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!