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I'm trying to count all the double coverings of the double torus. I know that the fundamental group of the double torus is

$$\pi_1(X)=\langle a,b,c,d;[a,b][c,d]\rangle $$

where $[a,b]=aba^{-1}b^{-1}$. I also know from the classification theorem for covering spaces that in fact it suffices to count subgroups of index 2. I'm not sure how to do this however.

Here's an idea I've just had - it suffices to count surjective homomorphisms $\pi_1(X)\rightarrow C_2$. These are precisely the maps which send at least one of $a,b,c,d$ to the generator $r$ of $C_2$. Indeed it's easily checked that any such map is a homomorphism. But there are exactly $2^4 -1=15$ such maps. Hence there are 15 double covers of the double torus. Does this sound plausible? To me $15$ seems a bit large...!

Many thanks in advance.

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    That is not the fundamental group of the double torus.2012-05-13
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    Your counting argument still works, though.2012-05-13
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    Sorry - my mistake, I'll edit the question so the fundamental group is right.2012-05-13
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    Yes thanks - my brain is clearly not working at this time of night! Thanks for verifying the counting argument though!2012-05-13
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    Your count of 15 is correct if what you are counting is pointed covering spaces (up to point preserving covering transformation). If you ignored base point, you would be counting instead the set of conjugacy classes of index 2 subgroups (the kernels of your 15 homomorphisms), so you would indeed get a number smaller than 15.2012-05-22
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    Ah interesting - that is now intuitively more satisfying!2012-05-22
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    @LeeMosher: but the subgroups are index 2; doesn't that mean they're automatically normal, hence stabilised by conjugation?2012-06-02
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    @LeeMosher Might I suggest you turn your comment(s) into an answer. This may help to remove this question from the unanswered queue. If you do, you can post a link to your answer [here](http://chat.stackexchange.com/rooms/9141/the-crusade-of-answers) to draw some attention to it and possibly get some upvotes.2017-03-09
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    Hah, that's a funny page. Okay, I might do that.2017-03-09

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