Let $$S_g = \langle a_1,b_1,...,a_g,b_g \mid \prod_{i=1}^g[a_i,b_i] = 1 \rangle$$ be the fundamental group of a genus $g$ orientable surface. Why is $S_g \ast S_h \cong S_{g+h}$, and is there a nice canonical isomorphism?


1 Answer 1


Those groups are not isomorphic, for example the free product has infinitely many ends and the surface group has one end since it is quasi-isometric to the hyperbolic plane (surface groups act geometrically on the hyperbolic plane you can apply Svarc-Milnor) which is one ended. You might be interested in Stallings theorem on ends.

  • $\begingroup$ Thanks - sorry I miss understood something - thanks for the help! $\endgroup$
    – user101010
    Nov 3, 2016 at 1:03
  • $\begingroup$ @user101010 Not a problem, it is a somewhat cute problem to prove false, and to see some applications of basic geometric group theory. $\endgroup$
    – user29123
    Nov 3, 2016 at 1:32
  • 1
    $\begingroup$ Nice answer! Here's a perhaps more topological way to do it, however: surfaces are $K(G, 1)$-spaces, so an isomorphism $S_g * S_h \cong S_{g+h}$ gives a homotopy equivalence $\Sigma_g \vee \Sigma_h \cong \Sigma_{g+h}$, where $\Sigma_\bullet$ denotes a surface of appropriate genus. But the 2nd homology groups don't agree. In short, the 2nd group (co)homology of the two groups don't agree. $\endgroup$ Nov 29, 2016 at 10:19

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