If \(\Gamma\) is a finite graph of finite groups with injective edge homomorphisms, show that the graph product of the groups has a free subgroup of finite index by constructing a suitable finite-sheeted covering space of \(K\) from universal covers of the mapping cylinders in \(K\). [The converse is also true: A finitely generated group having a free subgroup of finite index is isomorphic to such a graph product. For a proof of this see [Scott & Wall 1979], Theorem 7.3.]
Solution
Let \(\Gamma\) be a finite connected graph of finite groups with injective edge homomorphisms. Write \(G_v\) for the vertex groups and \(G_e\) for the edge groups. Choose \[ N \] to be a common multiple of the orders of all vertex and edge groups.
1. Over each vertex space \(K(G_v,1)\) take\[ \frac{N}{|G_v|} \]disjoint copies of its universal cover\[ E G_v\to K(G_v,1). \]The resulting map over the vertex space has total degree \(N\).
Each universal cover has degree \(|G_v|\) because its deck group is the finite group \(G_v\). Hence \[ \frac{N}{|G_v|}\cdot |G_v|=N. \]
2. Over each edge mapping cylinder take\[ \frac{N}{|G_e|} \]copies of the universal lifted edge cylinder\[ E G_e\times I. \]
The edge classifying space \(K(G_e,1)\) has universal cover \(E G_e\) of degree \(|G_e|\). Thus the displayed number of copies again gives total degree \(N\) over the interior of the edge cylinder.
3. The boundary components of these lifted edge cylinders can be matched exactly with the inverse images of the edge subspaces in the chosen vertex covers.
2.Hence the boundary copies can be paired and glued.The same count holds at the other endpoint of the edge.
4. After performing these gluings for every edge, one obtains a finite-sheeted covering\[ \widehat K\to K\Gamma \]of degree \(N\).
By <1>1–<1>3 the local degree is \(N\) on every vertex space, edge-cylinder interior, and attaching region. The pieces are glued by lifts of the original attaching maps, so the resulting map is locally a homeomorphism and has exactly \(N\) points in every fiber. Thus it is an \(N\)-sheeted covering.
5. Every connected component of \(\widehat K\) has free fundamental group.
Upstairs, every vertex space \(E G_v\) and every edge space \(E G_e\) is contractible. Collapsing each contractible vertex space to a point and each edge cylinder to an interval gives a homotopy equivalence from a component of \(\widehat K\) to its underlying incidence graph. The fundamental group of a graph is free. Equivalently, the lifted graph of groups has trivial vertex and edge stabilizer groups, so Exercise 3 applies.
6. Therefore the original graph product contains a free subgroup of finite index.
Choose any connected component \[ \widehat K_0\subseteq\widehat K. \] Since \(K\Gamma\) is connected, the restriction \[ \widehat K_0\to K\Gamma \] is a connected finite-sheeted covering, of some finite degree \(d\le N\). The corresponding subgroup \[ H=(p_*)\pi_1(\widehat K_0) \le\pi_1(K\Gamma) \] has index \(d\), and by 5 the group \(H\) is free. Hence \[ \boxed{\pi_1(K\Gamma)\text{ is virtually free}.} \]