What am I misunderstanding about this isomorphism?

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?