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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. –  Chris Eagle May 13 '12 at 23:20
    
Your counting argument still works, though. –  Chris Eagle May 13 '12 at 23:22
    
Sorry - my mistake, I'll edit the question so the fundamental group is right. –  Edward Hughes May 13 '12 at 23:33
    
Yes thanks - my brain is clearly not working at this time of night! Thanks for verifying the counting argument though! –  Edward Hughes May 13 '12 at 23:35
1  
@LeeMosher: but the subgroups are index 2; doesn't that mean they're automatically normal, hence stabilised by conjugation? –  Ben Millwood Jun 2 '12 at 18:01
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