reserve G for Group,
  a,b for Element of G,
  m, n for Nat,
  p for Prime;

theorem Th7:
  for G being Group, N being Subgroup of G
  for a,b being Element of G holds
  N is normal & b in N implies
  for n ex g being Element of G st g in N & (a * b)|^ n = a |^ n * g
proof
  let G be Group;
  let N be Subgroup of G;
  let a,b be Element of G;
  assume
A1: N is normal & b in N;
   defpred P[Nat] means for n ex g being Element of G
   st g in N & (a * b)|^ $1 = a |^ $1 * g;
A2: (a * b)|^ 0 = 1_G by GROUP_1:25;
    a |^ 0 * 1_G = 1_G * 1_G by GROUP_1:25
                .= 1_G by GROUP_1:def 4;
    then
A3: P[0] by A2,GROUP_2:46;
A4: now let n;
    assume P[n];
    then consider g1 be Element of G such that
A5: g1 in N & (a * b)|^ n = a |^ n * g1;
A6: (a * b)|^ (n + 1) = a |^ n * g1 * (a * b) by A5,GROUP_1:34
                     .= a |^ n * (g1 * (a * b)) by GROUP_1:def 3
                     .= a |^ n * ((g1 * a) * b) by GROUP_1:def 3;
    a * N = N * a by A1,GROUP_3:117;
    then g1 * a in a * N by A5,GROUP_2:104;
    then consider g2 be Element of G such that
A7: g1 * a = a * g2 & g2 in N by GROUP_2:103;
 (g1 * a) * b = a * (g2 * b) by A7,GROUP_1:def 3;
then  a |^ n * ((g1 * a) * b) = a |^ n * a * (g2 * b) by GROUP_1:def 3
                            .= a |^ (n + 1) * (g2 * b) by GROUP_1:34;
    hence P[n+1] by A1,A6,A7,GROUP_2:50;
  end;
  for n holds P[n] from NAT_1:sch 2(A3,A4);
  hence thesis;
end;
