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Brauer Induction and Elementary Subgroups — Examples

1 · Prerequisites

2 · Summary

These examples distinguish direct from nontrivial semidirect products, give an integral $S_{3}$ calculation, and show why cyclic subgroups cannot replace elementary ones integrally.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Small elementary and hyperelementary groups

Example

For distinct primes $p, q$ , $C_{p q} ≅ C_{q} \times C_{p}$ is $p$ -elementary. For odd $p$ , $D_{2 p} = C_{p} ⋊ C_{2}$ is $2$ -hyperelementary and is $2$ -elementary exactly when the action is trivial. Thus $S_{3} = C_{3} ⋊ C_{2}$ is $2$ -hyperelementary but not $2$ -elementary. Likewise, the nontrivial $C_{7} ⋊ C_{3}$ is $3$ -hyperelementary but not $3$ -elementary.

Facts & Assumptions

[F1]

The cited prerequisite is $p$ -elementary and $p$ -hyperelementary finite groups.

Verification

Given: the displayed semidirect products use their indicated conjugation actions.

1.1

In each case the first factor is cyclic of order prime to the displayed prime and the second is a $p$ -group.

F1given

2.1

Directness is equivalent to trivial conjugation; the reflections in $S_{3}$ invert $C_{3}$ , and the chosen action of $C_{3}$ on $C_{7}$ is nontrivial, establishing the two non-elementary assertions. ∎

step 1.1

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Brauer induction for $S_{3}$

Example

Let $T ≅ C_{2}$ and $A ≅ C_{3}$ in $S_{3}$ , let $ϵ$ be the sign character, and let $ρ$ be the degree-two irreducible. If $ω$ is either nontrivial linear character of $A$ , then $ρ = {Ind}_{A}^{S_{3}} ω,$ $1_{S_{3}} = {Ind}_{T}^{S_{3}} 1_{T} - {Ind}_{A}^{S_{3}} ω, ϵ = {Ind}_{T}^{S_{3}} {sgn}_{T} - {Ind}_{A}^{S_{3}} ω .$

Facts & Assumptions

[F1]

The cited prerequisite is Frobenius' formula for the character of an induced representation.

Verification

Given: the values of $1, ϵ, ρ$ on the classes $1, (12), (123)$ .

1.1

Frobenius' formula gives ${Ind}_{T} 1_{T} = 1 + ρ$ , ${Ind}_{T} {sgn}_{T} = ϵ + ρ$ , and ${Ind}_{A} ω = ρ$ by evaluating on those three classes.

F1given

2.1

Subtract the final equality from the first two. Both $T$ and $A$ are elementary, so these are integral Brauer-induction expressions. ∎

step 1.1

ExampleConstruction: Literature-sourcedVerification: Literature-sourcedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Trivial factors in an elementary group

Example

For every prime $p$ , every finite $p$ -group is $p$ -elementary by taking $C = 1$ ; every cyclic group of order prime to $p$ is $p$ -elementary by taking $P = 1$ ; and $1 = 1 \times 1$ is $p$ -elementary and $p$ -hyperelementary.

Facts & Assumptions

[F1]

The cited prerequisite is $p$ -elementary and $p$ -hyperelementary finite groups.

Verification

Given: the trivial group is cyclic and has order $1$ .

1.1

The order $1$ is prime to every prime, and it is also $p^{0}$ .

F1given

2.1

Therefore each displayed choice satisfies both factor conditions in the definition, including the simultaneous trivial-factor case. ∎

step 1.1

CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedjudge pass (gpt-5.6-terra)audited 2026-09-06Open item page →

Cyclic subgroups do not suffice for integral induction

Statement refuted

Every virtual character is an integral linear combination of characters induced from cyclic subgroups.

Facts & Assumptions

[F1]

The cited prerequisite is Frobenius' formula for the character of an induced representation.

Counterexample

Take $G = A_{5}$ . The trivial character $1_{G}$ is not in the integral cyclic induction subgroup.

Given: cyclic subgroups of $A_{5}$ have orders $1, 2, 3,$ or $5$ .

1.1

If $K \leq A_{5}$ is cyclic and $λ$ is linear, then ${Ind}_{K}^{A_{5}} λ (1) = [A_{5} : K]$ , which is respectively $60, 30, 20,$ or $12$ .

F1given

2.1

Every integral combination of such induced characters has even degree at $1$ , whereas $1_{G} (1) = 1$ . Thus $1_{G}$ cannot be such a combination. ∎

step 1.1

Brauer Induction and Elementary Subgroups — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Small elementary and hyperelementary groups

Example

Facts & Assumptions

Verification

Brauer induction for S3S3​

Example

Facts & Assumptions

Verification

Trivial factors in an elementary group

Example

Facts & Assumptions

Verification

Cyclic subgroups do not suffice for integral induction

Statement refuted

Facts & Assumptions

Counterexample

Sources

Brauer induction for $S_{3}$