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We know that mapping class group (MCG) $\Gamma_1$ for genus 1 closed surface is generated by two elements: $U$ of order 6 and $S$ of order 4. There is a defining relation that totally fixed the MCG $\Gamma_1$: $U^3=S^2$. (Is this correct?)

In arXiv:math/0309299, Korkmaz showed that mapping class group $\Gamma_2$ for genus 2 closed surface is also generated by two elements: $U_1=A_4A_3A_2A_1$ of order 10 and $U_2=A_5A_4A_3A_2A_1$ of order 6.

Do we know the defining relations among the two generators $U_1$ and $U_2$ that will allow us to define $\Gamma_2$?

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In Section 6 of the Korkmaz paper he uses the Wajnryb presentation to derive a presentation on his two generators, for genus $g>2.$ For genus $2$ one can use the exact same method, but applied to Birman-Hilden's (1971) presentation of the mapping class group of genus $2.$

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  • $\begingroup$ It seems the section 6 is about $g\geq 3$. $\endgroup$
    – Xiao-Gang Wen
    Commented May 21, 2015 at 22:15
  • $\begingroup$ @Xiao-GangWen true, I was careless in reading. For $g=2$ use the Birman-Hilden (really Bergau-Mennike) presentation instead of Wajnryb's, as now stated. $\endgroup$
    – Igor Rivin
    Commented May 21, 2015 at 22:56
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    $\begingroup$ With your hint, I found the following paper www-irma.u-strasbg.fr/~massuyea/talks/MCG.pdf. Theorem 3.3 answer my question above, but for the case of five generators. Thanks! $\endgroup$
    – Xiao-Gang Wen
    Commented May 22, 2015 at 16:49

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