Let $F_2$ be the free group on two elements $a,b$ and $C_2=\{0,1\},$ the cyclic group of order $2.$
Let $H$ be the kernel of the unique homomorphism $\phi:F_2\to C_2^2$ with $\phi(a)=(1,0)$ and $\phi(b)=(0,1).$
Any subgroup of a free group is free, by this theorem.
In this answer I prove that $H$ is generated by: $$a^2,b^2,(ab)^2,(ba)^2, (a^2b)^2, (ab^2)^2$$
Is $H$ free on these generators?
This is equivalent to asking if the group is free on:
$$a^2,b^2,(ab)^2,(ba)^2,ab^2a,ba^2b,$$
which might be easier, because these are all “words” of length at most $4.$
But I have no idea how to show there is no reduce the set of generators.
Best Answer
No. The first five of your six generators are free generators of $H$. Note that $$ba^2b = (ba)^2 (ab^2a)^{-1} (ab)^2.$$
In fact by the Schreier formula, the free rank of a subgroup of index $r$ in $F_n$ is $n(r-1)+1$, which gives $5$ for the free rank of $H$.