By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on $H_1(S, \mathbb{Z}))$ is symplectic (preserves the symplectic form, and the image under the Torelli map is $SP(2g, Sp(2g, \mathbb{Z}).$ mathbb{Z})).$ On the other hand, the action of the outer automorphism group of $F_{2g}$ on its abelianization is the whole $SL(2g, \mathbb{Z})$ of which $SP(2g, Sp(2g, \mathbb{Z})$ is a proper subgroup. So, since $Out(\pi_1(S_{g})) \neq Out(F_{2g})$ the groups themselves are not isomorphic.
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By Baer's (I believe) theorem, the mapping class group of a closed surface is actually the outer automorphism group of the fundamental group. On the other hand, the action of this group on $H_1(S, \mathbb{Z}))$ is symplectic (preserves the symplectic form, and the image under the Torelli map is $SP(2g, \mathbb{Z}).$ On the other hand, the action of the outer automorphism group of $F_{2g}$ on its abelianization is the whole $SL(2g, \mathbb{Z})$ of which $SP(2g, \mathbb{Z})$ is a proper subgroup. So, since $Out(\pi_1(S_{g})) \neq Out(F_{2g})$ the groups themselves are not isomorphic. |
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