reserve G for strict Group,
  a,b,x,y,z for Element of G,
  H,K for strict Subgroup of G,
  p for Element of NAT,
  A for Subset of G;
reserve G for Group;
reserve H, B, A for Subgroup of G,
  D for Subgroup of A;

theorem
  for G being Group st G is finite for C being Subgroup of G
  for A,B being Subgroup of C
  for D being Subgroup of A st D = A /\ B
  for E being Subgroup of B st E = A /\ B
  for F being Subgroup of C st F = A /\ B holds
  index (C,A), index (C,B) are_coprime
  implies index (C,B) = index (A,D) & index (C,A) = index (B,E)
proof
  let G such that
A1: G is finite;
  let C be Subgroup of G;
  let A,B be Subgroup of C;
  let D be Subgroup of A such that
A2: D = A /\ B;
  let E be Subgroup of B such that
A3: E = A /\ B;
  let F be Subgroup of C such that
A4: F = A /\ B;
  assume that
A5: index A, index B are_coprime;
  index F = index E * index B by A1,A3,A4,GROUP_2:149;
  then
A6: index E * index B = index D * index A by A1,A2,A4,GROUP_2:149;
  then (index B qua Integer) divides
  ((index A qua Integer) * (index D qua Integer)) by INT_1:def 3;
  then
A7: (index B qua Integer) divides (index D qua Integer) by A5,INT_2:25;
  ex n being Element of NAT st index D = (index B) * n
  proof
    consider i being Integer such that
A8: index D = (index B qua Integer) * i by A7,INT_1:def 3;
    0 <= i by A1,A8,Th20;
    then reconsider i as Element of NAT by INT_1:3;
    take i;
    thus thesis by A8;
  end;
  then
A9: index B divides index D by NAT_D:def 3;
A10: index(C,B) >= index(A,D) by A1,A2,Th19;
A11: index B = index D
  proof
    assume
A12: index B <> index D;
    index D > 0 by A1,Th20;
    then index B <= index D by A9,NAT_D:7;
    hence contradiction by A10,A12,XXREAL_0:1;
  end;
  (index A qua Integer) divides
  ((index B qua Integer) * (index E qua Integer)) by A6,INT_1:def 3;
  then
A13: (index A qua Integer) divides index E by A5,INT_2:25;
  ex n being Element of NAT st index E = (index A) * n
  proof
    consider i being Integer such that
A14: index E = (index A qua Integer) * i by A13,INT_1:def 3;
    0 <= i by A1,A14,Th20;
    then reconsider i as Element of NAT by INT_1:3;
    take i;
    thus thesis by A14;
  end;
  then
A15: index A divides index E by NAT_D:def 3;
A16: index A >= index E by A1,A3,Th19;
  index A = index E
  proof
    assume
A17: index A <> index E;
    index E > 0 by A1,Th20;
    then index A <= index E by A15,NAT_D:7;
    hence contradiction by A16,A17,XXREAL_0:1;
  end;
  hence thesis by A11;
end;
