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The character table of $Q_{8}$

Example

The character table of $Q_{8} = {1, - 1, i, - i, j, - j, k, - k}$ , with columns $1$ , $- 1$ , $i$ , $j$ , $k$ (sizes $1$ , $1$ , $2$ , $2$ , $2$ ), is

$\begin{array}{cccccc} 1 & - 1 & i & j & k \\ χ_{1} & 1 & 1 & 1 & 1 & 1 \\ χ_{2} & 1 & 1 & 1 & - 1 & - 1 \\ χ_{3} & 1 & 1 & - 1 & 1 & - 1 \\ χ_{4} & 1 & 1 & - 1 & - 1 & 1 \\ ψ & 2 & - 2 & 0 & 0 & 0 \end{array}$

Both orthogonality relations hold.

Facts & Assumptions

Given: The quaternion group $Q_{8} = {1, - 1, i, - i, j, - j, k, - k} \subseteq H^{\times}$ .

[F1]

$Q_{8}$ has order $8$ and $- 1$ is its only element of order $2$ ( $Q_{8}$ is a subgroup of $H^{\times}$ with eight elements, and $- 1$ is its only element of order $2$ ).

[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]

The abelianization $G^{a b} = G / G^{'}$ is abelian and homomorphisms to abelian groups factor uniquely through it (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]

In $Q_{8}$ , $i^{2} = j^{2} = k^{2} = - 1$ and $i j = k$ ; the commutator $[i, j] = i j i^{- 1} j^{- 1}$ equals $- 1$ ; the conjugates of $i$ are $\pm i$ , of $j$ are $\pm j$ , and of $k$ are $\pm k$ , so the classes are ${1}$ , ${- 1}$ , ${\pm i}$ , ${\pm j}$ , ${\pm k}$ .

[A2]

A homomorphism $φ : Q_{8} \to C^{\times}$ satisfies $φ (i)^{2} = φ (j)^{2} = φ (i) φ (j) φ (k) = 1$ , and every assignment $i \mapsto \pm 1$ , $j \mapsto \pm 1$ extends to one.

Verification

technique · direct

1.1

By [A1] the conjugacy classes of $Q_{8}$ are ${1}$ , ${- 1}$ , ${\pm i}$ , ${\pm j}$ , ${\pm k}$ , of sizes $1$ , $1$ , $2$ , $2$ , $2$ .

A1given

1.2

Since $[i, j] = - 1$ by [A1], $- 1 \in Q_{8}^{'}$ ; the quotient $Q_{8} / {\pm 1}$ has order $4$ by [F1] and every element of it squares to $1$ , so it is abelian, and [F3] gives $Q_{8}^{'} = {\pm 1}$ with abelianization of order $4$ .

A1F1F3given

2.1

By [F2] and [F3], the degree-one characters are the homomorphisms factoring through the abelianization of step 1.2; by [A2] they are exactly the four assignments $i \mapsto ε$ , $j \mapsto δ$ for $ε, δ \in {\pm 1}$ , with $k = i j$ sent to $ε δ$ . Their values are the first four rows of the table.

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 $- 1$ is orthogonal to the column of $1$ : $1 + 1 + 1 + 1 + 2 \overline{ψ (- 1)} = 0$ , so $ψ (- 1) = - 2$ . The column of $i$ is orthogonal to the column of $1$ : $1 + 1 - 1 - 1 + 2 \overline{ψ (i)} = 0$ , so $ψ (i) = 0$ ; likewise $ψ (j) = 0$ and $ψ (k) = 0$ .

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 $8$ , $8$ , $4$ , $4$ , $4$ , equal to the centralizer sizes $8$ , $8$ , $4$ , $4$ , $4$ , and distinct columns are orthogonal.

F6step 1.1step 4.1algebra∎

Depends on

Used by

FALSE: distinct irreducible complex characters of a finite group have distinct degrees False statement
FALSE: nonisomorphic finite groups always have different character tables False statement

Dependency tree · two levels

36 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, Example 3.15 (standard reference, not scraped)

The character table of Q8Q8​