# Free product of surface groups

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

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.

• Thanks - sorry I miss understood something - thanks for the help! – user101010 Nov 3 '16 at 1:03
• @user101010 Not a problem, it is a somewhat cute problem to prove false, and to see some applications of basic geometric group theory. – Paul Plummer Nov 3 '16 at 1:32
• 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. – Balarka Sen Nov 29 '16 at 10:19