Suppose that $\mathbb{Z}$ is a flat $\mathbb{Z}G$-module for a group $G$.
Question: Is $G$ the trivial group ?
Nb. I know that the question can be answered affirmatively if $G$ is finitely generated.
Edit: I think the following lemma solves the problem. I would be grateful if someone could have a look on its proof and give some feedback whether it looks ok. Thanks.
Lemma: Let $R \le S$ be rings with unit such that $S$ is flat as left $R$-module. Then every flat left $S$-module is also flat as left $R$-module.
Now let $G\neq 1$ be any group. If $G$ is abelian then $\mathbb{Z}$ isn't flat by Georges' argument. If $G$ is not abelian, we can find an abelian subgroup $1 \neq H \le G$. Now, if $\mathbb{Z}$ were flat as $\mathbb{Z}G$-module, it would also be flat as $\mathbb{Z}H$-module by the lemma. But we just saw that this isn' true. Hence $\mathbb{Z}$ isn't flat as $\mathbb{Z}G$-module. We have therefore shown:
For a group $G$ the following is equivalent:
- $G=1$
- $H_i(G,-)=0\,$ for all $i > 0$
- $H^i(G,-)=0\,$ for all $i > 0$
Proof of the Lemma: Let $E$ be a flat left $S$-module and let $i: M \to N$ be an embedding of right $R$-modules. We have to show that $i \otimes id_E: M \otimes_R E \to N \otimes_R E$ is also an embedding.
Since $S$ is a flat left $R$-module, tensoring with $S$ from the right yields an embedding $i \otimes id_S: M \otimes_R S\to N \otimes_R S$ of right $S$-modules. Similarly, as $E$ is a flat left $S$-module, we obtain the embedding $$(i \otimes id_S)\otimes id_E: (M \otimes_R S)\otimes_S E \to (N \otimes_R S) \otimes_S E$$ which, by associativity of the tensor product, is equivalent to $$i \otimes (id_S\otimes id_E): M \otimes_R (S\otimes_S E) \to N \otimes_R (S \otimes_S E)$$ which, by the natural isomorphism $S \otimes_S E \cong E$ is equivalent to $$i \otimes id_E: M \otimes_R E \to N \otimes_R E.$$ Hence $i \otimes id_E$ is an embedding and thus $E$ is flat as left $R$-module. QED