reserve G for Group,
  a,b for Element of G,
  m, n for Nat,
  p for Prime;

theorem Th17:
  for G being finite Group st for a being Element of G holds a is p-power
  holds G is p-group
proof
  let G be finite Group;
  assume
A1: for a being Element of G holds a is p-power;
  assume not G is p-group;
  then for r being Nat holds card G <> p |^ r;
  then consider s be Element of NAT such that
A2: s is prime & s divides card G and
A3: s <> p by Th1;
  consider b be Element of G such that
A4: ord b = s by A2,GROUP_10:11;
  b is p-power by A1;
  then ex r1 be Nat st ord b = p |^ r1;
  hence contradiction by A2,A4,A3,NAT_3:6;
end;
