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Inducing the trivial character of a subgroup of order two in gives plus an irreducible degree-two character
Example
Let . Then is the permutation character on the three left cosets of , so it has values on the class types , transpositions, and -cycles. Subtracting the trivial character gives the irreducible degree-two character .
Facts & Assumptions
Given: The subgroup and its trivial character .
Inducing the trivial character gives the permutation representation on (Inducing the trivial representation gives the permutation representation on ).
The character of a permutation representation counts fixed points (The character of a permutation representation counts fixed points).
A complex character is irreducible if and only if its self-inner-product is (A complex character is irreducible if and only if its self-inner-product is ).
Verification
By [F1] and [F2], the induced character counts fixed cosets of the left action on the three cosets of . The identity fixes all three cosets, a transposition fixes exactly one coset, and a -cycle fixes none, so .
Subtracting the trivial character gives the class function . Its self-inner-product is , so [F3] makes it irreducible; its value at the identity is , so it has degree .
Therefore with irreducible of degree .
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)