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

theorem Th20:
  for G being finite Group, H,N being Subgroup of G holds
  N is normal Subgroup of G &
  H is p-group & N is p-group
  implies ex P being strict Subgroup of G st the carrier of P = H * N &
  P is p-group
proof
  let G be finite Group;
  let H,N be Subgroup of G;
  assume that
A1: N is normal Subgroup of G and
A2: H is p-group & N is p-group;
  H * N = N * H by A1,GROUP_5:8;
  then consider P be strict Subgroup of G such that
A3: the carrier of P = H * N by GROUP_2:78;
A4: for a being Element of P holds a is p-power
  proof
    let a be Element of P;
    a in H * N by A3;
    then consider b,c be Element of G such that
A5: a = b * c & b in H & c in N by GROUP_11:6;
    b is p-power by A5,A2,Th15;
    then consider n be Nat such that
A6: ord b = p |^ n;
A7: b |^(p |^ n) = 1_G by A6,GROUP_1:41;
    consider d be Element of G such that
A8: d in N & (b * c) |^(p |^ n) = b |^(p |^ n) * d by A1,A5,Th7;
    d is p-power by A2,A8,Th15;
    then consider m be Nat such that
A9: ord d = p |^ m;
A10: d |^(p |^ m) = 1_G by A9,GROUP_1:41;
A11:a |^(p |^ (n + m)) = (b * c) |^(p |^ (n + m)) by A5,GROUP_8:4
                      .= (b * c) |^(p |^ n * (p |^ m)) by NEWTON:8
                      .= (b * c) |^(p |^ n) |^ (p |^ m) by GROUP_1:35
                      .= 1_G by A7,A8,A10,GROUP_1:def 4;
    reconsider a1 = a as Element of G by GROUP_2:42;
    a1 |^(p |^ (n + m)) = 1_G by A11,GROUP_8:4;
    then consider r be Nat such that
A12: ord a1 = p |^ r & r <= n + m by Th2,GROUP_1:44;
 ord a = p |^ r by A12,GROUP_8:5;
    hence thesis;
  end;
  take P;
  thus thesis by A3,A4,Th17;
end;
