Schreier's index formula A finitely generated group G is said to satisfy Schreier's index formula if for every subgroup H of index k in G we have: d(H) - 1 = k(d(G) - 1). For example, a finitely generated free group satisfies Schreier's index formula.
For a group G we denote by G' its commutator subgroup.
Let F be a free group on a finite number of generators. Does F/F'' satisfy Schreier's index formula? What about the next terms in the derived series of F?
 A: I claim that a residually finite, finitely generated group $G$ satisfies Schreier's formula iff it is free. 
Here is the idea. Let $X$ be the generating set of $G$ of size $d(G)$. Suppose $G$ is not free on that generating set. Suppose $K$ is the presentation complex for $G$ wrt generators $X$ and some set of relations. 
Then the Cayley graph of $G$ wrt to $X$ contains a simple loop $\ell$ at 1.  Here simple means it visits no vertex twice except when the loop closes. 
By residual finiteness we can find a finite cover $K'$ of $K$ in which the image $\ell'$ of $\ell$ is still a simple loop. I claim the finite index subgroup of $G$ represented by $K'$ based at the start of $\ell'$ fails Schreier's formula. Take a spanning tree for the 1-skeleton of $K'$ that contains all edges of the simple loop $\ell'$ except one. Then the usual method to compute the fundamental group of a graph would include $\ell'$ as part of a basis. But $\ell'$ is trivial in the fundamental group of $K'$ being the image of a loop in the Cayley graph of $G$. This shows the rank of the corresponding finite index subgroup of $G$ dropped.  
A: Benjamin Steinberg has now answered this in general. I used a computer calculation to get an explicit counterexample for $F/F''$. Let $F$ be free on $\{a,b\}$ and let $K$ be the kernel of the homomorphism $F \to S_3$ with $a \mapsto (1,2,3)$, $b \mapsto (1,2)$. So $|F:K|=6$ and $d(K)=7$.
Then $$[[a,b],[a^{-1},b^{-1}]], a^3,
b^2,
a^{-1}  b  a^{-1}  b^{-1},
(a  b)^2,
b^{-1}  a^{-1} b  a  b  a^{-1} b^{-1},
b^{-1} a^{-1} b  a  b  a  b  a  b  a^{-1} b^{-1}]$$ is a free basis of $K$, and since its first element lies in $F''$, it needs fewer than $7$ generators modulo $F''$.
