A consequence of Schreier's formula is that if G is a subgroup of the free group F of rank n < 1 and rank G ≦ n, then G = F or G is of infinite index in F. However, if S is a free sovlvable group of derived length I < 1 and H is a subgroup of S which is free solvable of the same length, then the rank of H does not exceed the rank of S. These observations led G. Baumslag to conjecture that if H is of finite index in S then H = S. In fact, we have sharper results in two directions. If H and S are free solvable of the same length, not only is H of infinite index in S, but δ1−1(S)/δ1−1(H) is torsion-free. In another direction we need not assume that S is free solvable, only that s is torsion-free and of derived length l (l > 1) and that H is not cyclic. Thus Stallings' theorem [11] that a finitely generated torsionfree group with a free subgroup of finite index is itself free has an even stronger counterpart in the variety of groups solvable of length at most l (l > 1): a torsionfree group in that variety with a non-cyclic free subgroup of finite index coincides with this subgroup. The proof relies on the following theorem: If S is a free solvable group, J is the group of automorphisms of S which induce the identity on S/S', and I is the group of inner automorphisms of S, then J/I is torsion-free. The proofs of these theorems form the bulk of the first four sections.