My broad question is regarding the lengths of (reduced) words in a subgroup of a free group.

As motivation, consider the free group $Gp(S)$ where $|S|=n$, that is, a free group of rank $n$. Let $S=\{a_1,a_2,\dots,a_n\}$. Now take a subgroup $N$ of $Gp(S)$ of index $2$. It is automatically normal, so is the kernel of a surjective homomorphism from $Gp(S)$ to $C_2=\{1,-1\}$. This means that every subgroup $N$ of index $2$ can be obtained by starting with a function from $S$ to $\{1,-1\}$ and extending it to a homomorphism of $Gp(S)$.

While any such subgroup $N$ has index $2$ (which intuitively can be thought of as roughly half the elements of $Gp(S)$ are in $N$), this intuition does not hold if we work with lengths of reduced words. That is, it is not true that for any given length $k$, the number of reduced words of length $k$ contained in $N$ is exactly half the total number of reduced words of length $k$ in $Gp(S)$. For instance, we could map $a_1 \to -1$ and $a_j \to 1$ for every $2 \leq j \leq n$, and the normal subgroup this gives us, while of index $2$, is certainly not "balanced" with respect to each length.

And as MTyson pointed out in his comment, this is not even possible in most cases. For instance, consider $S=\{a_1,a_2\}$, then the mapping $a_1 \to 1$ and $a_2 \to -1$ would map $8$ reduced words of length $2$ to $-1$, and $4$ of them to $1$. More generally, suppose $S=\{a_1,a_2,\dots,a_n\}$ with even $n$. Then on mapping half the number of generators to $-1$ (and the remaining to $1$), we get, for odd $k$, an equal number of reduced words of length $k$ mapped to $1$ and to $-1$. For even $k$, there are slightly more reduced words of length $k$ mapped to $-1$ than to $1$, the precise values of which can be calculated inductively through straightforward combinatorial arguments.

My question is whether, and how, this approach can be extended to subgroups of $Gp(S)$. Suppose $H$ is a normal subgroup of $Gp(S)$ of index $m$. Then $H$ itself is a free group (Neilsen's theorem) of rank $mn-m+1$, and so has a basis (of reduced words over alphabet $S$) of cardinality $mn-n+1$.

Again, consider an index $2$ subgroup $N$ of $Gp(S)$. The intersection $N \cap H$ is either trivially the whole of $H$ (which we shall ignore), or is an index $2$ subgroup of the group $H$. We can now ask the same question as before for the index $2$ subgroup $N \cap H$ in $H$.

Given the normal subgroup $H$ (using a basis), does there exist an index $2$ subgroup $N$ of $Gp(S)$ such that the number of length $k$ reduced words in $N \cap H$ is roughly half the number of length $k$ reduced words in $H$?

The main difficulty here arises from the fact that length is still with respect to alphabet $S$ (even though $H$ is itself a free group), and so there can be cancellations making matters difficult. It is hard enough to count the number of reduced words of a given length in $H$ (the *word problem*), but that is *not* what we care about (or so I believe)! We just want to "bisect" $H$ at each length using an index $2$ subgroup.

The next natural followup question is about quantification: what can be provably achieved?

Given the normal subgroup $H$ (using a basis) and an index $2$ subgroup $N$, let $|N \cap H|_{k}$ and $|H|_{k}$ denote the number of reduced words of length exactly $k$ in $N \cap H$ and $H$ respectively. We are then interested in $N$ that minimizes (asymptotically as a function of $n,m,k$) $$\left| \frac{1}{2} - \frac{|N \cap H|_{k}}{|H|_{k}} \right|$$ for a fixed $H$.

While this is a rather broad question, I'd be satisfied even with pointers to relevant tools and references that could possibly help. As of now, I am just staring at the wall without ideas on where to even begin.

EDIT: MTyson points out that we could ask the same question for a more restricted class of words: cyclically reduced words, and modulo cyclic permutations. Exact balance seems achievable at least for some simple hand-written examples.