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The character table of $Dih (C_{4})$

Example

For $G = Dih (C_{4}) = ⟨ r, s : r^{4} = s^{2} = 1, s r s^{- 1} = r^{- 1} ⟩$ , the conjugacy classes are represented by $1$ , $r^{2}$ , $r$ , $s$ , $s r$ (sizes $1$ , $1$ , $2$ , $2$ , $2$ ), and the character table is

$\begin{array}{cccccc} 1 & r^{2} & r & s & s r \\ χ_{1} & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & 1 & - 1 & 1 & - 1 \\ χ_{3} & 1 & 1 & 1 & - 1 & - 1 \\ χ_{4} & 1 & 1 & - 1 & - 1 & 1 \\ ψ & 2 & - 2 & 0 & 0 & 0 \end{array}$

Both orthogonality relations hold.

Facts & Assumptions

Given: The dihedral group $G = Dih (C_{4}) = C_{4} ⋊ C_{2}$ with $C_{4} = ⟨ r ⟩$ and $C_{2} = ⟨ s ⟩$ .

[F1]

$G$ has order $8$ , presentation $r^{4} = s^{2} = 1$ , $s r s^{- 1} = r^{- 1}$ , and every element has the unique form $r^{i}$ or $r^{i} s$ ( $Dih (C_{n}) = C_{n} ⋊ C_{2}$ with inversion action has order $2 n$ and the dihedral relations).

[F2]

Degree-one representations correspond to homomorphisms $G \to C^{\times}$ (Equivalence classes of degree-one representations are exactly homomorphisms $G \to k^{\times}$ ; equivalently they factor through $G / G^{'}$ , and they form an abelian group).

[F3]

Homomorphisms to abelian groups factor uniquely through the abelianization $G^{a b} = G / G^{'}$ (The derived subgroup is characteristic and the abelianization is universal).

[F4]

The squared degrees of the irreducible characters sum to $∣ G ∣$ (The regular character gives a second proof of the sum-of-squares formula).

[F5]

Row orthogonality: irreducible characters are orthonormal (The first orthogonality relation for irreducible complex characters).

[F6]

Column orthogonality: distinct columns are orthogonal and a column has squared norm the centralizer size (The second orthogonality relation for irreducible complex characters).

[A1]

From [F1]'s relations: $r^{2}$ is central; $s r s^{- 1} = r^{- 1}$ conjugates $r$ to $r^{3}$ ; $r s r^{- 1} = s r^{2}$ conjugates $s$ to $s r^{2}$ ; and $r (s r) r^{- 1} = s r^{3}$ conjugates $s r$ to $s r^{3}$ . Hence the classes are ${1}$ , ${r^{2}}$ , ${r, r^{3}}$ , ${s, s r^{2}}$ , ${s r, s r^{3}}$ .

[A2]

The commutator $[r, s] = r s r^{- 1} s^{- 1}$ equals $r^{2}$ , the quotient $G / ⟨ r^{2} ⟩$ has order $4$ and is abelian, and every assignment $r \mapsto ε$ , $s \mapsto δ$ with $ε, δ \in {\pm 1}$ extends to a homomorphism $G \to C^{\times}$ .

Verification

technique · direct

1.1

By [A1] the classes are ${1}$ , ${r^{2}}$ , ${r, r^{3}}$ , ${s, s r^{2}}$ , ${s r, s r^{3}}$ , of sizes $1$ , $1$ , $2$ , $2$ , $2$ .

A1given

1.2

By [A2], $r^{2} \in G^{'}$ , and the quotient $G / ⟨ r^{2} ⟩$ of order $4$ is abelian; by [F3] the homomorphisms to abelian groups factor through it, so $G^{'} = ⟨ r^{2} ⟩$ .

A2F3given

2.1

By [F2] and [F3], the degree-one characters are the homomorphisms factoring through the abelianization of step 1.2, namely the four assignments of [A2]: $χ (r) = ε$ , $χ (s) = δ$ . Their values are the first four rows of the table (with $χ (s r) = ε δ$ and $χ (r^{2}) = 1$ ).

F2F3A2step 1.2given

3.1

By [F4], the remaining irreducible degree $d$ satisfies $1 + 1 + 1 + 1 + d^{2} = 8$ , so $d = 2$ .

F4step 2.1algebra

4.1

By [F6], the column of $r^{2}$ is orthogonal to the column of $1$ : $1 + 1 + 1 + 1 + 2 \overline{ψ (r^{2})} = 0$ , so $ψ (r^{2}) = - 2$ . The columns of $r$ , $s$ , and $s r$ are orthogonal to the column of $1$ , giving $1 - 1 + 1 - 1 + 2 \overline{ψ (r)} = 0$ , $1 + 1 - 1 - 1 + 2 \overline{ψ (s)} = 0$ , and $1 - 1 - 1 + 1 + 2 \overline{ψ (s r)} = 0$ , so $ψ (r) = ψ (s) = ψ (s r) = 0$ .

F6step 2.1step 3.1algebra

5.1

The five rows form the displayed table. By [F5], $⟨ ψ, ψ ⟩ = \frac{1}{8} (4 + 4) = 1$ and $ψ$ is orthogonal to each degree-one row, so the table is complete; the degrees sum correctly.

F5step 2.1step 4.1algebra

6.1

By [F6], the column squared norms are $8$ , $8$ , $4$ , $4$ , $4$ ; the norm $8$ of the first column is the group order $∣ G ∣ = 8$ of [F1], equal to the centralizer size of the identity, and the remaining norms are the centralizer sizes. Distinct columns are orthogonal.

F6F1step 1.1step 4.1algebra∎

Depends on

Used by

FALSE: nonisomorphic finite groups always have different character tables False statement

Dependency tree · two levels

29 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

Shani Meynet and Robert Moscrop, McKay quivers and decomposition, Section 4.1 (standard reference, not scraped)

The character table of Dih⁡(C4)Dih(C4​)