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Let $X$ be a complex manifold and let $G$ and $H$ be two finite subgroups of its automorphism group $Aut(X)$. Suppose we are given that $X/G$ and $X/H$ are bi-holomorphic complex manifolds. What can we say about $G$ and $H$?

Is it the case that $G$ and $H$ have to be isomorphic as subgroups?

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    $\begingroup$ The quotient of $\mathbb P^1$ by any finite subgroup of its automorphism group is biholomorphic to $\mathbb P^1$, so in this case we can say absolutely nothing. $\endgroup$
    – Will Sawin
    Commented Aug 26, 2019 at 14:02
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    $\begingroup$ @WillSawin Could you please explain your statement. If we take the quotient of $\mathbb{P}^1$ by a finite order rotation than the quotient is an orbifold not necessarily a manifold. So what is the meaning of biholomorphism on an orbifold. $\endgroup$
    – Cusp
    Commented Aug 27, 2019 at 3:28
  • $\begingroup$ @Cusp One can take a geometric quotient in the category of schemes (i.e. the coarse space of the quotient orbifold). Because vikram didn't mention anything about orbifolds, I assumed they were thinking about this type of quotient. $\endgroup$
    – Will Sawin
    Commented Aug 27, 2019 at 13:48
  • $\begingroup$ If we're taking the quotient in the category of orbifolds, we only get bi-holomorphic complex manifold if the group actions $G$ and $H$ are free. (In which case it would be odd not to mention this at the beginning.) Then it seems hard but I expect G and H still need not be isomorphic. $\endgroup$
    – Will Sawin
    Commented Aug 27, 2019 at 13:50

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