
theorem
  for p be Safe Prime ex q be Sophie_Germain Prime,
      H1,H2,Hq,H2q be strict Subgroup of Z/Z*(p) st
   card H1 = 1 & card H2 = 2 & card Hq = q & card H2q = 2*q &
   for H be strict Subgroup of Z/Z*(p) holds
   H = H1 or H = H2 or H = Hq or H = H2q
proof
  let p be Safe Prime;
  consider q be Sophie_Germain Prime such that
A1: card (Z/Z*(p)) = 2*q by Th13;
A2: Z/Z*(p) is cyclic Group by INT_7:31;
  then consider a being Element of (Z/Z*(p)) such that
A3: ord a = 2 and
  gr {a} is strict Subgroup of (Z/Z*(p)) by A1,Th14;
  consider b being Element of (Z/Z*(p)) such that
A4: ord b = q and
  gr {b} is strict Subgroup of (Z/Z*(p)) by A1,A2,Th14;
A5: card gr {b} = q by A4,GR_CY_1:7;
  card (1).Z/Z*(p) = 1 by GROUP_2:69;
  then consider H1 be strict Subgroup of Z/Z*(p) such that
A6: card H1 = 1;
  Z/Z*(p) is strict Subgroup of Z/Z*(p) by GROUP_2:54;
  then consider H2q be strict Subgroup of Z/Z*(p) such that
A7: card H2q = 2*q by A1;
  card gr {a} = 2 by A3,GR_CY_1:7;
  then consider H2,Hq be strict Subgroup of Z/Z*(p) such that
A8: card H2 = 2 and
A9: card Hq = q by A5;
  take q,H1,H2,Hq,H2q;
  now
    let H be strict Subgroup of Z/Z*(p);
    consider G1 be strict Subgroup of Z/Z*(p) such that
    card G1 = card H and
A10: for G2 being strict Subgroup of Z/Z*(p) st card G2 = card H holds
    G2 = G1 by A1,A2,GROUP_2:148,GR_CY_2:22;
A11: G1 = H by A10;
    now
      per cases by A1,Th1,GROUP_2:148,INT_2:28;
      suppose
        card H = 1;
        hence H = H1 or H = H2 or H = Hq or H = H2q by A6,A10,A11;
      end;
      suppose
        card H = 2;
        hence H = H1 or H = H2 or H = Hq or H = H2q by A8,A10,A11;
      end;
      suppose
        card H = q;
        hence H = H1 or H = H2 or H = Hq or H = H2q by A9,A10,A11;
      end;
      suppose
        card H = 2*q;
        hence H = H1 or H = H2 or H = Hq or H = H2q by A7,A10,A11;
      end;
    end;
    hence H = H1 or H = H2 or H = Hq or H = H2q;
  end;
  hence thesis by A7,A8,A9,A6;
end;
