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The character table of $S_{4}$ and the normal subgroups it reveals

Example

The character table of $S_{4}$ , with columns $1$ , $(12)$ , $(123)$ , $(1234)$ , $(12) (34)$ (sizes $1$ , $6$ , $8$ , $6$ , $3$ ), is

$\begin{array}{cccccc} 1 & (12) & (123) & (1234) & (12) (34) \\ 1 & 1 & 1 & 1 & 1 & 1 \\ ε & 1 & - 1 & 1 & - 1 & 1 \\ χ_{3} & 3 & 1 & 0 & - 1 & - 1 \\ ε χ_{3} & 3 & - 1 & 0 & 1 & - 1 \\ χ_{2} & 2 & 0 & - 1 & 0 & 2 \end{array}$

The normal subgroups of $S_{4}$ are exactly ${1}$ , $V_{4}$ , $A_{4}$ , and $S_{4}$ .

Facts & Assumptions

Given: The group $S_{4}$ with class representatives $1$ , $(12)$ , $(123)$ , $(1234)$ , $(12) (34)$ .

[F1]

$S_{4}$ has five conjugacy classes, of sizes $1$ , $6$ , $8$ , $6$ , $3$ ( $S_{4}$ has five conjugacy classes of sizes $1$ , $6$ , $8$ , $6$ , and $3$ ).

[F2]

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).

[F3]

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

[F4]

Characters multiply on tensor products (Characters add on direct sums, multiply on tensor products, and conjugate on duals).

[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 $1$ ).

[F6]

The irreducible complex characters form an orthonormal basis of the class functions (The irreducible complex characters form an orthonormal basis of $c f (G)$ ).

[F7]

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

[F8]

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

[F9]

Normal subgroups are exactly intersections of kernels of irreducible characters (The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters).

[A1]

The kernel of a character of a representation is ${g : χ (g) = χ (1)}$ .

[A2]

The kernel of a direct sum of representations is the intersection of the kernels of the summands, and a permutation of a set of four letters with exactly two fixed points is a transposition.

[A3]

For class functions on $S_{4}$ , the standard inner product is $⟨ f, h ⟩ = \frac{1}{24} (f (1) \overline{h (1)} + 6 f ((12)) \overline{h ((12))} + 8 f ((123)) \overline{h ((123))} + 6 f ((1234)) \overline{h ((1234))} + 3 f ((12) (34)) \overline{h ((12) (34))}) .$

Verification

technique · direct

1.1

By [F2] the sign row is $(1, - 1, 1, - 1, 1)$ on the representatives: values $+ 1$ on even permutations and $- 1$ on odd ones. By [F3], the permutation character of $S_{4}$ has values $4$ , $2$ , $1$ , $0$ , $0$ (fixed points), so the standard character has values $3$ , $1$ , $0$ , $- 1$ , $- 1$ .

F2F3given

2.1

By [F4], the sign twist $ε χ_{3}$ has values $(3, - 1, 0, 1, - 1)$ .

F4step 1.1algebra

3.1

The trivial and sign characters are one-dimensional, hence irreducible. Using [A3] and the values from steps 1.1 and 2.1 gives $⟨ χ_{3}, χ_{3} ⟩ = \frac{1}{24} (9 + 6 + 0 + 6 + 3) = 1, ⟨ χ_{3}, 1 ⟩ = \frac{1}{24} (3 + 6 + 0 - 6 - 3) = 0.$ Because $∣ ε (g) ∣ = 1$ for every $g \in S_{4}$ , the same computation gives $⟨ ε χ_{3}, ε χ_{3} ⟩ = 1$ , and multiplying one factor by $ε$ preserves orthogonality with $1$ and with $ε$ . Therefore $1$ , $ε$ , $χ_{3}$ , and $ε χ_{3}$ are four pairwise orthogonal irreducible characters, the last two by [F5].

F5step 1.1step 2.1A3algebra

4.1

By [F6], irreducible characters form an orthonormal basis of the $5$ -dimensional class-function space of $S_{4}$ , so after the four orthogonal irreducibles of step 3.1 there is exactly one remaining irreducible character, call it $χ_{2}$ . By [F7], its degree $d$ satisfies $1 + 1 + 9 + 9 + d^{2} = 24$ , so $d = 2$ .

F6F7F1step 3.1algebra

5.1

By [F8], each column is orthogonal to the first column $(1, 1, 3, 3, 2)$ , so reading off the first four entries gives the fifth entry: at $(12)$ , $1 - 1 + 3 - 3 + 2 \overline{χ_{2} ((12))} = 0$ , so $χ_{2} ((12)) = 0$ ; at $(123)$ , $1 + 1 + 0 + 0 + 2 \overline{χ_{2} ((123))} = 0$ , so $χ_{2} ((123)) = - 1$ ; at $(1234)$ , $1 - 1 - 3 + 3 + 2 \overline{χ_{2} ((1234))} = 0$ , so $χ_{2} ((1234)) = 0$ ; at $(12) (34)$ , $1 + 1 - 3 - 3 + 2 \overline{χ_{2} ((12) (34))} = 0$ , so $χ_{2} ((12) (34)) = 2$ .

F8step 1.1step 4.1algebra

6.1

The five rows from steps 1.1, 2.1, and 5.1 now form an orthonormal basis of the class functions: step 3.1 already handles the first four rows, and $⟨ χ_{2}, χ_{2} ⟩ = \frac{1}{24} (4 + 0 + 8 + 0 + 12) = 1.$ Since $χ_{2}$ is orthogonal to the first four rows by construction from step 5.1, this is the displayed character table.

F6step 3.1step 5.1A3algebra

6.2

The kernels, by [A1]: $\ker 1 = S_{4}$ ; $\ker ε = A_{4}$ (the even permutations); $\ker χ_{3} = {1}$ , because $χ_{3} (σ) = 3$ means $fix (σ) = 4$ , i.e. $σ = 1$ ; $\ker ε χ_{3} = {1}$ ; $\ker χ_{2} = V_{4}$ , because $χ_{2} (σ) = 2$ exactly at the identity and the double transpositions, and [A2] identifies the class with exactly two fixed points as the transpositions.

A1A2F2F3step 5.1algebra

7.1

By [F9], the normal subgroups are exactly the intersections of the five kernels of step 6.2; by [A2] these intersections are ${1}$ , $V_{4}$ , $A_{4}$ , and $S_{4}$ .

F9A2step 6.2algebra∎

Depends on

Used by

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Sources

Peter Webb, A Course in Finite Group Representation Theory, Example 3.3.5 (standard reference, not scraped)
Pavel Etingof et al., Introduction to Representation Theory, Example 3.15 (standard reference, not scraped)

The character table of S4S4​ and the normal subgroups it reveals