$\mathbb{Z}^2 \ast \mathbb{Z}^2$ 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.

• To no subgroup... of what? – DonAntonio May 25 at 21:02
• 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. – Rylee Lyman May 25 at 21:20
• Sorry, it wasn't clear: subgroup of itself. I'll edit the question. – pizzaroll May 25 at 21:23
• 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 $\chi(T) + \chi(T) - \chi(pt) = -1$. – user670344 May 25 at 23:20
• @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 $X$. – Hempelicious May 26 at 18:24

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 $$H=G$$.

• Why does $G$ acting freely mean $H = G$? – pizzaroll May 25 at 22:36
• 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 $G=H$. – Lee Mosher May 25 at 23:27