My question is (a modification of) the following:
Let $N=\cap M$, where $M$ is maximal normal of finite index in the free group $G=F_2$. Then what is the group $G/N$?
Currently written, this is a flawed question - $N$ is necessarily trivial, because of cyclic groups of prime order, hence $F_2/N=F_2$. So...having thought about this a bit, I am pretty sure the following is the question I want to ask (as simple groups are generated by an involution and...some other element):
Let $N=\cap M$, where $M$ is maximal normal of finite index in the free product $G=\mathbb{Z}_2\ast\mathbb{Z}$ and the group $G/M$ is non-cyclic. Then what is the group $G/N$?
A different, but related question is:
Let $N=\cap_I M_i$, where $M_i$ is maximal normal of finite index in the free product $G=\mathbb{Z}_2\ast\mathbb{Z}$, the group $G/M_i$ is non-cyclic and $G/M_i\cong G/M_j$ for all $i, j\in I$. Then what is the group $G/N$?