 reserve G, A for Group;
 reserve phi for Homomorphism of A,AutGroup(G);
 reserve G, A for Group;
 reserve phi for Homomorphism of A,AutGroup(G);
reserve G1,G2 for Group;

theorem Th102:
  for n being non zero Nat
  for g1 being Element of INT.Group n st g1 = 1
  for x being Element of Dihedral_group n
  st x = <*g1,1_(INT.Group 2)*>
  for k being Nat st k <> 0 & k < n
  holds x |^ k <> 1_(Dihedral_group n)
proof
  let n be non zero Nat;
  let g1 be Element of INT.Group n;
  assume A0: g1 = 1;
  let x be Element of Dihedral_group n;
  assume A1: x = <*g1,1_(INT.Group 2)*>;
  let k be Nat;
  assume A2: k <> 0;
  assume A3: k < n;
  defpred P[Nat] means
  ex g being Element of INT.Group n st g = $1 mod n & g = g1 |^ $1;
  A4: P[0]
  proof
    take 1_(INT.Group n);
    thus 1_(INT.Group n) = 0 by GR_CY_1:14
                        .= 0 mod n by NAT_D:26;
    thus 1_(INT.Group n) = g1 |^ 0 by GROUP_1:25;
  end;
  A5: for j being Nat st P[j] holds P[j + 1]
  proof
    let j be Nat;
    assume B1: P[j];
    per cases;
    suppose B2: (j + 1) mod n = 0;
      ex g being Element of INT.Group n st
        g = (j + 1) mod n & g = g1 |^ (j + 1)
      proof
        take g = 1_(INT.Group n);
        thus g = 0 by GR_CY_1:14
              .= (j + 1) mod n by B2;
        ex t being Nat st (j + 1 = n * t + 0 & 0 < n) by B2,NAT_D:def 2;

        then consider t being Nat such that
        B2a: j + 1 = n * t;
        thus g1 |^ (j + 1) = g1 |^ (n * t) by B2a
                          .= (g1 |^ t) |^ n by GROUP_1:35
                          .= (g1 |^ t) |^ (card (INT.Group n))
                          .= g by GR_CY_1:9;
      end;
      hence P[j + 1];
    end;
    suppose B3: (j + 1) mod n <> 0;
      ex g being Element of INT.Group n st
        g = (j + 1) mod n & g = g1 |^ (j + 1)
      proof
        consider g0 being Element of INT.Group n such that
        B4: g0 = j mod n & g0 = g1 |^ j by B1;
        take g = g0 * g1;

        n <> 1
        proof
          assume B5: n = 1;
          (j + 1) = (1 * (j + 1)) + 0;
          then (j + 1) mod 1 = 0 by NAT_D:def 2;
          hence contradiction by B3,B5;
        end;
        then n > 1 by NAT_1:53;
        then B6: 1 in Segm n & (j mod n) in Segm n by NAT_1:44, NAT_D:1;

        thus g = (addint n).(g0, g1) by Th75
              .= (addint n).(j mod n, 1) by A0, B4
              .= ((j mod n) + 1) mod n by B6, GR_CY_1:def 4
              .= (j + 1) mod n by NAT_D:22;
        thus g = (g1 |^ j) * g1 by B4
              .= g1 |^ (j + 1) by GROUP_1:34;
      end;

      hence P[j + 1];
    end;
  end;
  A6: for j being Nat holds P[j] from NAT_1:sch 2(A4, A5);
  consider gk being Element of INT.Group n such that
  A7: gk = k mod n & gk = g1 |^ k by A6;
  k mod n = k by A3, NAT_D:24;
  then gk <> 0 by A2, A7;
  then gk <> 1_(INT.Group n) by GR_CY_1:14;
  then A8: g1 |^ k <> 1_(INT.Group n) by A7;
  x |^ k = <* (g1 |^ k), 1_(INT.Group 2) *> by A1, Th25;
  then x |^ k <> <* 1_(INT.Group n), 1_(INT.Group 2) *>
    by A8, FINSEQ_1:77;
  hence x |^ k <> 1_(Dihedral_group n) by Th17;
end;
