Does the same thing holds for arithmetic groups?

User109249, what you said is true. It is one way of defining group cohomology.

You might find Tom Goodwillie and Dustin Clausen's answer on the following link helpful: mathoverflow.net/questions/121340/…

Hi algori, can you be more clear about what do you mean by free complex of abelian groups? A finite chain complex of free abelian groups is certainly not enough. For example, one need to assume $H_0(C_\ast)$ is zero in you case?