Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-08-30
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Inducing the trivial character of a subgroup of order two in S3 gives 1 plus an irreducible degree-two character

Example

Let H=(12)S3. Then IndHS31H 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)S3 and its trivial character 1H.

[F1]

Inducing the trivial character gives the permutation representation on S3/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
1.1

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 IndHS31H=(3,1,0).

F1F2given
2.1

Subtracting the trivial character (1,1,1) gives the class function (2,0,1). Its self-inner-product is (1/6)(22+302+2(1)2)=1, so [F3] makes it irreducible; its value at the identity is 2, so it has degree 2.

F3step 1.1algebra
3.1

Therefore IndHS31H=1+χ2 with χ2 irreducible of degree 2.

step 2.1

Depends on

Used by

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