Let $G$ be a finitely generated nilpotent group and let $A\le G$ be a finite-index subgroup. I have two questions about $A$:


1. Is it true that the inclusion map $A \rightarrow G$ induces isomorphisms $H\_k(A;\mathbb{Q}) \rightarrow H\_k(G;\mathbb{Q})$ for all $k$, and thus that the Betti numbers of $A$ and $G$ are the same? This is true if $G$ is abelian and I suspect that it is true in general, but I can't seem to prove it or find an appropriate reference.
2. Letting $\gamma\_k$ denote the kth term of the lower central series, is it true that the abelian groups $\gamma\_k(A)/\gamma\_{k+1}(A)$ and $\gamma\_k(G)/\gamma\_{k+1}(G)$ have the same rank (i.e. become the same after tensoring with $\mathbb{Q}$) for all $k$?



Betti numbers and lower central series quotients of finite-index subgroups of nilpotent groups
Both statements are true. To prove them, we will need the following lemma:


**Lemma**: Let $G$ be a finitely generated nilpotent group of class $k$ and let $H$ be a subgroup of $G$.
Then $\gamma\_k(H)$ is a finite-index subgroup of $\gamma\_k(G)$.


**Proof**: Iterated $k$-fold commutators induce a surjective homomorphism
$\phi\colon \wedge^k G^{\text{ab}} \rightarrow \gamma\_k(G)$ of finitely generated abelian groups. Our hypotheses
imply that the image of $\wedge^k H^{\text{ab}}$ in $\wedge^k G^{\text{ab}}$ is finite-index, so the image of
the composition
$$\wedge^k H^{\text{ab}} \rightarrow \wedge^k G^{\text{ab}} \stackrel{\phi}{\rightarrow} \gamma\_k(G)$$
has finite-index. But this implies that $\gamma\_k(H)$ is finite-index in $\gamma\_k(G)$, as desired. $\square$


Here's the proof of 1:


**Theorem**: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then
$H\_n(H;\mathbb{Q}) \cong H\_n(G;\mathbb{Q})$ for all $n$.


**Proof**: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian,
and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for
$k$. Observe that $\gamma\_k(H) < \gamma\_k(G)$, and the lemma above says that
the abelian group $\gamma\_k(H)$ is a finite-index subgroup of the finitely generated abelian group $\gamma\_k(G)$. Setting
$\overline{H} = H / \gamma\_k(H)$ and $\overline{G} = G/\gamma\_k(G)$, we have a morphism between the central extensions
$$1 \rightarrow \gamma\_k(H) \rightarrow H \rightarrow \overline{H} \rightarrow 1$$
and
$$1 \rightarrow \gamma\_k(G) \rightarrow G \rightarrow \overline{G} \rightarrow 1.$$
We therefore get a morphism between the associated Hochschild-Serre spectral sequences. The map on
the $E^2$ page is of the form
$$H\_p(\overline{H};H\_q(\gamma\_k(H);\mathbb{Q})) \rightarrow H\_p(\overline{G};H\_q(\gamma\_k(G);\mathbb{Q})).$$
Since these are central extensions, this simplifies to a map
$$H\_p(\overline{H};\mathbb{Q}) \otimes H\_q(\gamma\_k(H);\mathbb{Q}) \rightarrow H\_p(\overline{G};\mathbb{Q}) \otimes H\_q(\gamma\_k(G);\mathbb{Q}).$$
By our inductive hypothesis, this is a tensor product of isomorphisms, so we conclude that
our two spectral sequences are the same and thus $H\_n(H;\mathbb{Q}) \cong H\_n(G;\mathbb{Q})$ for all $n$. $\square$


Here's the proof of 2:


**Theorem**: Let $G$ be a finitely generated nilpotent group and let $H < G$ be a finite-index subgroup. Then
the abelian groups $\gamma\_n(H) / \gamma\_{n+1}(H)$ and $\gamma\_n(G) / \gamma\_{n+1}(G)$ have the
same ranks for all $n$.


**Proof**: The proof will be by the degree $k$ of nilpotency of $G$. The base case $k=1$ means that $G$ is abelian,
and the theorem is easy. Assume now that the theorem is true for some $(k-1)$, and we will prove it for
$k$. Using our inductive hypothesis, it is enough to just deal with the bottom of the lower central
series and prove that the ranks of $\gamma\_k(H)$ and $\gamma\_k(G)$ are the same. But the lemma above
says that the abelian group $\gamma\_k(H)$ is a finite-index
subgroup of $\gamma\_k(G)$, so their ranks are definitely the same. $\square$


By the way, I can't help but point out a connection between your two questions. In his beautiful paper


Stallings, John,
Homology and central series of groups.
J. Algebra 2 (1965), 170-181.


Stallings proves the following theorem. Let $\gamma\_k^{tf}$ be the torsion-free lower central series you define in your [other](https://mathoverflow.net/q/385052/317) question.


**Theorem** (Stallings): Let $f\colon H \rightarrow G$ be a group homomorphism inducing an isomorphism on $H\_1(-;\mathbb{Q})$ and a surjection on $H\_2(-;\mathbb{Q})$. Then $f$ induces isomorphisms between $(\gamma^{tf}\_k(H)/\gamma^{tf}\_{k+1}(H)) \otimes \mathbb{Q}$ and
$(\gamma^{tf}\_k(G)/\gamma^{tf}\_{k+1}(G)) \otimes \mathbb{Q}$ for all $k$. $\square$


This can be applied to the inclusion of a finite-index subgroup by the first theorem above, and together with the positive answer to your other question you can deduce a positive answer to your second one (but one that is obviously way over-powered!).


