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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!

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    What $\sharp$ means?2018-01-22

1 Answers 1

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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 $\beta_2$.

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    @JiangnanYu: That comment should be at the question level, not at this answer level, no? It was **there** that SteveD talked about covers.2012-06-14