Abelain Group

anuj — Fri, 28 Feb 2025 14:33:23 +0000

Z(p^∞) = { z ∈ ℂ | z^pk = 1 for some integer k ≥ 1 }

Proof that Z(p^∞) is an Abelian Group

We define the set:

Z(p^∞) = { z ∈ ℂ | z^{p^k} = 1 for some integer k ≥ 1 }

We will verify the group axioms under multiplication.

If z₁, z₂ ∈ Z(p^∞), then there exist integers k₁, k₂ such that:

z₁^{p^k₁} = 1 and z₂^{p^k₂} = 1.

Let k = max(k₁, k₂), then p^k is a multiple of both p^k₁ and p^k₂. Thus,

(z₁ z₂)^{p^k} = z₁^{p^k} z₂^{p^k} = 1,

so z₁ z₂ ∈ Z(p^∞).

Multiplication in ℂ is associative, so for any z₁, z₂, z₃ ∈ Z(p^∞),

(z₁ z₂) z₃ = z₁ (z₂ z₃).

The number 1 is a root of unity since 1^{p^k} = 1 for all k. Thus, 1 ∈ Z(p^∞).

For any z ∈ Z(p^∞), there exists some k such that z^{p^k} = 1.

The inverse of z is z^-1, which satisfies:

(z^-1)^{p^k} = (z^{p^k})^-1 = 1.

Thus, z^-1 ∈ Z(p^∞).

Since multiplication in ℂ is commutative,

z₁ z₂ = z₂ z₁ for all z₁, z₂ ∈ Z(p^∞).

Since Z(p^∞) satisfies closure, associativity, identity, inverses, and commutativity, it forms an abelian group under multiplication.