theorem Th11:
  for G being finite Group, G1 being Subgroup of G, a being
Element of G holds card G = n & G=gr {a} & card G1 = p & G1= gr{a|^k} implies n
  divides k*p
proof
  let G be finite Group, G1 be Subgroup of G, a be Element of G;
  assume that
A1: card G = n and
A2: G=gr {a} and
A3: card G1 = p and
A4: G1= gr{a|^k};
  set z=k*p;
  consider m,l being Nat such that
A5: z=(n*m)+l and
A6: l < n by A1,NAT_1:17;
  l=0
  proof
    a|^k in gr {a|^k} by Th2;
    then reconsider h=a|^k as Element of G1 by A4,STRUCT_0:def 5;
    assume
A7: l<>0;
    a|^z = a|^k|^p by GROUP_1:35
      .= h|^p by GROUP_4:1
      .= 1_G1 by A3,GR_CY_1:9
      .= 1_G by GROUP_2:44;
    then
A8: 1_G = (a|^(n*m))*(a|^l) by A5,GROUP_1:33
      .= (a|^n|^m)*(a|^l) by GROUP_1:35
      .= ((1_G)|^m)*(a|^l) by A1,GR_CY_1:9
      .= (1_G)*(a|^l) by GROUP_1:31
      .= a|^l by GROUP_1:def 4;
    a is not being_of_order_0 & ord a = n by A1,A2,GR_CY_1:6,7;
    hence contradiction by A6,A7,A8,GROUP_1:def 11;
  end;
  hence thesis by A5,NAT_D:def 3;
end;
