I am working over $\mathbb{Q}$. I have read that if $H$ is a finite index normal subgroup in $G$, and $H_{i}(H;\mathbb{Q})$ has finite $\mathbb{Q}$-dimension, then $H_{i}(G;\mathbb{Q})$ has also finite $\mathbb{Q}$-dimension.
Nevertheless, the opposite does not hold; that is, $H_{i}(G;\mathbb{Q})$ having finite $\mathbb{Q}$-dimension does not imply that $H_{i}(H;\mathbb{Q})$ has finite $\mathbb{Q}$-dimension.
I was wondering if $\dim H_{i}(G;\mathbb{Q})\leq \dim H_{i}(H;\mathbb{Q})$? At least in some cases, where $i=1$, and so we care about the abelianization?
I have tried to use the transfer map, but I do not obtain anything... Thanks in advance!
The Lyndon-Hochschild-Serre spectral sequence has $E^2_{p,q} = H_p(G/H, H_q(H,\mathbb Q)) \implies H_{p+q}(G,\mathbb Q)$.
Since $G/H$ is finite, the $E^2$ term is $0$ for $p>0$, so the sequence degenerates at that page. It follows that $H_{q}(G,\mathbb Q) \cong H_0(G/H, H_q(H,\mathbb Q)) = H_q(H,\mathbb Q)_{G/H}$ (the coinvariants)
It follows that $H_*(G,\mathbb Q)$ is a quotient of $H_*(H,\mathbb Q)$, so it has a smaller dimension.
The same goes for cohomology, using this time invariant instead of co-invariants.
This could also provide a way to construct counter-examples to the converse statement : find an infinite dimensional $G/H$-module whose co-invariants are finite dimensional, and try to realize it as a group homology.
Here's an idea :
have $\mathbb Z$ act on $F\mathbb Z$ (the free group on generators $\mathbb Z$) by translation on generators; and $G$ the associated semi-direct product : $G=F\mathbb Z \rtimes \mathbb Z$, $H= F\mathbb Z \subset G$, and $G/H\cong \mathbb Z$.
Then $H_1(H,\mathbb Q) = \bigoplus_{n\in \mathbb Z}\mathbb Q$ (one for each generator) and $G/H$ acts on that by translating those summands, so that the $1$ in position $n$ becomes equal to the $1$ in position $n+1$ in the co-invariants : $H_1(H,\mathbb Q)_{G/H} \cong \mathbb Q$, which is of course finite-dimensional.