ExampleConstruction: AI-adaptedVerification: AI-generatedverified 2026-09-26 (gpt-6-sol)

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Inducing the trivial character of a subgroup of order two in $S_{3}$ gives $1$ plus an irreducible degree-two character

Example

Let $H = ⟨ (12) ⟩ \leq S_{3}$ . Then ${Ind}_{H}^{S_{3}} 1_{H}$ is the permutation character on the three left cosets of $H$ , so it has values $(3, 1, 0)$ on the class types $e$ , transpositions, and $3$ -cycles. Subtracting the trivial character gives the irreducible degree-two character $(2, 0, - 1)$ .

Facts & Assumptions

Given: The subgroup $H = ⟨ (12) ⟩ \leq S_{3}$ and its trivial character $1_{H}$ .

[F1]

Inducing the trivial character gives the permutation representation on $S_{3} / H$ (Inducing the trivial representation gives the permutation representation on $G / H$ ).

[F2]

The character of a permutation representation counts fixed points (The character of a permutation representation counts fixed points).

[F3]

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

Verification

technique · direct

1.1F1F2given

By [F1] and [F2], the induced character counts fixed cosets of the left action on the three cosets of $H$ . The identity fixes all three cosets, a transposition fixes exactly one coset, and a $3$ -cycle fixes none, so ${Ind}_{H}^{S_{3}} 1_{H} = (3, 1, 0)$ .

2.1F3step 1.1algebra

In the coset permutation module $C^{3}$ , the constant vectors form an invariant trivial line and the coordinate-sum-zero plane is an invariant complement. Hence subtracting the trivial character $(1, 1, 1)$ gives the genuine character $(2, 0, - 1)$ of that plane. Its self-inner-product is $(1 / 6) (2^{2} + 3 \cdot 0^{2} + 2 \cdot (- 1)^{2}) = 1$ , so [F3] makes it irreducible; its value at the identity is $2$ , so it has degree $2$ .

3.1step 2.1∎

Therefore ${Ind}_{H}^{S_{3}} 1_{H} = 1 + χ_{2}$ with $χ_{2}$ irreducible of degree $2$ .

Depends on

Used by

Dependency tree · two levels

9 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 4.11 (standard reference, not scraped)

Inducing the trivial character of a subgroup of order two in S3S3​ gives 11 plus an irreducible degree-two character