Question

Let n > 1. Consider the monoid D_2n = mn〈 a, b ∣ a² = 1, bⁿ = 1, ab = bⁿ⁻¹a 〉, n > 1. Prove that the presentation is Church–Rosser and defines a group of order 2n. Show that D_2n is isomorphic to the group of all symmetries of the regular n-gon on the plane. This is the so-called dihedral group of order 2n.

Answer 1

Suppose first that σ 6∈ A. Let eσ = (λx. c) d. The set A may contain redexes in c and d. Reducing σ first, a copy of d replaces each free occurrence of x in c (see Fig. 1). If we then reduce the redexes in e θA(e) θσ(e) θC(θA(e)) = θB(θσ(e)) A σ C B Figure 3 these copies of d in some acceptable order, then reduce the remaining redexes in c in some acceptable order, this yields the same result as reducing the redexes in d and c in some acceptable order before reducing σ, then reducing σ. Formally, take B = {σγi | 1 ≤ i ≤ m} ∪ {σδiτj | 1 ≤ i ≤ k, 1 ≤ j ≤ n}, where A = {σ00γi | 1 ≤ i ≤ m} ∪ {σ1τj | 1 ≤ j ≤ n} and the free occurrences of x in c are located at {σ00δ1, . . . , σ00δk}. The elements of A of the form σ00γi represent the redexes in c, which after reducing σ become the elements of B of the form σγi . The elements of A of the form σ1τj represent the redexes in d, which after reducing σ become the elements of B of the form σδiτj representing the corresponding redexes in the copies of d that replaced the free occurrences of x in c. In Fig. 1, k = 2. If σ ∈ A, then it must appear last in any acceptable ordering of A. By the previous argument, there exists B ⊆ σ↓ such that θ∅ (θA(e)) = θσ(θA−{σ} (e)) = θB(θσ(e)).