# Can the Klein bottle cover the torus? Show the Klein bottle can't be covered by a space $X$ with $\pi_1(X)=\mathbb{Z}/3\mathbb{Z}$.

I'm brushing up on covering spaces for an upcoming exam, and I came across the following problems related to the Klein bottle:

1. Can there exist a cover of the torus by the Klein bottle?

and

1. Let $$X$$ be a connected and locally path connected space with $$\pi_1(X)=\mathbb{Z}/3\mathbb{Z}$$. Show the Klein bottle cannot be covered by $$X$$.

I'm mainly looking for hints here, not necessarily full solutions. For (1), should I be looking at the Galois correspondence for covers of the torus? And for (2), do I also want to look at the Galois correspondence, or would it be easier to show that $$\mathbb{Z}/3\mathbb{Z}$$ does not embed into the fundamental group of the Klein bottle $$\langle a,b\mid abab^{-1}\rangle$$?

Am I on the right track for these, or should I be doing something else?

• No because there is no injection $\Bbb Z/2\to\Bbb Z$ Jun 3, 2021 at 20:57
• Are you talking about (1)? Why does it follow from that? Jun 3, 2021 at 21:30
• @AndresMejia I know that, but I don't see what that has to do with $\mathbb{Z}/2$ and $\mathbb{Z}$. The fundamental group of the torus is not $\mathbb{Z}$, and the fundamental group of the Klein bottle is not $\mathbb{Z}/2$. Jun 3, 2021 at 22:51
• I meant $\Bbb Z^2$ (and any homomorphism must pass through the abelianization) Jun 3, 2021 at 23:00

The easiest way (in my mind) to see the answer to $$1$$ is "no" is that covering spaces of an orientable space are orientable. If you like, you can consider fundamental groups instead. If $$K$$ covered $$T$$, then by the galois correspondence we would need to have $$\pi_1 K$$ as a subgroup of $$\pi_1 T$$. But every map from $$\pi_1 K = \langle a, b \mid abab^{-1} \rangle$$ to $$\pi_1 T$$ sends $$a \mapsto 0$$. Do you see why?
For your second question, I think your idea with embeddings of fundamental groups is a good one. It's "well known" that $$\pi_1 K$$ is torsion free. In particular, there are no elements of order $$3$$. Can you show this? It's a bit tricky, so I'll leave a hint under the fold:
Can you show $$\pi_1 K = \mathbb{Z} \ltimes \mathbb{Z}$$? Is it clear that this is torsion free? It might be helpful to rewrite $$\pi_1 K = \langle a, b \mid bab^{-1} = a^{-1} \rangle$$.