 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 Th122:
  for n being even non zero Nat
  for k being Nat st n = 2*k & n > 2
  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) *>
  holds (for g being Element of Dihedral_group n holds
  g in center Dihedral_group n iff g = 1_(Dihedral_group n) or g = x |^ k )
proof
  let n be even non zero Nat;
  let k be Nat;
  assume A1: n = 2*k;
  assume A2: n > 2;
  let g1 be Element of INT.Group n;
  assume A3: g1 = 1;
  let x be Element of Dihedral_group n;
  assume A4: x = <* g1, 1_(INT.Group 2) *>;
  1 in INT.Group 2 by EltsOfINTGroup2;
  then reconsider a2=1 as Element of INT.Group 2;
  reconsider y = <* 1_(INT.Group n), a2 *> as Element of Dihedral_group n
  by Th9;

  let g be Element of Dihedral_group n;
  thus g in center Dihedral_group n implies
    g = 1_(Dihedral_group n) or g = x |^ k
  proof
    assume A5: g in center Dihedral_group n;
    assume A6: not g = 1_(Dihedral_group n);
    consider j being Nat such that
    A7: j < n and
    A8: g = (x |^ j) * y or g = x |^ j by A3, A4, Th108;
    per cases by A8;
    suppose A9: g = (x |^ j) * y;
      A10: x |^ 2 = 1_(Dihedral_group n)
      proof
        B1: g * x = ((x |^ j) * y) * x by A9
                 .= ((x |^ j) * y) * (x |^ 1) by GROUP_1:26
                 .= (x |^ (n + j - 1)) * y by A4,Th109
                 .= (x |^ (j + (n - 1))) * y
                 .= ((x |^ j) * (x |^ (n - 1))) * y by GROUP_1:33;
        B2: x * g = x * ((x |^ j) * y) by A9
                 .= (x |^ 1) * ((x |^ j) * y) by GROUP_1:26
                 .= ((x |^ 1) * (x |^ j)) * y by GROUP_1:def 3
                 .= (x |^ (1 + j)) * y by GROUP_1:33
                 .= ((x |^ j) * x) * y by GROUP_1:34;
        x * g = g * x by A5, GROUP_5:77
             .= ((x |^ j) * (x |^ (n - 1))) * y by B1;
        then ((x |^ j) * x) * y = ((x |^ j) * (x |^ (n - 1))) * y by B2;
        then (x |^ j) * x = (x |^ j) * (x |^ (n - 1)) by GROUP_1:6;
        then x = (x |^ (n - 1)) by GROUP_1:6;
        then x * x = (x |^ ((n - 1) + 1)) by GROUP_1:34
                  .= 1_(Dihedral_group n) by A4, Th101;
        hence thesis by GROUP_1:27;
      end;
      x |^ 2 <> 1_(Dihedral_group n) by A2,A3,A4,Th102;
      then contradiction by A10;
      hence g = x |^ k;
    end;
    suppose A12: g = x |^ j;
      A13: g |^ 2 = 1_(Dihedral_group n)
      proof
        B1: y * g = y * (x |^ j) by A12
                 .= (x |^ (n - j)) * y by A4,Th106;
        B2: g * y = (x |^ j) * y by A12;
        y * g = g * y by A5, GROUP_5:77
             .= (x |^ j) * y by B2;
        then (x |^ (n - j)) * y = (x |^ j) * y by B1;
        then x |^ j = (x |^ (n - j)) by GROUP_1:6
                   .= x |^ (n + (- j))
                   .= (x |^ n) * (x |^ (- j)) by GROUP_1:33
                   .= (1_(Dihedral_group n)) * (x |^ (- j)) by A4, Th101
                   .= x |^ (- j) by GROUP_1:def 4;
        then (x |^ j) * (x |^ j) = x |^ ((- j) + j) by GROUP_1:33
                                .= 1_(Dihedral_group n) by GROUP_1:25;
        hence 1_(Dihedral_group n) = (x |^ j) |^ 2 by GROUP_1:27
                                  .= g |^ 2 by A12;
      end;

      g1 |^ (2 * j) = g1 |^ 0
      proof
        g |^ 2 = (x |^ j) |^ 2 by A12
              .= x |^ (j * 2) by GROUP_1:35
              .= <* g1 |^ (2 * j), (1_(INT.Group 2)) *> by A4,Th25;
        then <* g1 |^ (2 * j), (1_(INT.Group 2)) *>
         = 1_(Dihedral_group n) by A13
        .= <* 1_(INT.Group n), 1_(INT.Group 2) *> by Th17;
        hence g1 |^ (2 * j) = 1_(INT.Group n) by FINSEQ_1:77
                           .= g1 |^ 0 by GROUP_1:25;
      end;
      then (2 * j) mod n = 0 mod n by A3,ThINTGroupOrd3
                        .= 0 by NAT_D:26;
      then A14: (2 * j) = (n * ((2 * j) div n)) + 0 by NAT_D:2;
      (2 * j) div n = 0 or (2 * j) div n = 1 by A7,Th121,NAT_1:23;
      then per cases;
      suppose (2 * j) div n = 0;
        then 2 * j = n * 0 + 0 by A14
                  .= 0;
        then A15: j = 0;
        g = x |^ j by A12
         .= x |^ 0 by A15
         .= 1_(Dihedral_group n) by GROUP_1:25;
        then contradiction by A6;
        hence thesis;
      end;
      suppose (2 * j) div n = 1;
        then 2 * j = n * 1 + 0 by A14
                  .= 2 * k by A1;
        hence g = x |^ k by A12;
      end;
    end;
  end;
  assume g = 1_(Dihedral_group n) or g = x |^ k;
  then per cases;
  suppose g = 1_(Dihedral_group n);
    hence thesis by GROUP_2:46;
  end;
  suppose g = x |^ k;
    hence g in center Dihedral_group n by A1,A3,A4,Th120;
  end;
end;
