# Covering of Klein Bottle by Torus

I have to show there is a covering of the Klein Bottle by the Torus. I realize this has been answered here: Two-sheeted covering of the Klein bottle by the torus.

However, by the Galois Correspondence we know that covering maps of the Klein Bottle correspond bijectively with subgroups of the fundamental group of the Klein Bottle. If we let $$T$$ denote the Torus and $$K$$ the Klein Bottle, then $$\pi_1(T) \cong \mathbb{Z} \times \mathbb{Z}$$ and $$\pi_1(K) \cong \langle a,b: abab^{-1} = 1 \rangle$$. To show that there is a covering of the Torus by the Klein Bottle would it be enough to show that $$\langle a,b: abab^{-1} = 1 \rangle$$ has a subgroup isomorphic to $$\mathbb{Z} \times \mathbb{Z}$$? Moreover, if this is the case, is this an easier problem to handle?

• That would only follow if spaces corresponding to certain subgroups were unique up to homeomorphism, which I don't believe is true. Jan 16 '20 at 0:10
• Doesn't the correspondence tell us that the spaces are unique up to homeomorphism?
– Mike
Jan 16 '20 at 0:12
• I'm trying to verify that right now. Jan 16 '20 at 0:13
• @CharlesHudgins It is true: see math.unl.edu/~mbrittenham2/classwk/872s07/lecnotes/… end of page 6 to beginning of page 7.
– Mike
Jan 16 '20 at 0:18
• One subgroup of $\langle a, b\ |\ abab^{-1} = 1\rangle$ that is isomorphic to $\mathbb{Z}\times \mathbb{Z}$ is the subgroup generated by $a$ and $b^2$ (indeed $ab^2= b a^{-1} b = b^2 a)$. Jan 16 '20 at 0:37

One subgroup of $$\langle a, b\ |\ abab^{-1} = 1 \rangle$$ that is isomorphic to $$\mathbb{Z}\times \mathbb{Z}$$ is the subgroup generated by $$a$$ and $$b^2$$. Indeed these elements commute: $$ab^2 = ba^{-1}b = b^2 a$$, and as this is a finite-index sub-group of a surface group it is also a surface group, so in particular it only has one relation. In general you could take $$a^m$$ and $$b^{2n}$$.
• I do have a little concern: While I see that $a,b^2$ commute, are you sure that there are no other relations between these two elements. I know that presentations work weird and sometimes there are "hidden" relations. Is there any way to guarantee that doesn't happen here?
• I think we're ok in this case. For any surface group $G\cong \pi_1(\Sigma)$ for some surface $\Sigma$, every finite-index subgroup is again a surface group, because it is the fundamental group of a covering space and a covering space of a surface is a surface. In particular, a finite-index subgroup of a surface group will be 1-relator Jan 16 '20 at 1:20