theorem
  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;
