Exercises from Cohomology of Groups by Kenneth S. Brown

Note that if ∑ agg ∈ I, then ∑ ag = 0 so that ∑ agg = ∑ agg− ∑ ag1 = ∑ ag(g−1), so that every element in I is a Z-linear combination of the elements g− 1 (g ∈ G, g 6= 1). Also, if ∑ ag(g− 1) = 0, it follows that ag = 0 for all g so that the elements g − 1 (g ∈ G, g 6= 1) are linearly independent. (b) Notice that the ideal generated by the elements s− 1 (s ∈ S) is the same as the ideal generated by the elements s − 1 (s ∈ S, e = ±1) because s−1 − 1 = −s−1(s − 1). We may thus assume that S is closed under inverses, so that any element of G may be written as g = s1s2 . . . sn. Then we have g − 1 = s1s2 . . . sn−1(sn − 1) + s1s2 . . . sn−2(sn−1 − 1) + . . .+ s1(s2 − 1) + (s1 − 1). By part (a), it follows that the elements s− 1 (s ∈ S) generate I as a left ideal. (c) For any g ∈ G, by hypothesis we may write g − 1 = ∑ ri(si − 1) for ri ∈ ZG. In other words, we have g − 1 = ∑ ai,hh(si − 1) = ∑ ai,h(hsi − h). Since the sums are finite, and the ai,h are integers, we can split up this sum and write