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