This might be a duplicate. This question also feels routine (it is also the execrise 10, page 88 in Hatcher). From Harvard qualification exam, 1990.

Let $X$ be figure eight.

1) How many 3-sheeted, connected covering space are there for $X$ up to isomorphism?

2) How many of these are normal (i.e Galois) covering spaces?

There are almost uncountably many covering spaces $Y$ for $X$ (one can check the corresponding page in Hatcher, page 58). The question is how to classify them nicely. I know that $p_{*}\pi_{1}(Y)$ has index 3 in $\pi_{1}(X)=\mathbb{Z}* \mathbb{Z}$(the free group generated by two generators). But I do not know how to find all index 3 subgroups of $\mathbb{Z}*\mathbb{Z}$. On the other hand if $H$ is normal in $\mathbb{Z}*\mathbb{Z}$, then the above question can be greatly simplified, but I still do not know how to solve it precisely. I tried to think in terms of deck transformations, etc but did not get anywhere.


Hint: Instead of thinking about index 3 subgroups of $\Bbb Z \star \Bbb Z$, consider what connected 3-fold covers of $S_1 \vee S_1$ look like. Any such cover is a connected graph on 3 vertices of valence 4 (why?), and there are only finitely many such graphs. Then use deck transformations to check if each cover is regular.

  • $\begingroup$ I see, need a pen and a piece of paper instead of thinking in front of the laptop screen. Thanks. $\endgroup$ – Bombyx mori Aug 6 '12 at 1:29
  • $\begingroup$ Thanks. The hint makes it easier to visualize. $\endgroup$ – Bombyx mori Aug 6 '12 at 1:50
  • $\begingroup$ So (it seems to me) there are only two normal subgroups, since one need to preserve the rotational invariance. There are a total 4 isomorphic classes, but I may be missing something. $\endgroup$ – Bombyx mori Aug 6 '12 at 2:10
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    $\begingroup$ @Ergin: Once you have all of the possible graphs described, normal subgroups correspond to the graphs on which the group of deck transformations act transitively. You can check that there are only two of these (up to isomorphism). $\endgroup$ – Brandon Carter Jun 20 '13 at 14:28
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    $\begingroup$ @Ergin: Say $\pi_1(S^1 \vee S^1) = \langle a, b \rangle$. Then there are only 2 normal subgroups up to swapping $a$ and $b$ (and in one case, swapping $b$ with $b^{-1}$). There are actually 4 normal subgroups. Some people will not distinguish, for example, between the 4th and 5th covers, as these are really only swapping $a$ and $b$. That was what was meant by the 2 isomorphism classes of regular covers. $\endgroup$ – Brandon Carter Jun 22 '13 at 18:25

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