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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.

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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.

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    I see, need a pen and a piece of paper instead of thinking in front of the laptop screen. Thanks.2012-08-06
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    Thanks. The hint makes it easier to visualize.2012-08-06
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    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.2012-08-06
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    @user32240: That's right.2012-08-06
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    Why are there only 2 normal subgroups? Can you explain it?2013-06-19
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    @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).2013-06-20
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    I see but I can't find all deck transformations of fourth and fifth 3-sheeted coverings of $S^{1}VS^{1}$ in the following link. http://math.ou.edu/~forester/6813F06/e1sol.pdf you can look at question 4. So I couldn't know whether their subgroup is normal2013-06-21
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    @Ergin: In both of those covers, rotation clearly gives one deck transformation (of order 3). Note that this tells you already that the corresponding subgroup is normal (you don't have to find _all_ deck transformations). On the other hand, if you pick a particular distinguished point (say the top left point), a deck transformation is completely distinguished by where this point is mapped to. If you don't see this immediately, you should work through it to convince yourself. Thus the group of deck transformations is $\mathbf{Z}/3\mathbf{Z}$.2013-06-21
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    Thanks. Then second and third covers are also normal by the same reason. Is it true?2013-06-22
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    @Ergin: Exactly.2013-06-22
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    But you said there are only 2 normal subgroups. Now we have found 4 normal subgroups. Right? I appologize for the long discussion.2013-06-22
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    @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.2013-06-22