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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 at 20:08

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