0PDD Lemma 8.2.12. Let α,β∈Gn\alpha,\beta\in G_{n}. We have q(α)=q(β)q(\alpha)=q(\beta) if and only if α∼β\alpha\sim\beta.
0PDE Proof. Lemma 8.2.8 shows that if α∼β\alpha\sim\beta, then q(α)=q(β)q(\alpha)=q(\beta). Assume now q(α)=q(β)q(\alpha)=q(\beta). There are α′,β′∈En\alpha^{\prime},\beta^{\prime}\in E_{n} with α′∼α\alpha^{\prime}\sim\alpha and β′∼β\beta^{\prime}\sim\beta (Lemma 8.2.11) and we have q(α′)=q(α)=q(β)=q(β′)q(\alpha^{\prime})=q(\alpha)=q(\beta)=q(\beta^{\prime}). It follows now from Lemma 8.2.9 that α′=β′\alpha^{\prime}=\beta^{\prime}, hence α∼β\alpha\sim\beta. ∎