2.50:

Suppose H is normal in G. For x, y in G consider [x,y]=xyx^-1y^-1. G/H is abelian iff (xN)(yN)=(yN)(xN) for all x,y, iff xyN = yxN, iff Nxy=Nyx, iff (xy)(yx)^-1 in N, for all x, y in G. Since (xy)(yx)^-1= xyx^-1y^-1= [x,y], this shows that G/H is abelian if and only if [x,y] is in H for all x, y in G. Since H is a subgroup and [G,G] is the subgroup generated by all commutators [x,y], the last statement is equivalent to the statement that [G,G] is a subgroup of H.