I understand that this topic have been discussed again and again, yet I'm posting this question because I haven't find an satisfying answer. Most answer didn't show how they construct the covering. So I'm posting this question again, and sharing my progresss. I couldn't finish it. Any correction or advice or hint is welcome.

My approach for this problem is something like that: Notice that $H:=\langle a,b\mid bab^{-1}a\rangle$ is isomorphic to $\mathbb{Z}\rtimes \mathbb{Z}$, from which there is a epimorphism to $D_n$. Fundamental group of torus is abelian, so we are looing for an abelian nonnormal subgroup of $H$, whose image is an abelian nonnormal subgroup of $D_n$. Also notice that all covering by torus factors through the classical double cover, so we'd better consider $D_{n}$ with $n$ even.

Now there's a classification theorem for $D_n$'s subgroup: $\langle a^d \rangle$ with $d\vert n$, which is always normal, and $\langle a^d, a^rb\rangle$, where $d\vert n$ and $0\leq r <d$. When $n$ is even, $\langle a^2,b\rangle$ and $\langle a^2,ab\rangle$ are also normal, and that's all normal subgroups. (See This article).

Now back to our question. Notice that $D_2$ and $D_4$ are themselves abelian(hence has no nonnormal subgroup), so we'd better start at $D_6$. (This is the reason why a nonnormal covering of torus has at least 6 sheets). We already know what does a nonnormal subgroup looks like, we want to find an abelian subgroup. The answer here shows the only noncyclic abelian subgroup is $D_2\cong V_4$, the Klein four group. So the subgroup should be $\langle a^3,b \rangle$.

Then we consider the pullback in $H$, and that's where I got stucked. It looks the pullback is just the covering by Klein bottle.

Where did I go wrong?

(Something might help for those not familiar with Klein bottle: This answer explains why the finite covering space must be tori or Klein bottles.)


I realized something must go wrong in my argument. An answer to this question is a six sheeted cover $\langle a^3, a^2b^2\rangle$, and the image in $D_6$ is, since $b^2=e$, $\langle a^3\rangle$. It is a normal subgroup of $D_6$.

I have no ideal what's going on...

Update on 9/19

I asked my professor and he said even if we had an abelian subgroup of $D_n$ , because $D_n$ does not act properly (sorry I can't recall the exact word) on $\ker\mathbb{Z}\rtimes \mathbb{Z}\to D_n$ its preimage might not be abelian. So no, this approach does not work.

Some other reference: Notice that $a\mapsto a^k$ gives a monomorphism, and when $k>2$ the image is not normal ( For $a^2 b a b^{-1}= a$ gives ​$a^2b=aba^{-1}$, and when $k>2$ $ab^2$ is not contained in the image). This is, at least part of, the systematical way of finding nonnormal covering of Klein bottle by Klein bottles. (The idea is given here in Chinese.)

I'll try to read materials of infinite groups and see if and I can find a satisfying answer; for now I don't think I can make any progress. Still, any hints/answers/references are welcomed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.