theorem Th26:
   N is Subgroup of H implies N ~ A c= H ~ A
proof
  assume
A1: N is Subgroup of H;
  let x be object;
  assume
A2: x in N ~ A;
  then reconsider x as Element of G;
  x * N meets A by A2,Th14;
  then x * H meets A by A1,GROUP_3:6,XBOOLE_1:63;
  hence thesis;
end;
