$F_2$ is free group with 2 generators. Now I want to know how many subgroup with index 2 it has. And what about index 3. For the latter case, how can I judge whether the subgroup is normal?
I am trying to solve this by considering covering space. For the index 2 case, I actually only get the wedge sum of 3 circles as covering space(or homotopy equivalent to this). I believe there should be others. For index 3 case, I find two kinds of covering space: wedge sum of 4 circles and 3 vertex with 3 circles. So which should be normal?
This only answers the index 2 case (and does not use covering spaces).
Since every subgroup of index $2$ is normal this can be obtained by calculating how many surjective homorphisms there are from $F_2$ to $\mathbb{Z}_2$. Since there are only four maps from a two element set $\{x,y\}$ to $\mathbb{Z}_2$ there are exactly four homorphisms from $F_2$ to $\mathbb{Z}_2$ and only three of these are surjective. The three normal subgroups they correspond to are:
$$S_1=\{(x^{\alpha_1}y^{\beta_1}...x^{\alpha_n}y^{\beta_n})\,|\, \alpha_1+...+\alpha_n\text{ is divisible by }2\}$$ $$S_2=\{(x^{\alpha_1}y^{\beta_1}...x^{\alpha_n}y^{\beta_n})\,|\, \beta_1+...+\beta_n\text{ is divisible by }2\}$$ $$S_3=\{(x^{\alpha_1}y^{\beta_1}...x^{\alpha_n}y^{\beta_n})\,|\, \alpha_1+...+\alpha_n+\beta_1+...+\beta_n\text{ is divisible by }2\}.$$
In a similar way one can find all normal subgroups of index 3 but not arbitrary subgroups of index 3.