$S^g$ is a closed surface with genus $g$, we know that the fundamental group $\pi_1(S^g)=\{a_1,a_2,\dots ,a_g,b_1,\dots,b_g|a_1b_1a_1^{-1}b_1^{-1}\dots a_gb_ga_g^{-1}b_g^{-1}=1\}$, how to calculate the center of $\pi_1(S^g)$? I think the center of $\pi_1(S^g)$ is trivial, if $g\ge 2$, but I cannot prove it.
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The fundamental group is the amalgamated product of free groups, whose centers are trivial, so that its center is trivial, too.
Dietrich Burde
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How to write the fundamental group in the way as the amalgamated product of free groups? – Yui May 02 '14 at 15:24
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See the decomposition in Theorem 1.2. here: http://math.nsc.ru/~bogopolski/Articles/decomp.pdf. – Dietrich Burde May 02 '14 at 15:34
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You can use the fact that these groups are small cancellation (for $g>1$) to prove that their centers are trivial. See the final chapter of Lyndon and Schupp's book Combinatorial group theory. Alternatively, you can use the fact that they are torsion-free hyperbolic (a generalisation of small cancellation). See Bridson and Haefliger's book Metric spaces of non-positive curvature. However, I am sure that there are many simpler, and more direct, ways.
An alternative method would be to use the fact that this is a one-relator group. See this answer of mine on an (essentially identical) question.
