Normal form theorem for graphs of groups. Let $\mathcal{G}$ be a graph of groups and $G=\pi_1(G)$.

1. Any $g \in G$ can be written as
$g = g_0 t_{e_1}^{\epsilon_1} g_1 \cdots t_{e_n}^{\epsilon_n} g_n\quad$
as before.
2. If $g = 1$, this expression includes `backtracking’, meaning that for some $i$, $e_i = e_{i + 1}$ with $\epsilon_i = - \epsilon_{i + 1}$, and furthermore that if $\epsilon_i = \pm 1$, then $g_i \in \partial_\pm ( G_{e_i} )$.

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree $T$ is a tree.

Proof. To simplify notation, set $t_i = t_{e_i}^{\epsilon_i}$, so

$g = g_0 t_1 g_1 \cdots t_n g_n$.

Fix base points $*_i$ in the vertex spaces $X_{v_i}$, which are chosen to coincide when the vertices do. Then $g_i$ is a loop in $X_{v_i}$ based at $*_i$, and $t_i$ is a path, crossing the corresponding edge space, from $*_i$ to $*_{i + 1}$. This allows us to consider $g$ as a loop in $X_{\mathcal{G}}$ based at $*_0$. (We may assume $v_0 = v_n$ by adding letters from a maximal tree.)

Consider the universal covering $\widetilde{X}_{\mathcal{G}}$ and fix a base point $\widetilde{*}_0$ over $*_0$ in $\widetilde{X}_{\widetilde{v}_0}$. Let $\widetilde{g}$ be the lift of $g$ based at $\widetilde{*}_0$ and $\gamma$ its image in the Bass–Serre tree $T$. We now analyze $\widetilde{g}$ and $\gamma$ closely.

Choose $\widetilde{e}_i$ adjoining $\widetilde{X}_{\widetilde{v}_0}$ and $\widetilde{X}_{\widetilde{v}_1}$ so that the edge traversed by $t_i$ when lifted at $\widetilde{*}_{i - 1}$ corresponds to the coset $1 \cdot G_{e_i} \subseteq G_{v_i} / G_{e_i}$.

Then $g_0$ lifts to a path in $\widetilde{X}_{\widetilde{v}_0}$ which terminates at $g_0 \widetilde{*}_0$. Similarly, $t_1$ lifts at $*_0$ to a path across the edge $\widetilde{e}_1$ to the vertex space $t_1\widetilde{X}_{\widetilde{v}_1}$ terminating at $t_1 \widetilde{*}_1$. Therefore, $g_0 t_1$ lifts at $\widetilde{*}_0$ to a path which crosses the edge space $g_0 \widetilde{e}_1$ and ends at $g_0 t_1 \widetilde{*}_1$.

Then, $g_1$ lifts at $\widetilde{*}_1$ to a path in $\widetilde{X}_{\widetilde{v}_1}$ ending at $\widetilde{*}_1$, and $t_2$ lifts at $\widetilde{*}_1$ to a path across the edge $\widetilde{e}_2$ into the vertex space $\widetilde{X}_{\widetilde{v}_2}$, and terminating at $t_2 \widetilde{*}_2$. Thus $g_0 t_1 g_1 t_2$ lifts at $\widetilde{*}_0$ to a path which crosses $g_0 \widetilde{e}_1$, through $g_0 \widetilde{X}_{\widetilde{v}_1}$, across $g_0 t_1 g_1 \widetilde{e}_2$, and ending at

$g_o t_1 g_1 t_2 \widetilde{*}_2 \in g_0 t_1 g_1 t_2 \widetilde{X}_{\widetilde{v}_2}$.

We continue this process until we have explicitly constructed $\widetilde{g}$. By hypothesis, $g = 1$, so $\widetilde{g}$ and $\gamma$ are both loops in $\widetilde{X}_{\mathcal{G}}$ and $T$, respectively. Since $T$ is a tree, $\gamma$ must backtrack.

This implies that $\widetilde{e}_i=\widetilde{e}_{i + 1}$ and that $\epsilon_i = -\epsilon_{i + 1}$. That is, by Lemma 18,

$g_{i + 1} \in \partial_\pm^{e_i} ( G_{e_i} )$.

Therefore, we have found a backtracking, and can accordingly shorten $g$. This proves the theorem. $\square$