
There is also a natural central series that it makes sense to call the "torsion-free lower central series", and is defined inductively as follows. First, define $\gamma_1^{tf}(G) = G$$\gamma_1^{\tf}(G) = G$. Now assume that $\gamma_k^{tf}(G)$$\gamma_k^{\tf}(G)$ has been defined for some $k \geq 1$. We then have a finitely generated abelian group $$V_k = \gamma_k^{tf}(G) / [G,\gamma_k^{tf}(G)].$$$$V_k = \gamma_k^{\tf}(G) / [G,\gamma_k^{\tf}(G)].$$ Let $V_k^{tor}$$V_k^{\tor}$ be the torsion subgroup of $V_k$, and define $\gamma_{k+1}^{tf}(G)$$\gamma_{k+1}^{\tf}(G)$ to be the pullback of $V_k^{tor}$$V_k^{\tor}$ under the projection $\gamma_k^{tf}(G) \rightarrow V_k$$\gamma_k^{\tf}(G) \rightarrow V_k$.
It follows from the definitions that each $\gamma_k^{tf}(G)/\gamma_{k+1}^{tf}(G)$$\gamma_k^{\tf}(G)/\gamma_{k+1}^{\tf}(G)$ is a finitely generated free abelian group. Moreover, $\gamma_k(G) < \gamma_k^{tf}(G)$$\gamma_k(G) < \gamma_k^{\tf}(G)$ for all $k$.
Question: Is it true that $\gamma_k^{tf}(G)/\gamma_{k+1}^{tf}(G)$$\gamma_k^{\tf}(G)/\gamma_{k+1}^{\tf}(G)$ and $\gamma_k(G)/\gamma_{k+1}(G)$ have the same rank (i.e. become isomorphic after tensoring with $\mathbb{Q}$) for all $k$?
