Quasi-isometric Classification of Free Products of Surface Groups

Let $$S_g$$ denote the compact surface with $$g$$ holes, and denote its fundamental group as $$$$\Sigma_g=\pi_1(S_g)=\left.$$$$ In particular, we have that $$\Sigma_1 = \mathbb{Z}^2$$.

There are two tools I use for the proof. First we have that for $$g\geq 2$$, the group $$\Sigma_g$$ is quasi-isometric to $$\Sigma_2$$ (this I already understand, see https://math.stackexchange.com/q/2818142 and my following remark about coverings). Secondly, if we can find that two groups are commensurable, then they are quasi-isometric (this means we can pass by finite index subgroups). An equivalent way to use this is by defining finite-degree covering maps between two surfaces. Here I shall mention the equality: $$\pi_1 (S)*\pi_1 (S')=\pi_1 (S\vee S').$$

I try to proof that a free product $$G = \Sigma_{g_1}*\dots \Sigma_{g_n}=\coprod_{i=1}^n \Sigma_{g_i}$$ is quasi-isometric to one of following five groups: $$\mathbb{Z}^2 , \Sigma_2, \mathbb{Z}^2* \mathbb{Z}^2, \mathbb{Z}^2* \Sigma_2, \Sigma_2* \Sigma_2$$.

This way we can show that

• $$\mathbb{Z}^2*...* \mathbb{Z}^2$$ is quasi-isometric to $$\mathbb{Z}^2* \mathbb{Z}^2$$.
• $$\Sigma_2*...* \Sigma_2$$ is quasi-isometric to $$\Sigma_2* \Sigma_2$$.
• $$\Sigma_2*...* \Sigma_2 * \mathbb{Z}^2*...* \mathbb{Z}^2$$ is quasi-isometric to $$\Sigma_2* \mathbb{Z}^2$$.

This is where I am stuck. These three equivalences should be proven by showing the two groups are finite index subgroups of each other (note that $$\Sigma_2$$ can be replaced by $$\Sigma_g$$ at any time). An indirect way to show these are finite index subgroups of each other, is by defining a finite-degree covering map from one to the other. If I remember well this is due to the Seifert–Van Kampen theorem.

This question and many insights of for a proof are based on the paper "Amenability, bilipschitz equivalence, and the von Neumann conjecture" by Kevin White.

Edit:

I imagine there are many ways to prove this. I wish to thank Lee Mosher for his solution, but I have the impression that Kevin White uses more elementary arguments. Probably, that is why he adds no reference to the proof. In my opinion, he seems to imply easy subgroups can be found. Here is his proof.

If $$G$$ and $$H$$ are each nontrivial free products of one-ended groups, then $$G$$ is quasi-isometric to $$H$$ if and only if the set of quasi-isometry types of free factors of $$G$$ equals the set of quasi-isometry types of free factors of $$H$$.
1. The quasi-isometry class of $$\mathbb Z^2$$, which equals the quasi-isometry class $$[\mathbb R^2]$$ of the Euclidean plane $$\mathbb R^2$$; and
2. The quasi-isometry class of the $$\Sigma_g$$'s with $$g \ge 2$$, which equals the quasi-isometry class $$[\mathbb H^2]$$ of the hyperbolic plane $$\mathbb H^2$$.
And your answer is the correct one, as one sees by applying the above theorem: the three quasi-isometry classes of free products in your question correspond to the three nonempty subsets of the set $$\{[\mathbb H^2],[\mathbb R^2]\}$$, namely $$\{[\mathbb H^2],[\mathbb R^2]\}, \quad \{[\mathbb H^2]\}, \quad \{[\mathbb R^2]\}$$