Suppose to have two groups $C$ and $G$ (not necessarily abelian) such that $C < G$ (subgroup, not necessarily proper). Let's fix an abelian group $A$ such that it is a trivial $G$-module (and consequently by restriction also a trivial $C$-module).
Is it true that $\forall n \in \mathbb{Z}$, $H_n(C,A) < H_n(G,A)$?
While I was solving some basic exercises on group homology, this hypothesis simply came naturally to my mind. I tried to find an answer insisting for enough time, but unfortunately without any progress.
Furthermore, I honesty don't know if the claim is likely to be true or false, and I don't manage to see any help on my book of reference (the Weibel's one). Does anyone have an idea, please? I hope not to ask something extremely trivial. In this case, my apologies, but my experience with group cohomology is of...a week (literally).
Any help/reference/hint if of course fully appreciated. Thank you
Ps: in order to avoid confusion, I underline that (since the context is the group homology, when I say $G$-module I refer to this definition [https://en.wikipedia.org/wiki/G-module]).
It's rarely true.
Take $A=\mathbb{Z}$. Then for any group $G$, $H_1(G,\mathbb{Z})$ is the abelianization of $G$: i.e., $G/[G,G]$.
So if you take any group $G$ whose abelianization is trivial (for example, any non-abelian simple group), and any subgroup $C$ with non-trivial abelianization (for example, any non-trivial cyclic subgroup), then $H_1(C,\mathbb{Z})$ is not a subgroup of $H_1(G,\mathbb{Z})$.