Graphs of Groups and Bass Serre Theory

Let $\mathrm{\Gamma }:={\pi }_1(\mathcal{G},T)$, and let $\tilde{X}$ be its Bass-Serre tree. By [L1], $\mathrm{\Gamma }$ acts on $\tilde{X}$ without inversions, the quotient graph is $X$, and for the standard lifts $G_v$ and $G_e$ from [L2] the stabilizers are exactly the chosen groups of $\mathcal{G}$. Thus the quotient graph of groups recovered from this action is the original graph of groups $\mathcal{G}$.

Apply [L4] to the action of $\mathrm{\Gamma }$ on $\tilde{X}$, but choose the maximal subtree $T^{\mathrm{\prime }}$ of the quotient graph. Step 1.1 identifies that quotient graph of groups with $\mathcal{G}$, so [L4] gives $\mathrm{\Gamma }\cong {\pi }_1(\mathcal{G},T^{\mathrm{\prime }})\mathrm{.}$ Since $\mathrm{\Gamma }={\pi }_1(\mathcal{G},T)$, the two relative fundamental groups are isomorphic.

If $X$ has no geometric edges, then ${\pi }_1(\mathcal{G},T)$ is the unique vertex group, so every element already has edge length $0$ and the identity clause is immediate. If $X$ has one geometric edge, then either that edge lies in $T$, in which case [L1] and [L2] present ${\pi }_1(\mathcal{G},T)$ as the corresponding amalgamated free product and [L4] gives the required normal form, or it lies outside $T$, in which case [L1] and [L2] present ${\pi }_1(\mathcal{G},T)$ as the corresponding HNN extension and [L5] gives the required normal form.

Assume the theorem holds for graphs of groups with at most $n$ geometric edges, and let $X$ have $n+1$. If $X$ has an edge $e$ outside $T$, remove that edge. The graph $X\setminus \{e\}$ stays connected because it still contains the spanning tree $T$, and [L1] and [L2] identify ${\pi }_1(\mathcal{G},T)$ with the HNN extension of ${\pi }_1(\mathcal{G}{\mathrm{\mid }}_{X\setminus \{e\}},T)$ having stable letter $e$ and associated subgroups the two images of $G_e$. Otherwise $X=T$ is a tree. Choose a leaf edge $e$ of $T$; removing it splits $X$ into connected components $X_1$ and $X_2$, and $T\setminus \{e\}$ restricts to maximal subtrees $T_1\subseteq X_1$ and $T_2\subseteq X_2$. Unwinding [L1] and [L2] then identifies ${\pi }_1(\mathcal{G},T)$ with the amalgamated free product of ${\pi }_1(\mathcal{G}{\mathrm{\mid }}_{X_1},T_1)$ and ${\pi }_1(\mathcal{G}{\mathrm{\mid }}_{X_2},T_2)$ over $G_e$.

In each smaller graph-of-groups fundamental group, the induction hypothesis supplies reduced closed representatives. Applying the amalgam normal form [L4] or the HNN normal form [L5] to the decomposition from step 1.2 and then expanding the smaller representatives back into the original generators gives a closed graph-of-groups word in which the only forbidden simplifications are exactly the backtracking patterns ruled out by [L3]. Hence every element of ${\pi }_1(\mathcal{G},T)$ has a reduced closed graph-of-groups representative.

The same normal-form theorems [L4] and [L5] say that a reduced amalgam word or Britton-reduced word is nonidentity whenever it has positive syllable length, and that syllable length $0$ gives the identity only when the remaining base-group coefficient is the identity. Under the identifications of step 1.2, this is exactly the statement that a reduced closed graph-of-groups word of positive edge length is nonidentity, and that an edge-length-$0$ reduced closed word is the identity only when its unique coefficient is the identity in the relevant vertex group.

A nonidentity element of $G_v$ is already a reduced graph-of-groups word of edge length $0$. The edge-length-$0$ clause of [L1] says that such a reduced word represents the identity only when its unique coefficient is the identity of $G_v$.

Therefore no nonidentity element of $G_v$ maps to the identity in ${\pi }_1(\mathcal{G},T)$. So the canonical map $G_v\to {\pi }_1(\mathcal{G},T)$ is injective.

The graph $\tilde{X}$ is connected. Indeed, a reduced graph-of-groups word for $\gamma \in {\pi }_1(\mathcal{G},T)$ records an edge path from the base coset $G_{v_0}$ to the vertex $\gamma G_v$, so every vertex is reached from a base vertex by some path in $\tilde{X}$.

A reduced closed path in $\tilde{X}$ based at a vertex coset determines a reduced graph-of-groups word of positive edge length representing the identity element, because following the path returns to the initial coset. That contradicts [L2]. Hence $\tilde{X}$ has no nontrivial reduced closed path.

Being connected and having no nontrivial reduced closed path, $\tilde{X}$ is a simplicial tree.

Left multiplication ${\gamma }^{\mathrm{\prime }}\cdot (\gamma G_v)=({\gamma }^{\mathrm{\prime }}\gamma )G_v$ and ${\gamma }^{\mathrm{\prime }}\cdot (\gamma G_e)=({\gamma }^{\mathrm{\prime }}\gamma )G_e$ is well defined on the cosets of [L1], and the formulas for origin and terminus in [L1] are preserved by that multiplication. So $\mathrm{\Gamma }$ acts by graph automorphisms on $\tilde{X}$.

The orbit of the base vertex coset $G_v$ is the set of all cosets $\gamma G_v$, so the quotient vertices are identified with the original vertices $v$; the same holds for edges. Hence the quotient graph is exactly $X$. An inversion would send some oriented edge coset to its reverse, which would force the two opposite orientations of one edge of $X$ into the same orbit; that does not happen in the quotient description.

The stabilizer of the base vertex coset $G_v$ is $G_v$ itself, and similarly for $G_e$. Therefore [L3] gives $\mathrm{Stab}(\gamma G_v)=\gamma G_v{\gamma }^{-1}$ and $\mathrm{Stab}(\gamma G_e)=\gamma G_e{\gamma }^{-1}$ for arbitrary cosets in the same orbits.

Combining steps 1.1, 2.1, and 2.2 with [L2] proves all four claims.

If the chosen lift $\tilde{e}$ is replaced by $h\tilde{e}$, then the edge stabilizer changes from $\mathrm{Stab}(\tilde{e})$ to $h\mathrm{Stab}(\tilde{e})h^{-1}$ by [L2], and the same conjugation change occurs for the endpoint stabilizers of $o(\tilde{e})$ and $t(\tilde{e})$.

The new connecting element can be taken as $h{g}_e{k}^{-1}$ when the terminal lift is changed by $k$. Therefore the new terminus map sends $hx{h}^{-1}$ to $k(g_e^{-1}x{g}_e)k^{-1}$. This is exactly the old boundary monomorphism conjugated by the vertex-group identifications from step 1.1.

Hence the boundary monomorphisms are independent of representative choices up to canonical conjugacy.

Choose the lift data from [L1] successively along the maximal subtree $T_0$ so that every tree edge has connecting element $g_e=1$. Send each chosen vertex stabilizer element to itself in $G$ and each oriented edge symbol to the corresponding $g_e$. The identity $g_{\bar{e}}=g_e^{-1}$ from [L1] respects the reversal relation, and the definition of the boundary map respects the conjugacy relation of [L2], so this defines a homomorphism from the path group to $G$. Because every tree edge maps to $1$, [L3] yields a homomorphism $\mathrm{\Phi }:{\pi }_1(\mathcal{G},T_0)⟶G\mathrm{.}$

Fix a chosen lift ${\tilde{v}}_0$ of some quotient vertex $v_0$. Let $g\in G$. The unique path in the tree $T$ from ${\tilde{v}}_0$ to $g{\tilde{v}}_0$ projects to an edge path $e_1,\dots ,e_n$ in the quotient graph. Inductively along that path, choose stabilizer elements $h_0,\dots ,h_n$ at the intermediate chosen lifts so that the $j$th lifted edge is $\mathrm{\Phi }(h_0{e}_1{h}_1\cdots e_j){\tilde{e}}_j$ and its terminal vertex is $\mathrm{\Phi }(h_0{e}_1{h}_1\cdots e_j{h}_j)\widetilde{t(e_j)}$. At the end one obtains $g=\mathrm{\Phi }(h_0{e}_1{h}_1\cdots e_n{h}_n)$. Hence $\mathrm{\Phi }$ is surjective.

Let $\tilde{X}$ be the Bass-Serre tree of $(\mathcal{G},T_0)$. Using [L4], define $\mathrm{\Psi }(\gamma G_v):=\mathrm{\Phi }(\gamma )\tilde{v},\mathrm{\Psi }(\gamma G_e):=\mathrm{\Phi }(\gamma )\tilde{e}\mathrm{.}$ This is well defined because the vertex and edge groups in [L1] are actual stabilizer subgroups of the chosen lifts, and the incidence formulas of [L4] match the connecting elements $g_e$ by construction.

The map $\mathrm{\Psi }$ is $\mathrm{\Phi }$-equivariant by definition. At a chosen vertex coset $G_v$, the incident edges of $\tilde{X}$ above an oriented quotient edge $e$ are the cosets $h{G}_e$ with $h\in G_v$, and $\mathrm{\Psi }$ sends them bijectively to the incident edges $h\tilde{e}$ at the chosen lift $\tilde{v}$. By equivariance the same holds at every vertex. Since [L6] says $\tilde{X}$ is a tree and step 2.1 shows that every translate of every chosen lift lies in the image, $\mathrm{\Psi }$ is a covering map from a connected tree onto the tree $T$, hence an isomorphism of graphs.

Let $\gamma \in {\pi }_1(\mathcal{G},T_0)$ satisfy $\mathrm{\Phi }(\gamma )=1$. By [L5], choose a reduced closed graph-of-groups word representing $\gamma$. Under the map $\mathrm{\Psi }$ of step 2.2, its closed edge path begins and ends over the same quotient vertex, and the endpoint is the $\mathrm{\Phi }(\gamma )$-translate of the starting point. Thus it traces a closed path in $T$. If the word had positive edge length, this would be a nontrivial reduced closed path in the tree $T$, impossible. Therefore the reduced representative has edge length $0$, so by the length-$0$ clause of [L5] it is just one coefficient from a chosen vertex stabilizer. But $\mathrm{\Phi }$ restricts on each vertex group to the actual inclusion into $G$, so $\mathrm{\Phi }(\gamma )=1$ forces that coefficient to be the identity. Hence $\gamma =1$, and $\mathrm{\Phi }$ is injective.

Steps 2.1 and 3.2 show that $\mathrm{\Phi }$ is an isomorphism, and step 3.1 gives the $G$-equivariant identification of the Bass-Serre tree with the original tree. This proves both claims.

Because the unique geometric edge belongs to the maximal subtree, [L2] kills the edge symbol. The only remaining generators are the two vertex groups, and the only remaining cross relation is that the two images of the edge group agree.

That is exactly the pushout presentation named in [L1], so the fundamental group of the one-segment graph of groups is $A{\ast }_{C}B$.

Since the quotient graph has one vertex and the loop edge is outside the maximal subtree, [L2] leaves one edge symbol $t$ together with the vertex group $A$. The defining relation of the path group is exactly $t\alpha (c)t^{-1}=\beta (c)$.

Therefore the resulting fundamental group is precisely the HNN extension described in [L1].

Because the action is free, every vertex and edge stabilizer in the quotient graph of groups is trivial. By [L1], it is therefore enough to compute the fundamental group of a graph of trivial groups.

With all stabilizers trivial, the path-group relations reduce to $\bar{e}=e^{-1}$ and there are no vertex-group generators. After killing the maximal-tree edges as in [L3], the remaining generators are exactly the non-tree oriented edges, with no further relations. That is the free-group universal property of [L2].

Hence $G$ is free on the non-tree edge generators of the quotient graph.

By [L1], the graph-of-groups fundamental group acts on its Bass-Serre tree. Because every stabilizer is trivial, this action is free and without inversions.

Applying [L2] to the action from step 1.1 shows that the fundamental group is free. If the quotient graph is finite, a maximal subtree uses all vertices and all but the non-tree geometric edges, so the free basis from the previous corollary has one generator for each such edge.

Let $X$ be the star-shaped graph with one central vertex of trivial group, one leaf vertex $v_i$ of group $G_i$ for each $i\in I$, and one trivial edge joining the center to $v_i$. The whole star is a maximal subtree, so [L2] and [L3] kill all edge symbols and leave only the leaf vertex groups with no cross-relations. By the universal property in [L1], this relative fundamental group is exactly ${\ast }_{i\in I}{G}_i$, so we may regard $G$ as the fundamental group of this graph of groups.

Let $T$ be the Bass-Serre tree of the star graph of step 1.1. The subgroup $H$ acts on $T$ without inversions, so [L5] identifies $H$ with the fundamental group of the quotient graph of groups $H\mathrm{\backslash }T$. By [L4], every edge stabilizer in $T$ is trivial, and every vertex stabilizer over the leaf orbit of $v_i$ has the form $H\cap g{G}_i{g}^{-1}$ for some $g\in G$.

Choose a maximal subtree $T_0$ of $H\mathrm{\backslash }T$. Because every edge group is trivial, the quotient graph-of-groups path group has no conjugacy relations, only the reversal relations from [L2]. Killing the edges of $T_0$ via [L3] therefore leaves a free product of the nontrivial vertex stabilizers together with a free group generated by the geometric edges outside $T_0$. Hence $H$ is a free product of the groups $H\cap g{G}_i{g}^{-1}$ that occur as nontrivial vertex stabilizers, together with some free group $F$.

A vertex of $T$ above the leaf vertex $v_i$ is a left coset $g{G}_i$. Two such vertices, $g{G}_i$ and $g^{\mathrm{\prime }}{G}_i$, lie in the same $H$-orbit exactly when $g^{\mathrm{\prime }}=hga$ for some $h\in H$ and $a\in G_i$, equivalently when $Hg{G}_i=H{g}^{\mathrm{\prime }}{G}_i$. Thus the $H$-orbits of vertices above $v_i$ are indexed by the double cosets $H\mathrm{\backslash }G\mathrm{/}{G}_i$, and choosing one representative from each double coset with nontrivial stabilizer gives exactly the index set in the statement. Combining this with step 3.1 proves the theorem.

Fix $j$. Apply [L1] to the subgroup $A_j\le G$ inside the decomposition $G\cong B_1\ast \cdots \ast B_n\ast F_s$. Because $A_j$ is freely indecomposable and not infinite cyclic, its Kurosh decomposition can contain neither a positive-rank free part nor two distinct nontrivial factors. By [L2], any subgroup of a conjugate of $F_s$ is free, so a nontrivial such subgroup would be either infinite cyclic or freely decomposable. Therefore the unique nontrivial Kurosh factor is $A_j$ itself, and it is contained in a conjugate of some $B_k$.

Let $B_k^{\mathrm{\prime }}=g{B}_k{g}^{-1}$ be a conjugate containing $A_j$. Now view $B_k^{\mathrm{\prime }}$ as a subgroup of the first decomposition $G\cong A_1\ast \cdots \ast A_m\ast F_r$ and apply [L1] again. The identity double coset for the factor $A_j$ contributes the nontrivial subgroup $A_j$ to the Kurosh decomposition of $B_k^{\mathrm{\prime }}$. Since $B_k^{\mathrm{\prime }}$ is also freely indecomposable and not infinite cyclic, its Kurosh decomposition has no second nontrivial factor and no free part. Hence $B_k^{\mathrm{\prime }}=A_j$. So every $A_j$ is conjugate to some $B_k$.

Apply the same argument with the roles of the two decompositions reversed: every $B_k$ is conjugate to some $A_j$. Also, applying [L1] to the subgroup $A_j$ inside its own decomposition shows that $A_j$ meets every conjugate of $A_{\mathrm{\ell }}$ for $\mathrm{\ell }\mathrm{\ne }j$ and every conjugate of $F_r$ trivially, because the identity double coset for $A_j$ already supplies the only possible nontrivial Kurosh factor. Therefore two distinct $A_j$ cannot both be conjugate to the same $B_k$. By symmetry the correspondence is bijective, so after permuting indices we get $m=n$ and each $A_j$ is conjugate to $B_j$.

After that permutation, quotient $G$ by the normal closure of the factors $A_1,\dots ,A_m$. In the first decomposition this kills the nonfree factors and leaves $F_r$; in the second decomposition it kills the conjugate factors $B_1,\dots ,B_m$ and leaves $F_s$. Thus $F_r\cong F_s$. Since both free groups have finite rank, [L3] gives $r=s$. This proves the uniqueness and rank statement.

By [L1], the definition is algebraic: it starts from generators, edge relations, and a quotient by tree-edge relations.

A topological realization may exist later, but it is not part of the definition recorded in step 1.1. Therefore the statement is false.

Consider the one-segment graph of groups with vertex groups $C_2$ and $C_3$ and trivial edge group. Its fundamental group is $C_2\ast C_3$ by [L2]. Let $a$ and $b$ be nonidentity elements of the two factors. By [L1], the vertex coset $b{C}_2$ has stabilizer $b{C}_2{b}^{-1}$.

If $ba{b}^{-1}$ lay in $C_2$, say $ba{b}^{-1}=c$, then $ba{b}^{-1}{c}^{-1}$ would be a reduced positive-length free-product word representing the identity, contrary to [L3]. Hence $b{C}_2{b}^{-1}\mathrm{\ne }{C}_2$: this vertex stabilizer is a conjugate of, but not literally equal to, the chosen vertex group.

If a graph of groups has a nontrivial vertex group $G_v$, then [L1] says that $G_v$ fixes the corresponding vertex of the Bass-Serre tree.

So this tree action has a nonidentity stabilizer and is not free. Therefore the statement is false.

Take a one-loop quotient graph. Giving it trivial vertex and edge groups produces the fundamental group $\mathbb{Z}$ by [L2]. Giving the same loop vertex group $C_2$ and trivial edge group produces $C_2\ast \mathbb{Z}$. By [L3], each fundamental group acts on its Bass-Serre tree with that same one-loop quotient graph.

The first group is free and hence torsion-free by [L4], while [L5] embeds the order-$2$ vertex subgroup $C_2$ in the second, so they are not isomorphic. Thus the quotient graph alone does not determine the acting group.

Let $G=C_2\ast C_3$ and let $H=C_2$ be the first embedded factor, whose embedding is supplied by [L2]. In the Kurosh decomposition of $H$, the identity double coset contributes the intersection $H\cap C_2=H$.

The subgroup $H=C_2$ contains a nonidentity element of order $2$, whereas [L3] says every free group is torsion-free. Thus $H$ is a subgroup of a free product that is not free, and the universal statement is false.