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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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Have you tried thinking about this geometrically, i.e., what surface covers this space with index 2? – user641 Jun 13 '12 at 0:43
@SteveD Yes,yes!A double cover of $\mathbb T^2 \sharp \mathbb T^2$ corresponds to a subgroup of index 2.But I'm sorry it's a little hard for me to find a double cover of it :( – Jiangnan Yu Jun 13 '12 at 1:17
And I'm still curious how to find it through groups – Jiangnan Yu Jun 13 '12 at 1:18
Dear Jiangnan, A subgroup of index two is normal, so why not try to find a homomorphism from $\pi_1$ to a group of order two, and then determine the kernel. Finding a homomorphism from a group described by generators and relations is normally easier than directly desribing subgroups inside, because the generators and relations tell you precisely what kind of homomorphic images are allowed! Regards, – Matt E Jun 13 '12 at 4:10
To add to Matt E's comment, every subgroup of index 2 arises this way, and once you've figured out your homomorphism, a Reidemeister-Schreier rewriting will give you a presentation for that subgroup. – user641 Jun 13 '12 at 12:55

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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I think you mean normally generated at the end there. – user641 Jun 13 '12 at 12:52
@SteveD: Oops. Yes. It's almost normal, one might say, but not quite. – Arturo Magidin Jun 13 '12 at 15:28
@JiangnanYu: no, that won't be a subgroup of index $2$. Every subgroup of index $2$ is generated by $6$ elements (at least). – user641 Jun 14 '12 at 1:26
@JiangnanYu: While that subgroup projects onto the one you want, it is not equal to the one you want because it does not contain the commutator subgroup (By the way, I've never seen the notation $C_G$ for the commutator subgroup, and it is likely to be misunderstood, since $C_G(K)$ stands for the centralizer). You are missing, among others, $\alpha_1^{-1}\alpha_2\alpha_1$, $\alpha_1\alpha_2\alpha_1^{-1}$, $\alpha_1^{-1}\beta_i\alpha_1$, and $\alpha_1\beta_i\alpha_1^{-1}$, $i=1,2$. – Arturo Magidin Jun 14 '12 at 2:37
@SteveD Hi,$\mathbb T^2$ is a double cover of $\mathbb T^2$,so $\mathbb T^2\# \mathbb T^2 \# \mathbb T^2$ is a double cover of $\mathbb T^2 \# \mathbb T^2$,am I right?:) – Jiangnan Yu Jun 14 '12 at 4:38

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