theorem Th25:
  for a holds (Conjugate a") * (Conjugate a) = Conjugate 1_G
proof
  let a;
  set f1 = (Conjugate a") * (Conjugate a);
  set f2 = Conjugate 1_G;
A1: for b holds f1.b = b
  proof
    let b;
    (Conjugate a").((Conjugate a).b) = (Conjugate a").(b |^ a) by Def6
      .= b |^ a |^ a" by Def6
      .= b by GROUP_3:25;
    hence thesis by FUNCT_2:15;
  end;
  for b holds f1.b = f2.b
  proof
    let b;
    thus f1.b = b by A1
      .= f2.b by Th23;
  end;
  hence thesis;
end;
