 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 Th107:
  for n being non zero Nat
  for g1 being Element of INT.Group n st g1 = 1
  for a2 being Element of INT.Group 2 st a2 = 1
  for x,y being Element of Dihedral_group n
  st x = <*g1,1_(INT.Group 2)*> & y = <*(1_(INT.Group n)),a2*>
  for z being Element of Dihedral_group n
  ex k being Nat st z = (x |^ k)*y or z = x |^ k
proof
  let n be non zero Nat;
  let g1 be Element of INT.Group n;
  assume A1: g1 = 1;
  let a2 be Element of INT.Group 2;
  assume A2: a2 = 1;
  let x,y be Element of Dihedral_group n;
  assume A3: x = <*g1,1_(INT.Group 2)*>;
  assume A4: y = <*(1_(INT.Group n)),a2*>;
  let z be Element of Dihedral_group n;
  consider g being Element of INT.Group n, a being Element of INT.Group 2
    such that
  A6: z = <* g, a *> by Th12;
  consider k being Nat such that
  A7: g = g1 |^ k & g = k mod n by A1, ThINTGroupOrd;
  take k;
  assume A8: z <> (x |^ k)*y;
  A9: x |^ k = <* g1 |^ k, 1_(INT.Group 2) *> by A3,Th25
             .= <* g, 1_(INT.Group 2) *> by A7;

  a in INT.Group 2;
  then a = 0 or a = 1 by EltsOfINTGroup2;
  then per cases;
  suppose A10: a = 0;
    x |^ k = <* g, 1_(INT.Group 2) *> by A9
          .= z by A6, A10, GR_CY_1:14;
    hence thesis;
  end;
  suppose a = 1;
    then a = a2 by A2;
    then (x |^ k) * y = z by A4,A6,A9,Th16;
    then contradiction by A8;
    hence thesis;
  end;
end;
