
theorem
  for G being finite Group, p being Prime st p divides
  card G ex g being Element of G st ord g = p
proof
  let G be finite Group;
  let p be Prime;
  reconsider p9 = p as prime Element of NAT by ORDINAL1:def 12;
A1: 1 < p9 by NAT_4:12;
  consider P be strict Subgroup of G such that
A2: P is_Sylow_p-subgroup_of_prime p by Th10;
  P is p-group by A2;
  then consider H be finite Group such that
A3: P = H and
A4: H is p-group;
  consider r be Nat such that
A5: card H = p |^ r by A4;
  assume
A6: p divides card G;
  now
    assume r=0;
    then card H = 1 by A5,NEWTON:4;
    then card G = 1 * index P by A3,GROUP_2:147;
    hence contradiction by A6,A2;
  end;
  then 0 + 1 < r + 1 by XREAL_1:6;
  then 1 <= r by NAT_1:13;
  then 1 - 1 <= r - 1 by XREAL_1:9;
  then reconsider r9 = r - 1 as Element of NAT by INT_1:3;
  0 + 1 < p9|^r9 + 1 by XREAL_1:6;
  then 1 <= p|^r9 by NAT_1:13;
  then
A7: 1*p <= p|^r9*p by XREAL_1:64;
  set H9 = (Omega).H;
A8: card H = card (Omega).H * 1;
  p |^ r = p |^(r9+1) .= p|^r9 * p by NEWTON:6;
  then card H9 > 1 by A5,A7,A1,XXREAL_0:2;
  then consider g be Element of H9 such that
A9: g <> 1_H9 by GR_CY_1:11;
  reconsider H99 = gr {g} as strict Group;
A10: H99 is cyclic Group by GR_CY_2:4;
  reconsider H99 as finite strict Group;
  set n = card H99;
A11: now
    assume n = 1;
    then ord g = 1 by GR_CY_1:7;
    hence contradiction by A9,GROUP_1:43;
  end;
  n >= 1 by GROUP_1:45;
  then n > 1 by A11,XXREAL_0:1;
  then p divides n by A5,A8,Lm8,GROUP_2:148;
  then consider H999 be strict Subgroup of H99 such that
A12: card H999 = p9 and
  for G2 being strict Subgroup of H99 st card G2=p9 holds G2=H999 by A10,
GR_CY_2:22;
  consider h9 be Element of H999 such that
A13: ord h9 = p by A12,GROUP_8:1;
  H99 is Subgroup of G by A3,GROUP_2:56;
  then reconsider H999 as strict Subgroup of G by GROUP_2:56;
  reconsider h9 as Element of H999;
  reconsider h=h9 as Element of G by GROUP_2:42;
  take h;
  thus thesis by A13,GROUP_8:5;
end;
