$\newcommand\tf{\text{tf}}\newcommand\tor{\text{tor}}$Let $G$ be a finitely generated group. Let $\gamma\_k(G)$ denote the $k$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 $\gamma\_1^{\tf}(G) = G$. Now assume that $\gamma\_k^{\tf}(G)$ has been defined for some $k \geq 1$. We then have a finitely generated abelian group
$$V\_k = \gamma\_k^{\tf}(G) / [G,\gamma\_k^{\tf}(G)].$$
Let $V\_k^{\tor}$ be the torsion subgroup of $V\_k$, and define $\gamma\_{k+1}^{\tf}(G)$ to be the pullback of $V\_k^{\tor}$ under the projection $\gamma\_k^{\tf}(G) \rightarrow V\_k$.


It follows from the definitions that each $\gamma\_k^{\tf}(G)/\gamma\_{k+1}^{\tf}(G)$ is a finitely generated free abelian group. Moreover, $\gamma\_k(G) < \gamma\_k^{\tf}(G)$ for all $k$.


**Question**: Is it true that $\gamma\_k^{\tf}(G)/\gamma\_{k+1}^{\tf}(G)$ and $\gamma\_k(G)/\gamma\_{k+1}(G)$ have the same rank (i.e. become isomorphic after tensoring with $\mathbb{Q}$) for all $k$?



Lower central series vs torsion-free lower central series
You can prove this by repeatedly applying the following lemma:


**Lemma**:
Let $G$ be a group and let $A,B \lhd G$ be finitely generated normal subgroups. Assume that $A$ and
$B$ are commensurable. Then $[G,A]$ and $[G,B]$ are commensurable and
$$\left(A / [G,A]\right) \otimes \mathbb{Q} \cong \left(B / [G,B]\right) \otimes \mathbb{Q}.$$


At the kth stage of the argument, you've proven that $\gamma\_k(G)$ is a finite-index subgroup of $\gamma^{tf}\_k(G)$. Applying the lemma (with $G$ replaced with $G/\gamma\_{k+1}(G)$ —- this doesn’t change the relevant subquotients, and since it makes the group finitely generated nilpotent it make all its subgroups finitely generated) we deduce that
the ranks of $\gamma\_k(G)/\gamma\_{k+1}(G)$ and $\gamma^{tf}(G)/[G,\gamma^{tf}\_{k}(G)]$ are the same, and also that $\gamma\_{k+1}(G)$ is a finite-index subgroup of $[G,\gamma^{tf}\_{k+1}(G)]$. By construction, the ranks of $\gamma^{tf}(G)/[G,\gamma^{tf}\_{k}(G)]$ and $\gamma^{tf}(G)/\gamma^{tf}\_{k+1}(G)$ are the same (giving one conclusion!), and also that $\gamma^{tf}\_{k+1}(G)$ is commensurable with $[G,\gamma^{tf}\_{k}(G)]$ (letting us continue the induction!).


**Proof of lemma**:
We can assume without loss of generality that $A$ is a finite-index subgroup of $B$. The purported isomorphism can be rewritten as
$$H\_1(A;\mathbb{Q})\_G \cong H\_1(B;\mathbb{Q})\_G,$$
where the subscripts indicate that we are taking the $G$-coinvariants. Since $B/A$ is a finite group, the Hochschild-Serre spectral sequence of the extension
$$1 \rightarrow A \rightarrow B \rightarrow B/A \rightarrow 1$$
degenerates to show that
$$H\_k(B;\mathbb{Q}) \cong H\_0(B/A;H\_k(A;\mathbb{Q})) = H\_k(A;\mathbb{Q})\_B \quad \text{for all $k \geq 0$},$$
and taking the $G$-convariants of this gives that
$$H\_k(A;\mathbb{Q})\_G = \left(H\_k(A;\mathbb{Q})\_B\right)\_G \cong H\_k(B;\mathbb{Q})\_G \quad \text{for all $k \geq 0$}.$$
The special case $k=1$ of this is what we were supposed to prove.


To see that $[G,A]$ and $[G,B]$ are
commensurable, we will need the **Hirsch index** of a general group $\Gamma$, which is the supremum of the $n$ such that there exists a chain
$$\Gamma\_0 < \Gamma\_1 < \cdots < \Gamma\_n$$
of subgroups of $\Gamma$ such that $\Gamma\_i$ is an infinite-index subgroup of $\Gamma\_{i+1}$ for all $0 \leq i < n$ (this is usually only defined for polycyclic groups, but it is not hard to see that this reduces to the usual definition in that case; see [this](https://math.stackexchange.com/a/1000546) answer for more details).


Let $r$ be the common dimension of $\left(A / [G,A]\right) \otimes \mathbb{Q}$ and $\left(B / [G,B]\right) \otimes \mathbb{Q}$.
Since $A$ is a finite-index normal subgroup of $B$, we can apply the additivity result proved [here](https://math.stackexchange.com/a/1000546) to the short exact sequence
$$1 \longrightarrow A/[G,A] \longrightarrow B/[G,A] \longrightarrow B/A \longrightarrow 1$$
to deduce that
$$h(B/[G,A]) = h(A/[G,A]) + h(B/A) = r + 0 = r.$$
Applying this additivity now to the short exact sequence
$$1 \longrightarrow [G,B]/[G,A] \longrightarrow B/[G,A] \longrightarrow B/[G,B] \longrightarrow 1,$$
we see that
$$r = h(B/[G,A]) = h([G,B]/[G,A]) + h(B/[G,B]) = h([G,B]/[G,A]) + r.$$
It follows that $h([G,B]/[G,A])=0$, so $[G,A]$ is a finite-index subgroup of $[G,B]$, as desired.


