Normal subgroups of F2

Question

How many distinct normal subgroups H there are in the free group of rank 2 - $F_2$, so that $F_2/H \cong V_4$, where $V_4$ stands for the Klein four-group?

Answer 1

Let $x$ and $y$ be the generators and let $H$ be any such group. In the quotient, all elements are $2$-torsion, so $x^2 \in H$, $y^2 \in H$, $(xy)^2 \in H$. The quotient is abelian, so $xyx^{-1}y^{-1} \in H$. But $$ \{x, y \hspace{2pc} | \hspace{1pc} x^2, y^2, (xy)^2, xyx^{-1}y^{-1}\} $$ is already the Klein $4$ group, so there is only one such subgroup $H$.

Answer 2

There is exactly one such subgroup $H$ of $F_2$, because for any two surjective morphisms $f,g\colon F_2\to C_2\times C_2=V_4$ we have $\ker(f)\cong \ker(g)$, and $F_2/\ker(f)\cong F_2/\ker(g)\cong V_4$. So then $F_2/H\cong V_4$ implies $H\cong \ker(f)$ for any such $f$.