TITLE: Subgroup of $\pi_1(\mathbb T^2 \sharp \mathbb T^2 )$ with index $2$
QUESTION [5 upvotes]: We know $\pi_1(\mathbb {T^2} \sharp  \mathbb T^2)=\langle\alpha_1,\beta_1,\alpha_2,\beta_2|\alpha_1 \beta_1 \alpha_1^{-1} \beta_1^{-1}\alpha_2\beta_2\alpha_2^{-1}\beta_2^{-1}=1\rangle$.
My question is:How to find a subgroup of it with index $2$?
I think we need to find a subgroup H of $\pi_1(\mathbb {T^2} \sharp  \mathbb T^2)$ which has two generator.And $aH \cup bH=\pi_1(\mathbb {T^2} \sharp  \mathbb T^2)$ for $a,b\in \pi_1(\mathbb {T^2} \sharp  \mathbb T^2）$.
But how to deal with the equivalent relation?Thank you!

REPLY [2 votes]: Let $G=\pi_1(\mathbb{T}^2\sharp\mathbb{T}^2)$. Then $G^{\rm ab}\cong\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}\times\mathbb{Z}$, with generators being the images of $\alpha_1$, $\beta_1$, $\alpha_2$, and $\beta_2$. Any subgroup of index two must contain the commutator subgroup of $G$, and so it just corresponds to maps from $G^{\rm ab}$ onto $C_2$. Maps from $G^{\rm ab}$ to $C_2$ are in bijection with elements of $C_2^4$, so there are example $15$ subgroups of index $2$, corresponding to how you map the $\alpha_i$.
For example, the map given by mapping $\alpha_1$ to the generator of $C_2$ and all other generators to the identity gives the subgroup of index $2$ that is the normal closure of the subgroup generated by $\alpha_1^2$, $\alpha_2$, $\beta_1$, and $\beta_2$.