theorem Th32:
  A c= B implies gr A is Subgroup of gr B
proof
  assume
A1: A c= B;
  now
    let a;
    assume a in gr A;
    then consider F,I such that
A2: len F = len I and
A3: rng F c= A and
A4: Product(F |^ I) = a by Th28;
    rng F c= B by A1,A3;
    hence a in gr B by A2,A4,Th28;
  end;
  hence thesis by GROUP_2:58;
end;
