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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.

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  • $\begingroup$ To no subgroup... of what? $\endgroup$ – DonAntonio May 25 at 21:02
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    $\begingroup$ I think it should be easier to use covering space theory. $\mathbb Z^2 * \mathbb Z^2$ is the fundamental group of a wedge of two tori, call it $X$, by Van Kampen. if $H$ is a subgroup of index $n$, there is a finite-sheeted cover $X_H$ of $X$. The only finite covers of tori are again tori, so it should be possible to show that any possible cover is homotopy equivalent to a space built by wedging tori together in pairs. (a bunch of tori arranged into a ring, for instance, should be possible) A proper subgroup requires at least two lifts of the wedge point, hence at least three tori. $\endgroup$ – Rylee Lyman May 25 at 21:20
  • $\begingroup$ Sorry, it wasn't clear: subgroup of itself. I'll edit the question. $\endgroup$ – pizzaroll May 25 at 21:23
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    $\begingroup$ Suppose that $G = \pi_1(X)$ for a CW complex $X$, and that there is an inclusion $G \hookrightarrow G$ with index $\infty > d > 1$. This implies there is a degree $d$ covering map $X \rightarrow X$, and thus $\chi(X) = d \chi(X) = 0$. But the Euler characteristic of $T \wedge T$ is $\chi(T) + \chi(T) - \chi(pt) = -1$. $\endgroup$ – user670344 May 25 at 23:20
  • $\begingroup$ @user670344: that's a nice approach. You probably need to assume $X$ is finite-dimensional. You definitely need to assume $X$ is acyclic. And you can only say the covering is homotopy equivalent to $X$. $\endgroup$ – Hempelicious May 26 at 18:24
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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 $H=G$.

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  • $\begingroup$ Why does $G$ acting freely mean $H = G$? $\endgroup$ – pizzaroll May 25 at 22:36
  • $\begingroup$ Pick $g \in G$ and let $g \cdot E = E'$. Since $E$ is also a fundamental domain for the action of $H$, it follows that there exists $h \in H$ such that $h \cdot E = E'$. Therefore $h^{-1}g \cdot E = E$. Since $G$ acts freely on edges, $h^{-1}g = \text{Id}_G$, and so $h=g$. This proves that $G < H$ and so $G=H$. $\endgroup$ – Lee Mosher May 25 at 23:27

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