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I am trying to construct a non-normal covering of Klein bottle with itself and by a torus. For Klein bottle to Klein bottle, I got a three sheeted covering, just glue three Klein bottles, which is non-normal but whatever covering I am constructing by torus it is coming normal. Any help will be appreciated.

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1 Answer 1

Consider the fundamental group of the Klein bottle, $K=\langle x,y\mid y^{-1}xy=x^{-1}\rangle$.

You are looking for a finite index subgroup of $K$ (call it $H$), that is abelian and non-normal. Note that any such covering factors through the orientable double cover, corresponding to the subgroup $\langle x, y^2\rangle$.

Consider $H=\langle xy^{-2}, y^6\rangle$. This is abelian, non-normal, and of index $6$ in $K$. It can be described by the diagram below:

enter image description here

To see that this covering is non-normal, here is a picture of the resulting torus (right before the final gluing): the different colored regions are left invariant by the deck transformations (you can also see there are two distinct paths traced by the black arrows, which border the different colors):

enter image description here

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I should mention that $6$ is the smallest number of sheets you can have for a non-normal covering of the torus to the Klein bottle. – user641 Dec 5 '12 at 4:44
How do you see that the torus diagram above corresponds to the subgroup $H=\langle y^6,xy^{-2}\rangle$? If $\tilde{X}$ is the torus cover, then $\pi_1(\tilde{X})=\langle S,T|STS^{-1}T^{-1}\rangle$ where $S$ denotes the left vertical side of the square and T the top horizontal side. Clearly $p_{\ast}(S)=y^6$. But why $p_{\ast}(T)=xy^{-2}$? – user54631 Aug 13 '14 at 20:08
@user54631 In this torus diagram the four corners are not being identified, unlike in the standard square - in particular the endpoints of the blue arrow are not the same. However, the initial points of the two initial red arrows are the same, and the loop first going backwards along two green arrows (getting $y^{-2}$) and then forwards along blue (getting $x$) is a representative for the generator $T$. So $p_*(T) = xy^{-2}$. – Ben Dyer Nov 18 '14 at 21:59

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