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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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    $\begingroup$ Intersection of two finite index subgroups is again a finite index subgroup, so the conclusion holds for all infinite groups. $\endgroup$
    – Misha
    Mar 13, 2013 at 13:48
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    $\begingroup$ @Misha: Don't you mean the conclusion doesn't hold, since otherwise $|S_g|=(S_g:1)=(S_g:A\cap B))< \infty$ ? $\endgroup$
    – Ralph
    Mar 13, 2013 at 13:55
  • $\begingroup$ Ralph, yes, that's what I meant. $\endgroup$
    – Misha
    Mar 13, 2013 at 13:56

1 Answer 1

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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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