Finite index subgroups in Amalgamated Free products

Question

Are there any results along the following lines:

Let $\Gamma_1$ and $\Gamma_2$ be groups with respective finite index subgroups $\Gamma_0^i$ for $i=1,2$. If $\Gamma_1 \cap \Gamma_2 \leq \Gamma_0^i$ for $i=1,2$ can we conclude that $[\Gamma_1 \ast_{\Gamma_1 \cap \Gamma_2} \Gamma_2: \Gamma_0^1 \ast_{\Gamma_1 \cap \Gamma_2} \Gamma_0^2] < \infty$.

Answer 1

This is not true, and in fact the exact opposite holds in most cases.

Just as an example, if $\Gamma_1$ and $\Gamma_2$ are infinite cyclic groups, and if just one of $\Gamma^1_0$ or $\Gamma^2_0$ is a proper subgroup, then the index is infinite.

Answer 2

The following example shows that this cannot hold: let $\Gamma = <a_1, a_2 ; >$, the free group on $\{a_1,a_2\}$. Let $\Gamma_i = gp<a_i>$ for $i=1,2$, where $gp<S>$ means the subgroup of $\Gamma$ generated by $S \subset \Gamma$. Let $\Gamma_0^i = gp<a_i^2>$ for $i = 1, 2$. Then $\Gamma_1 \cap \Gamma_2 = \Gamma_0^1 \cap \Gamma_0^2 = \{1\}$ and $[\Gamma_i:\Gamma_0^i] = 2$ for $i=1,2$. But $\Gamma = \Gamma_1 * \Gamma_2$ and $gp<a^2,b^2> \cong \Gamma_0^1 * \Gamma_0^2$. The latter group has infinite index in $\Gamma$; for instance, no positive power of $ab$ belongs to $gp<a^2,b^2>$.

From a more geometric point of view, proper finite index subgroups of an amalgamated free product should not be amalgamated free products themselves. Suppose $\Gamma_i = \pi_1(X_i)$, where $X_i$ is a nice space. For simplicity, assume $\Gamma_1 \cap \Gamma_2 = \{1\}$. Then $\pi_1(X_1 \cup [1,2] \cup X_2) = \Gamma_1 * \Gamma_2$, where the segment $[1,2]$ is glued to a basepoint of $X_i$ at the endpoint $i$. A finite index subgroup of $\Gamma_1 * \Gamma_2$ will give rise to a finite sheeted covering of $X_1 \cup [1,2] \cup X_2$. But such a covering will have finitely many copies of the segment $[1,2]$. As such, the space most naturally decomposes as an amalgam of more than two subgroups. Think about the case where $X_1 = X_2 = S^1$.