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Identify the compact surfaces $X$ for which there exist a proper subgroup $G$ of $\pi_1(X)$ such that $G\cong \pi_1(X)$.

EDIT: Suggestions?

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    one possibility is a torus2012-11-29
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    Sorry, but I would rather not identify such surfaces for you until you've shown where your own efforts broke down.2012-11-29
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    @ Neal: Yea sorry, I should have been more careful! I am not looking just for an answer at all, I want to know how to get to the answer. But I am having trouble figuring out how to tackle this question, so any hints would be appreciated?2012-11-29

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Here's another way to rephrase the question. Every subgroup of $\pi_1(M)$ corresponds to a connected cover of $M$. Further, $\pi_1(M)$ characterizes closed surfaces, in the sense that if two closed surfaces have isomorphic fundamental groups, then they are diffeomorphic. (This is something very special to closed surfaces!)

So, another way to recast your question is the following: Which closed surfaces cover themselves in a nontrivial way? (Nontrivial means more than 1-sheeted).

As a further hint: study the Euler Characteristic.

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    Thanks for your comment on my deleted post. To answer it: because was being stupid.2012-11-29
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    I've found that if I haven't been stupid at least one time each day, I haven't been thinking about math enough ;-).2012-11-29
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    @JasonDeVito How would the Euler Characteristic help me in this case?2012-11-29
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    Projectiv space $RP(2)$ would be another example I think?2012-11-29
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    @susan: How does Euler Characteristic behave with respect to covers? Also, $\mathbb{R}P^2$ doesn't work - it only covers itself in the trivial way.2012-11-29
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    @JasonDeVito I think it multiplies, but i am not sure exactly how? Can you elaborate?2012-11-29
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    You're right that it multiples. So if $M\rightarrow N$ is a $k$-sheeted covering, then $\chi(M) = k\chi(N)$. What does that tell you if you also know $M\cong N$?2012-11-29
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    @JasonDeVito: By $M\cong N$ do you mean their fundamental groups?2012-11-29
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    @JasonDeVito I am tempted to say that the Euler characteristic of the surface has to be zero, but I don't have a very strong argument??2012-11-29
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    By $M\cong N$ I meant that $M$ and $N$ are diffeomorphic. You're right about the Euler characteristic - try to come up wit ha strong argument ;-). Incidentally, you must still prove that the two surfaces of Euler characteristic $0$ (the torus and Klein bottle) actually have such nontrivial coverings.2012-11-29