Question

Let U₁​={z∈C∣∣z∣=1}⊂C* (a) Prove that U₁​ is a subgroup of C*. (b) Prove that C≅U₁​×R+

Answer 1

To prove that U₁​ is a subgroup of C*, we need to show that it satisfies closure, identity, and inverse. After satisfying these conditions, we can establish that U₁​ is a subgroup of C*. To prove that C≅U₁​×R+, we need to show the existence of an isomorphism between C and U₁​×R+ that preserves the group structure.

Step-by-step explanation:

(a) Prove that U₁​ is a subgroup of C*:

To prove that U₁​ is a subgroup of C*, we need to show that it satisfies the three conditions of a subgroup: closure, identity, and inverse.

1. Closure: Let z₁, z₂ ∈ U₁​. This means that |z₁| = 1 and |z₂| = 1. We have to prove that |z₁z₂| = 1. Since |z₁z₂| = |z₁||z₂| = 1 * 1 = 1, closure is satisfied.

2. Identity: The identity element of U₁​ is 1, since |1| = 1. For any z ∈ U₁​, we have z * 1 = z. Thus, the identity element is present.

3. Inverse: For every z ∈ U₁​, there exists an inverse element z⁻¹ such that z * z⁻¹ = 1. Since |z| = 1, the inverse of z is its complex conjugate. Thus, the inverse element is present.

Since U₁​ satisfies all three conditions, it is a subgroup of C*.

(b) Prove that C≅U₁​×R+:

To prove that C≅U₁​×R+, we need to show that there exists an isomorphism between C and U₁​×R+, which preserves the group structure.

Let φ be the mapping such that φ(z) = (z/|z|, |z|) for all z ∈ C. We can show that φ is an isomorphism:

1. φ is well-defined: Since the modulus of z, |z|, is a positive real number, z/|z| gives us a point on U₁​, and |z| gives us a positive real number. Therefore, φ(z) ∈ U₁​×R+ for all z ∈ C.
2. φ preserves group operation: Let z₁, z₂ ∈ C. We need to prove that φ(z₁ * z₂) = φ(z₁) * φ(z₂). Using the definition of φ, we have φ(z₁ * z₂) = (z₁ * z₂)/|z₁ * z₂|, |z₁ * z₂|) and φ(z₁) * φ(z₂) = (z₁/|z₁|, |z₁|) * (z₂/|z₂|, |z₂|) = (z₁ * z₂)/(|z₁| * |z₂|, |z₁ * z₂|). Since |z₁ * z₂| = |z₁| * |z₂|, we can conclude that φ(z₁ * z₂) = φ(z₁) * φ(z₂), thus preserving the group operation.
3. φ is bijective: φ is injective because if φ(z₁) = φ(z₂), then (z₁/|z₁|, |z₁|) = (z₂/|z₂|, |z₂|). This implies that z₁/|z₁| = z₂/|z₂| (modulus equality) and |z₁| = |z₂| (real number equality). From this, we can deduce that z₁ = z₂, proving injectivity. φ is surjective because for any (u, r) ∈ U₁​×R+, we can find z = ur in C such that φ(z) = (u, r).

Since φ is a bijective mapping that preserves the group structure, we can conclude that C≅U₁​×R+.