# 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?

• 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