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I know that the fundamental group of the double torus is $\pi_1(M)=\langle a,b,c,d;a^{-1}b^{-1}abc^{-1}d^{-1}cd\rangle$.

How can I calculate its center subgroup $C$? Is $C$ trivial?

Let $p$ be the quotient map from $\pi_1(M)$ to $H_1(M)$, maybe it's easy to prove that $p(C)$ is $0$ in $H_1(M).$ That will also solve my problem.

Thanks, Yan

  • 0
    According to a theorem of Preissman, every nontrivial subgroup of the fundamental group of a compact negatively curved Riemannian manifold is isomorphic to $\mathbb{Z}$. This doesn't quite prove that the center of the fundamental group of a surface is trivial, but that would be my guess.2012-11-06
  • 3
    Like all of the surface groups (apart from the torus itself, where it is free abelian of rank 2), this one is torsion-free and word-hyperbolic. Its centre must be trivial, since word-hyperbolic groups do not contain non-cyclic free abelian subgroups.2012-11-06
  • 1
    Two quick ways: (1) $G$ is an amalgamated product of two free groups (whose centers are trivial), so it itself has trivial center. (2) As Derek Holt remarked, all fundamental groups of hyperbolic surfaces have trivial centers. This can be seen "geometrically" by allowing $a$ and $b$ in $G$ to commute; then the isometries they represent in the hyperbolic plane have the same fixed points, and using discreteness, imply $a$ and $b$ are powers of the same word in $G$. This says centralizers are cyclic, and it is easy to go from there to trivial center (since $G$ itself is non-cyclic).2012-11-06

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