Let $M$ be a connected orientable two-dimensional orbifold with only cone points as singular points. Assume that $M$ has genus $\geq 1$. Let $\alpha$ be a loop around an order $p$ cone point. Can we conclude that $\alpha$ has order $p$ in the orbifold fundamental group $\pi_1^{\mathrm{orb}}(M)$?
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$\begingroup$ This is correct yes. For instance, you can say that $M$ admits an orbifold metric with non-positive (actually negative if $p>1$) curvature thanks to uniformization. This tells you that $M$ is developable hence the local isotropy groups inject in the fundamental group. I imagine that there is probably a better, direct way to see this though. $\endgroup$– HenriCommented Mar 8 at 13:20
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$\begingroup$ @Henri: existence of a non-positively curved metric is much weaker than the full strength of uniformisation (which says that each conformal class of metrics contains a constant-curvature representative). $\endgroup$– HJRWCommented Mar 8 at 13:29
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As mentioned in comments, the answer is "yes" and there are many ways to see it. I would refer you to §2 of Peter Scott's survey paper "The geometries of 3-manifolds", in which he discusses 2-dimensional orbifolds. His Theorem 2.3 gives a complete list of bad orbifolds, i.e. orbifolds that are not developable. None of them are hyperbolic, so your orbifold $M$ is good, i.e. developable. This means that $M$ arises as a quotient $G\backslash S$, where $S$ is a surface, $G$ is the orbifold fundamental group, and the cone points correspond precisely to the orbits of points with non-trivial stabiliser. In particular, every cone point corresponds to an element in $G$ of the correct order.
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1$\begingroup$ In connection with the OP's hypothesis that $M$ has genus $\ge 1$, it is perhaps more relevant that the list of bad orbifolds has none of genus $\ge 1$. $\endgroup$ Commented Mar 12 at 15:16
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$\begingroup$ @LeeMosher: You're quite right. I had misremembered exactly what the OP had written. $\endgroup$– HJRWCommented Mar 12 at 15:59