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$S_{3}$ has three irreducible complex characters of degrees $1$ , $1$ , and $2$

Example

The symmetric group $S_{3}$ has exactly three irreducible complex characters: the trivial character $1$ , the sign character $sgn$ , and the standard character $χ_{s t d}$ ; their degrees are $1$ , $1$ , and $2$ .

Facts & Assumptions

Given: The symmetric group $S_{3}$ acting on ${1, 2, 3}$ .

[F1]

The conjugacy classes of $S_{n}$ are indexed by cycle types (The conjugacy classes of $S_{n}$ are indexed by the tuples $(c_{1}, \dots, c_{n})$ with $\sum k c_{k} = n$ ).

[F2]

The degrees of all irreducible characters satisfy $\sum_{i} n_{i}^{2} = ∣ G ∣$ (The regular character gives a second proof of the sum-of-squares formula).

[F3]

The sign representation is the one-dimensional representation in which $σ$ acts by $sgn (σ)$ (The sign representation of $S_{n}$ and the restriction ${Res}_{H}^{G} (V)$ of a representation to a subgroup).

[F4]

The standard character of $S_{n}$ is $χ_{s t d} (σ) = fix (σ) - 1$ (The standard representation of $S_{n}$ has character equal to the number of fixed points minus $1$ ).

[F5]

A complex character is irreducible exactly when its self-inner-product is $1$ (A complex character is irreducible if and only if its self-inner-product is $1$ ).

[A1]

The standard inner product of class functions on a group of order $n$ is $⟨ f, h ⟩ = \frac{1}{n} \sum_{g} f (g) \overline{h (g)}$ .

Verification

technique · direct

1.1

By [F1] the cycle types of $S_{3}$ are the identity type, the type of a transposition, and the type of a $3$ -cycle, so $S_{3}$ has three conjugacy classes; a direct count of each type gives class sizes $1$ , $3$ , and $2$ .

F1given

1.2

By [F3] the trivial and sign characters are distinct one-dimensional characters, hence irreducible: a one-dimensional space has no proper nonzero subspaces. Their degrees are $1$ and $1$ .

F3given

2.1

By [F4], the standard character has values $2$ , $0$ , and $- 1$ on the three classes of step 1.1 (fixed points $3$ , $1$ , and $0$ minus $1$ ). Using [A1], $⟨ χ_{s t d}, χ_{s t d} ⟩ = \frac{1}{6} (4 + 3 \cdot 0 + 2 \cdot 1) = 1$ , so by [F5] the standard character is irreducible, of degree $χ_{s t d} (1) = 2$ .

F4A1F5step 1.1algebra

3.1

The three irreducible characters of steps 1.2 and 2.1 have squared degrees $1 + 1 + 4 = 6 = ∣ S_{3} ∣$ . By [F2] the sum over all irreducible characters is also $6$ , so no further irreducible character exists.

F2step 1.2step 2.1algebra∎

Depends on

Used by

The character table of S₃ Example

Dependency tree · two levels

15 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

Peter Webb, A Course in Finite Group Representation Theory, Example 3.1.2 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Section 3.5 (standard reference, not scraped)

S3S3​ has three irreducible complex characters of degrees 11, 11, and 22