$\mathbb{Z}^2 \ast \mathbb{Z}^2$ is isomorphic to no finite index proper subgroup of itself I want to prove that $G = \mathbb{Z}^2 \ast \mathbb{Z}^2$ is isomorphic to no finite index proper subgroup of itself
Here is my partial attempt:
Consider the Bass-Serre tree $T$ that $G$ acts on in the standard way, with vertices given by the cosets of each copy of $\mathbb{Z}^2$ and trivial edge stabilisers.
Any subgroup also acts freely on the edges. Kurosh's theorem says that any subgroup $H \leq G$ is of the form $\ast_i H_i \ast F$ where $F$ is free and the $H_i$ are subgroups of conjugates of $\mathbb{Z}^2$. (Can also see this by looking at how $H$ acts on the tree $T$).
At this point I become a bit unsure and handwavy, this argument is almost certainly incorrect. If we collect all the free parts in $F$, I think the $H_i$ can only be isomorphic to $\mathbb{Z}^2$, since $\mathbb{Z}^2$ doesn't have any non-free proper subgroups, only lines which are $\mathbb{Z}$, which we collected in $F$. Then we need at least two $H_i$ to be $\mathbb{Z}^2$ otherwise we're not finite index, then we're done?
Any help with this argument (or a better argument) is welcome.
 A: Here's how you can proceed once you've applied the Kurosh subgroup theorem.
Starting from the expression $H = (*_i H_i) * F$ for your subgroup $H \le G$, break the first term into 
$$(*_{i \in I_0} H_i) * (*_{i \in I_1} H_i) * (*_{i \in I_2} H_i) * F
$$
where if $i \in I_k$ then $H_i$ is a free abelian group of rank $k$ ($k = 0,1,2$). 
We can then collect terms of the free product and rewrite this as
$$(*_{i \in I_2} H_i) * \underbrace{\left(F * (*_{i \in I_1} F_i) \right)}_{\text{free of rank $R_1 = \text{rank}(F) + \left| I_1 \right|$}}
$$ 
So what you have is a free product of $R_2 = \left| I_2 \right|$ rank 2 free abelian groups and a free group of rank $R_1$. Using Grushko's Theorem, you can prove that such a group is determined, up to isomorphism, by the the ordered pair $(R_1,R_2)$.
So, your subgroup $H$ is isomorphism to the whole group $G$ if and only if $\left|I_2\right|=2$ and $\text{rank}(F) = \left|I_1\right|=0$. 
Since your subgroup $H$ has finite index in $G$, the quotient graph of groups $T/H$ is a finite tree. Since a vertex of $T/H$ labelled with the trivial group would have to have countably infinite valence, there can be no such vertices. And there are no vertices labelled with a rank $1$ abelian group. All remaining vertices of $T/H$ are therefore labelled with rank $2$ abelian groups. Since there are only two such vertices, the the graph of groups $T/H$ is forced to be of exactly the same type as $T/G$: two $\mathbb Z^2$ vertices and one edge labelled by the trivial group. It follows that any single edge $E$ of the original Bass-Serre tree is simultaneously a fundamental domain for the whole group $G$ and for its subgroup $H$. Since $G$ acts freely on edges, it follows that $H=G$.
