(a) Let G be a finite group generated by two elements u and v of order 2 each. Show that G is isomorphic to the dihedral group of…

(a) G is the dihedral group of order 2n Set-up. G = ⟨u, v⟩ is finite, with o(u) = o(v) = 2, so u² = v² = e, u⁻¹ = u and v⁻¹ = v. Put r = uv. Since G is finite, r has finite order, and we are given o(r) = n.

Step 1: G = ⟨r, u⟩. Because u⁻¹ = u, we have v = u(uv) = ur. So every word in u and v is a word in u and r, and G = ⟨r, u⟩.

Step 2: the dihedral relation. Using u⁻¹ = u and v⁻¹ = v:

u r u⁻¹ = u(uv)u = u²·vu = vu = v⁻¹u⁻¹ = (uv)⁻¹ = r⁻¹.

So ur = r⁻¹u, and more generally u rᵏ = r⁻ᵏ u for every integer k.

Step 3: |G| = 2n. By the relation ur = r⁻¹u, any product of the generators r and u can be rewritten with all powers of r moved to the left. Hence every element of G has the form rᵏ or rᵏu (0 ≤ k ≤ n − 1). So G = ⟨r⟩ ∪ ⟨r⟩u and |G| ≤ 2n.

Next we show u ∉ ⟨r⟩. Suppose u = rᵏ. Then u commutes with r, so r = uru⁻¹ = r⁻¹, i.e. r² = e and n ≤ 2.

• If n = 1, then r = e and u ∈ ⟨r⟩ = {e}, which contradicts o(u) = 2.
• If n = 2, then u ∈ {e, r}. So u = r = uv, which gives v = e, contradicting o(v) = 2.

Hence u ∉ ⟨r⟩, and the two cosets ⟨r⟩ and ⟨r⟩u are disjoint, each with n elements. Therefore |G| = 2n and the 2n elements rᵏ, rᵏu are distinct.

Step 4: the isomorphism. The dihedral group of order 2n has the presentation D₂ₙ = ⟨a, b | aⁿ = b² = e, bab⁻¹ = a⁻¹⟩, and it has exactly 2n elements aᵏ, aᵏb. Define φ on generators by φ(a) = r and φ(b) = u. The defining relations are preserved: rⁿ = e (because o(r) = n), u² = e, and uru⁻¹ = r⁻¹ (Step 2). By von Dyck's theorem, φ therefore extends to a homomorphism D₂ₙ → G. Explicitly, φ(aᵏbᵉ) = rᵏuᵉ. φ is onto because r and u generate G. A surjection between finite sets of the same size 2n is also one-one. Hence φ is an isomorphism and G ≅ D₂ₙ.

Degenerate cases. For n = 1, u = v and G = {e, u} is cyclic of order 2, which is D₂. For n = 2, uv = vu and G = {e, u, v, uv} is the Klein four-group, which is D₄. For n ≥ 3, G is the group of symmetries of a regular n-gon: r is the rotation through 2π/n and u is a reflection.

(b) Convergence of the recursive sequence Here uₙ₊₁ = √(7 + uₙ) and u₁ = √7. Let L = (1 + √29)/2 ≈ 3.1926 be the positive root of x² − x − 7 = 0, so that L² = L + 7 and L = √(7 + L).

Bounded above by L (induction). First, u₁ = √7 ≈ 2.6458 < L. If uₙ < L, then uₙ₊₁ = √(7 + uₙ) < √(7 + L) = L. So 0 < uₙ < L for all n.

Monotone increasing. Since uₙ > 0,

uₙ₊₁² − uₙ² = 7 + uₙ − uₙ² = −(uₙ² − uₙ − 7) = −(uₙ − L)(uₙ − L′),

where L′ = (1 − √29)/2 < 0 is the other root. For 0 < uₙ < L we have uₙ − L < 0 and uₙ − L′ > 0, so uₙ₊₁² > uₙ². As all terms are positive, uₙ₊₁ > uₙ. (Numerically: u₁ = 2.6458, u₂ = 3.1058, u₃ = 3.1790, u₄ = 3.1904, …)

Existence of the limit. (uₙ) is increasing and bounded above by L. By the monotone convergence theorem it converges to some l, with √7 ≤ l ≤ L.

Identifying the limit. The function t ↦ √(7 + t) is continuous. Letting n → ∞ in uₙ₊₁ = √(7 + uₙ) gives l = √(7 + l), so l² − l − 7 = 0 and l = (1 ± √29)/2. The negative root (1 − √29)/2 ≈ −2.19 is impossible, because l ≥ √7 > 0.

Conclusion. lim uₙ = (1 + √29)/2 ≈ 3.1926, the positive root of x² − x − 7 = 0. The limit was taken only after its existence had been established.

(c) Contour integration Let I = ∫₀²π sin²θ/(a + b cos θ) dθ, with a > b > 0.

Substitution. On the unit circle C: |z| = 1, put z = e^(iθ). Then dθ = dz/(iz), cos θ = (z² + 1)/(2z) and sin θ = (z² − 1)/(2iz), so sin²θ = −(z² − 1)²/(4z²). Also a + b cos θ = (bz² + 2az + b)/(2z). Hence

I = ∮_C [−(z² − 1)²/(4z²)]·[2z/(bz² + 2az + b)]·dz/(iz) = (i/2) ∮_C f(z) dz,

where f(z) = (z² − 1)²/[z²(bz² + 2az + b)] and −1/(2i) = i/2.

Poles. f has a double pole at z = 0. It also has simple poles at the roots of bz² + 2az + b = 0:

α = (−a + √(a² − b²))/b, β = (−a − √(a² − b²))/b, with αβ = 1 and α − β = 2√(a² − b²)/b.

Location. |α| = (a − √(a² − b²))/b. The inequality a − √(a² − b²) < b is equivalent to a − b < √(a² − b²), i.e. (a − b)² < (a − b)(a + b), i.e. a − b < a + b. This is true since b > 0, and squaring was legitimate because a − b > 0. So |α| < 1, and |β| = 1/|α| > 1. Only z = 0 and z = α lie inside C.

Residue at z = 0 (double pole). Res = d/dz[(z² − 1)²/(bz² + 2az + b)] at z = 0 = [4z(z² − 1)(bz² + 2az + b) − (z² − 1)²(2bz + 2a)]/(bz² + 2az + b)² at z = 0 = (0 − 2a)/b² = −2a/b².

Residue at z = α (simple pole). Here bz² + 2az + b = b(z − α)(z − β), so

Res = (α² − 1)²/[α²·b(α − β)] = (α − 1/α)²/[b(α − β)].

Since 1/α = β, this equals (α − β)²/[b(α − β)] = (α − β)/b = 2√(a² − b²)/b².

Residue theorem. I = (i/2)·2πi·[−2a/b² + 2√(a² − b²)/b²] = −π·(2/b²)[√(a² − b²) − a],

so I = (2π/b²){a − √(a² − b²)}, as required.

Check. The result is positive, as it must be for a positive integrand. For a = 3, b = 2 it gives (π/2)(3 − √5) ≈ 1.2000, which agrees with direct numerical integration.