reserve i for Element of NAT;

theorem
  for G being Group for F2 being Subgroup of G
  for F1 being normal Subgroup of F2
  for a,b being Element of F2 holds (a*F1)*(b*F1) = a*b*F1
proof
  let G be Group;
  let F2 be Subgroup of G;
  let F1 be normal Subgroup of F2;
  let a,b be Element of F2;
A1: (nat_hom F1).a=(a*F1) & (nat_hom F1).b=(b*F1) by GROUP_6:def 8;
A2: @((nat_hom F1).a)=(nat_hom F1).a & @((nat_hom F1).b)=(nat_hom F1).b by
GROUP_6:def 5;
  thus a*b*F1 =(nat_hom F1).(a*b) by GROUP_6:def 8
    .=((nat_hom F1).a) *((nat_hom F1).b) by GROUP_6:def 6
    .=(a*F1 )*(b*F1) by A1,A2,GROUP_6:19;
end;
