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Let $\mathcal{S}_g$ denote the fundamental group of an oriented surface of genus $g\ge 2$.

Does $\mathcal{S}_g$ contain subgroups $A$ and $B$ of finite index such that $A\cap B = \lbrace e\rbrace$?

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closed as too localized by Benjamin Steinberg, Derek Holt, Misha, Alain Valette, Ian Agol Mar 13 '13 at 19:56

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Intersection of two finite index subgroups is again a finite index subgroup, so the conclusion holds for all infinite groups. –  Misha Mar 13 '13 at 13:48
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@Misha: Don't you mean the conclusion doesn't hold, since otherwise $|S_g|=(S_g:1)=(S_g:A\cap B))< \infty$ ? –  Ralph Mar 13 '13 at 13:55
    
Ralph, yes, that's what I meant. –  Misha Mar 13 '13 at 13:56
    
Thanks everybody. Sorry for the trivial question. –  Mark Grant Mar 13 '13 at 17:18

1 Answer 1

up vote 3 down vote accepted

As Misha says in a comment, for any group $G$ with subgroups $A,B$, we have

$|G:A\cap B|\leq |G:A||G:B|$

(exercise). In particular, if $G$ is infinite (as here) then $A\cap B$ is non-trivial.

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