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The character table of a finite cyclic group over $C$

Example

For $G = ⟨ g ⟩$ cyclic of order $n$ and $ζ$ a primitive $n$ -th root of unity, the character table has rows $χ_{k}$ and columns $g^{m}$ with entry $ζ^{k m}$ ( $0 \leq k, m \leq n - 1$ ). Both orthogonality relations hold, and the squared degrees sum to $n = ∣ G ∣$ .

Facts & Assumptions

Given: A cyclic group $G = ⟨ g ⟩$ of order $n \geq 1$ and a primitive $n$ -th root of unity $ζ$ .

[F1]

The irreducible characters are $χ_{k} (g^{m}) = ζ^{k m}$ for $0 \leq k \leq n - 1$ , pairwise distinct (The irreducible complex characters of a finite cyclic group are the $n$ powers of a primitive $n$ th root).

[F2]

Vieta's formulas: if the monic $f (t) = t^{n} + a_{1} t^{n - 1} + \dots + a_{n}$ splits as $\prod_{i = 1}^{n} (t - α_{i})$ , then $a_{1} = - \sum_{i} α_{i}$ (Vieta's formulas identify the coefficients of a split monic polynomial with elementary symmetric functions of its roots).

[F3]

The first orthogonality relation reads $⟨ χ_{i}, χ_{j} ⟩ = δ_{i j}$ (The first orthogonality relation for irreducible complex characters).

[F4]

The second orthogonality relation reads $\sum_{i} χ_{i} (g) \overline{χ_{i} (h)} = ∣ C_{G} (g) ∣$ when $h$ is conjugate to $g$ and $0$ otherwise (The second orthogonality relation for irreducible complex characters).

[A1]

In $C [x]$ , $t^{n} - 1 = \prod_{m = 0}^{n - 1} (t - ζ^{m})$ , because the two sides have the same degree, the same roots, and are monic.

Verification

technique · direct

1.1

By [F1] the entry in row $χ_{k}$ and column $g^{m}$ is $ζ^{k m}$ , so the table has the stated shape.

F1given

1.2

If $n = 1$ , then $G$ is trivial and $\sum_{m = 0}^{n - 1} ζ^{k m} = 1$ for the unique value $k = 0$ . Assume now that $n > 1$ . Then the polynomial $t^{n} - 1$ has coefficient $0$ at $t^{n - 1}$ , so [A1] and [F2] give $\sum_{m = 0}^{n - 1} ζ^{m} = 0$ . More generally, if $k \equiv 0 (m o d n)$ then $\sum_{m = 0}^{n - 1} ζ^{k m} = n$ ; if $0 < k < n$ , the element $η = ζ^{k}$ has order $n / d$ , where $d = \gcd (k, n)$ , so $\sum_{m = 0}^{n - 1} ζ^{k m} = d \sum_{r = 0}^{n / d - 1} η^{r} .$ Here $n / d > 1$ , and the inner sum vanishes by the same Vieta argument applied to $t^{n / d} - 1$ . Thus $\sum_{m = 0}^{n - 1} ζ^{k m}$ is $n$ when $k \equiv 0 (m o d n)$ and $0$ otherwise.

F2A1givenalgebra

2.1

Row orthogonality: by [F3] and step 1.1, $δ_{k l} = ⟨ χ_{k}, χ_{l} ⟩ = \frac{1}{n} \sum_{m} ζ^{k m} \overline{ζ^{l m}} = \frac{1}{n} \sum_{m} ζ^{(k - l) m}$ , and the last sum is $n$ when $k \equiv l (m o d n)$ and $0$ otherwise by step 1.2, which matches $δ_{k l}$ .

F3step 1.1step 1.2algebra

2.2

Column orthogonality: by [F4] and step 1.1, $\sum_{k} χ_{k} (g^{m}) \overline{χ_{k} (g^{m^{'}})} = \sum_{k} ζ^{k (m - m^{'})}$ , which is $n$ when $m \equiv m^{'} (m o d n)$ and $0$ otherwise by step 1.2; since $G$ is abelian every centralizer is $G$ , so this is exactly $∣ C_{G} (g) ∣ δ_{m m^{'}}$ .

F4step 1.1step 1.2algebra

3.1

All $n$ degrees are $1$ , so $\sum_{k} n_{k}^{2} = n = ∣ G ∣$ .

F1step 1.1algebra∎

Depends on

Used by

FALSE: every complex class function with self-inner-product 1 is a character False statement

Dependency tree · two levels

19 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Pavel Etingof et al., Introduction to Representation Theory, Section 3.3 (standard reference, not scraped)
Peter Webb, A Course in Finite Group Representation Theory, Section 4.1 (standard reference, not scraped)

The character table of a finite cyclic group over CC