From Rotman's Algebraic Topology:
If $K$ is a connected simplicial complex with basepoin $p$, then $\pi(K,p) \simeq G_{K,T}$, where $T$ is a maximal tree in $K$ and $G_{K,T}$ represents the group with presentation $$ \langle \text{all edges $(p,q)$ in $G$} | (p,q) = 1 \text{ if $(p,q)$ is an edge in $T$ and $(p,q)(q,r) = (p,r)$ if $\{p,q,r\}$ is a simplex in $K$}\rangle$$
The isomorphism is defined as $\bar \phi : G_{K,T} \rightarrow \pi (K,p)$ by $(u,v)R \mapsto \phi(u,v) = [\alpha_u(u,v)\alpha_v^{-1}]$ where $\alpha_x$ is the unique path from $p$ to $x$ in $T$.
$(1.)$ Let $K$ be defined as follows: Draw a hexagon. Add the center point (p). For each vertex (not the center) draw a line from the vertex through the center point (p) and connect it to the opposing vertex. Let $K$ be the simplicial complex consisting of all the verticies, lines, and faces created from triangulating this hexagon.
$(2.)$ Define $T$ in $K$ as the tree created by drawing a line from $p$ to each vertex of the hexagon. This is maximal as it reaches all vertices of $K$.
Let $(u,v)$ and $(v,w)$ be an edges along the outside of the hexagon. Then $(u,v)R \neq (v,w)R$ but $\bar \phi (u,v)R = \bar \phi (v,w)R$ because $\bar \phi (u,v)R = [\alpha_u (u,v) \alpha_v^{-1}] = [\alpha_v (v,w) \alpha_w^{-1}] = \bar \phi (v,w)R$. This is true because $\alpha_u (u,v) \alpha_v^{-1}$ and $\alpha_v (v,w) \alpha_w^{-1}$ are homotopic in $K$.
But then it must be the case that $(u,v)R = (v,w)R$ since $\bar \phi$ is an isomorphism.
Why is it the case then that $(u,v)R = (v,w)R$? Or am I missing something else in my understanding of this isomorphism?
For an edge $(u,v)$ along the outside of the hexagon, you always have the simplex $\{u,v,p\}$ in $K$. Hence $(u,v)(v,p)R = (u,p)R$. Now both $(v,p)$ and $(u,p)$ are edges in $T$ so that $(v,p)R=(u,p)R=1$ which also results in $(u,v)R=1$.
Indeed this implies that any edge $(u,v)$ of $K$ satisfies $(u,v)R=1$ (either because it is in $T$ or because of the above argument) so that $G_{K,T} = \{1\}$ is the trivial group.