The braid group $B_n(S^2)$ on the sphere possesses a finite element of order which generates a cyclic group $M$ in $B_n(S^2)$. My question is this: What is the index of $H$ in $B_n (S^2)$? Same question for the braid group on the projective plane.
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2$\begingroup$ Please use tex for formatting mathematics, and add some background information. Probably "element of finite order" instead of "finite element of order"? Also: are $M$ and $H$ different objects? $\endgroup$– Sebastian GoetteFeb 21, 2016 at 17:03
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$\begingroup$ Thank you Sebastian, ok i would to ask " element of finite order" $\endgroup$– Nouar DegaichiFeb 22, 2016 at 10:49
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$\begingroup$ I'm not sure whether your $H$ is the same as your $M$, which is the cyclic group generated by the only nontrivial torsion element of $B_{n}(S^{2})$. If $n = 2$, then $M$ is all of $B_{2}(S^{2}) \cong \mathbb{Z} / 2 \mathbb{Z}$. Otherwise, $M$ has infinite index in $B_{n}(S^{2})$. $\endgroup$– Jeff YeltonFeb 22, 2016 at 11:07
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$\begingroup$ yes for n=2 it's ok but if n=3 , B_{3}(S²) is also finite and so H is of finite indexe, if n greather than 3, B_{n}(S²) is infinite $\endgroup$– Nouar DegaichiFeb 22, 2016 at 11:30
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1$\begingroup$ sorry Jeff, but B₃(S²) has order 12 because if we set a=σ₁σ₂σ₁ and b=σ₁σ₂ Then we have B₃(S²)=〈a,b / a²=b³=(ab)² 〉 So a⁴=b⁶=1 a²=(ab)² then a=b⁻¹ab⁻¹ and hence σ₁⁴=a⁴=1 it's similiar that σ₂ is of order 4 $\endgroup$– Nouar DegaichiFeb 23, 2016 at 10:09
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