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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 torus –  i. m. soloveichik Nov 29 '12 at 17:17
Sorry, but I would rather not identify such surfaces for you until you've shown where your own efforts broke down. –  Neal Nov 29 '12 at 17:24
@ 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? –  susan Nov 29 '12 at 17:51
+1, I don't see how this deserves 4 down votes. –  Matt N. Nov 29 '12 at 18:12

1 Answer 1

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. –  Matt N. Nov 29 '12 at 18:15
I've found that if I haven't been stupid at least one time each day, I haven't been thinking about math enough ;-). –  Jason DeVito Nov 29 '12 at 18:39
@JasonDeVito How would the Euler Characteristic help me in this case? –  susan Nov 29 '12 at 19:18
Projectiv space $RP(2)$ would be another example I think? –  susan Nov 29 '12 at 19:25
@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. –  Jason DeVito Nov 29 '12 at 20:54

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