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A cyclic group of order n has exactly n additive characters

Statement

Write Cn=⟨g⟩ additively, so jg denotes the sum of j copies of g and ng=0. If Cn has order n≥1, then its set Cn^ of additive characters has cardinality n. Each character is uniquely determined by χ(g), and the possible values are exactly the n distinct complex nth roots of unity. For each such root z, the corresponding character is χz(jg)=zj for every j∈Z.

Facts & Assumptions

Given: A cyclic group Cn=⟨g⟩ of finite order n≥1.

[F2]

The finite-order case of the cyclic classification identifies Z/nZ with Cn by the map [j]↦jg; hence every element is jg, and jg=kg exactly when n divides j−k (Every cyclic group is isomorphic to (Z,+) or to (Z/n,+) for its finite order n≥1).

[F3]

For n≥1, the complex nth roots of unity are exactly n distinct values (The n-th roots of a complex number and the n distinct roots of unity for every n≥1).

[F4]

The notation Cn^ is the set of additive characters of Cn (Dual group of a finite abelian group).

Proof

1.1F1F2F3F4

If χ∈Cn^ as defined in [F4], then χ(g)n=χ(ng)=χ(0)=1 by [F1] and the given relation ng=0, so evaluation at g maps this character set into the set in [F3]. Every element is jg by [F2], and [F1] gives χ(jg)=χ(g)j; thus evaluation at g is injective.

1.2F1F2F3F4F5construct

Conversely, for each nth root of unity z, define χz(jg):=zj for j∈Z. If jg=kg, [F2] gives j−k=qn for some integer q, so zj=zk(zn)q=zk by [F3, F5]; hence the value is well defined. Also χz(jg+kg)=zj+k=χz(jg)χz(kg) by [F5], so χz is an additive character in [F4], and χz(g)=z.

2.1step 1.1step 1.2F3F4∎

Steps 1.1 and 1.2 make evaluation at g a bijection from the character set [F4] to the n distinct roots in [F3]. This also covers n=1: the quotient is trivial and the sole root is 1. Therefore ∣Cn^∣=n.

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