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M-groups are solvable (Taketa)

Statement

Let $G$ be a finite $M$ -group (Monomial representations, monomial characters, and M-groups) and let $1 = f_{1} < f_{2} < \dots < f_{r}$ be the distinct degrees of its irreducible complex characters, so that $r = ∣ c d (G) ∣$ . Then for every $k$ with $1 \leq k \leq r$ and every irreducible complex character $χ$ of $G$ with $χ (1) = f_{k}$ the $k$ -th derived subgroup of $G$ satisfies $G^{(k)} \leq \ker χ$ (The derived series, solvable groups, and derived length, The kernel of a complex character). In particular $G^{(r)} = 1$ : every finite $M$ -group is solvable, and its derived length satisfies $d l (G) \leq r = ∣ c d (G) ∣$ .

Facts & Assumptions

Given: A finite $M$ -group $G$ with distinct irreducible character degrees $1 = f_{1} < f_{2} < \dots < f_{r}$ , an index $k \in {1, \dots, r}$ , and an irreducible complex character $χ$ of $G$ with $χ (1) = f_{k}$ .

[F1]

Every irreducible complex character of an $M$ -group is monomial: $χ = {Ind}_{H}^{G} λ$ for some subgroup $H \leq G$ and some linear character $λ$ of $H$ . (Monomial representations, monomial characters, and M-groups).

[F2]

The derived series is $G^{(0)} = G$ , $G^{(m + 1)} = [G^{(m)}, G^{(m)}]$ , hence $G^{(m + 1)} \leq G^{(m)}$ ; $G$ is solvable when $G^{(n)} = 1$ for some $n$ , and the derived length is the least such $n$ . (The derived series, solvable groups, and derived length).

[F3]

$G^{'}$ is characteristic, hence normal, in $G$ , and every homomorphism from $G$ into an abelian group has $G^{'}$ in its kernel; characteristic subgroups are normal and characteristicness is transitive. (The derived subgroup is characteristic and the abelianization is universal, Characteristic subgroups are normal, and characteristicity is transitive).

[F4]

$⟨ {Ind}_{H}^{G} ψ, η ⟩_{G} = ⟨ ψ, {Res}_{H}^{G} η ⟩_{H}$ for complex characters $ψ$ of $H$ and $η$ of $G$ . (Frobenius reciprocity for complex characters).

[F5]

The irreducible complex characters of a finite group form an orthonormal basis of its class functions, and for an honest character $π$ the multiplicities in $π = \sum_{θ} m_{θ} θ$ satisfy $m_{θ} = ⟨ π, θ ⟩ = \dim {Hom}_{G} (V_{π}, V_{θ}) \in Z_{\geq 0}$ . (The irreducible complex characters form an orthonormal basis of $c f (G)$ , The class-function inner product $⟨ χ_{V}, χ_{W} ⟩$ equals $\dim {Hom}_{G} (W, V)$ ).

[F6]

$\dim_{C} {Ind}_{H}^{G} W = [G : H] \dim_{C} W$ for a finite-dimensional complex $H$ -representation $W$ , and the character of an induced module is the induced character. (The dimension of an induced finite-dimensional representation is $[G : H] \dim W$ , The induced character ${Ind}_{H}^{G} χ$ of a complex character, The induced $R$ -linear $G$ -module ${Ind}_{H}^{G} W$ as $H$ -covariant functions on $G$ ).

[F7]

For a nonzero finite-dimensional complex $H$ -representation $W$ one has $\ker ({Ind}_{H}^{G} W) = ⋂_{g \in G} g (\ker W) g^{- 1}$ , and for $W = 1_{H}$ this kernel is ${Core}_{G} (H) \leq H$ . (Kernel of an induced representation is the intersection of the conjugates of the kernel of the inducing representation).

[F8]

The intersection of the kernels of all irreducible complex characters of a finite group $G$ is trivial, and the trivial character $1_{G}$ is an irreducible character of degree $1$ . (The normal subgroups of a finite group are exactly the intersections of kernels of irreducible complex characters, An irreducible complex character, Subrepresentations, direct sums of representations, and irreducibility).

[A1]

The kernel of a direct sum of representations is the intersection of the kernels of its summands; hence a subgroup lying in the kernel of every irreducible constituent of an honest character lies in the kernel of that character.

Proof

technique · induction on $k$

1.1

Base case $k = 1$ : then $χ$ is a linear character, that is, a homomorphism $G \to C^{\times}$ into the abelian group $C^{\times}$ , so $G^{(1)} = G^{'} \leq \ker χ$ by [F3].

F3givenbase

1.2

Induction hypothesis: for every $j$ with $1 \leq j < k$ and every irreducible character $χ^{'}$ of $G$ with $χ^{'} (1) = f_{j}$ , one has $G^{(j)} \leq \ker χ^{'}$ .

1.3

Assume now $k \geq 2$ . By [F1] the irreducible character $χ$ of degree $f_{k}$ is monomial, say $χ = {Ind}_{H}^{G} λ$ with $H \leq G$ and $λ$ a linear character of $H$ . By [F6], $f_{k} = χ (1) = [G : H] \cdot 1 = [G : H] \geq 2$ , so $H$ is a proper subgroup. Let $π = {Ind}_{H}^{G} 1_{H}$ be the induced trivial character; again $π (1) = [G : H] = f_{k}$ , and by Frobenius reciprocity [F4], $⟨ π, 1_{G} ⟩_{G} = ⟨ 1_{H}, 1_{H} ⟩_{H} = 1$ .

F1F4F6given

2.1

Writing the honest character $π$ in the orthonormal basis of irreducible characters as $π = \sum_{θ} m_{θ} θ$ with multiplicities $m_{θ} = ⟨ π, θ ⟩ \in Z_{\geq 0}$ by [F5], step 1.3 gives $m_{1_{G}} = 1$ ; evaluating at $1 \in G$ gives $f_{k} = π (1) = \sum_{θ} m_{θ} θ (1)$ , so $\sum_{θ \neq 1_{G}} m_{θ} θ (1) = f_{k} - 1$ and every constituent $θ \neq 1_{G}$ of $π$ has degree $θ (1) \leq f_{k} - 1 < f_{k}$ , hence $θ (1) = f_{j}$ for a unique index $j < k$ .

F5step 1.3

3.1

For every constituent $θ \neq 1_{G}$ of $π$ , step 2.1 gives $θ (1) = f_{j}$ with $j \leq k - 1$ , so the induction hypothesis of step 1.2 yields $G^{(j)} \leq \ker θ$ , and since $j \leq k - 1$ the series is descending by [F2], so $G^{(k - 1)} \leq G^{(j)} \leq \ker θ$ . The trivial constituent satisfies $G^{(k - 1)} \leq G = \ker 1_{G}$ . By [A1] the kernel of $π$ is the intersection of the kernels of its constituents, so $G^{(k - 1)} \leq \ker π$ .

A1F2step 1.2step 2.1

4.1

The kernel formula [F7] for the induced trivial representation gives $\ker π = {Core}_{G} (H) \leq H$ , so $G^{(k - 1)} \leq H$ by step 3.1.

F7step 3.1

5.1

Therefore $G^{(k)} = [G^{(k - 1)}, G^{(k - 1)}] \leq [H, H] = H^{'}$ by [F2], and $H^{'} \leq \ker λ$ because $λ$ is a homomorphism into the abelian group $C^{\times}$ , by [F3].

F2F3step 4.1

6.1

Each term of the derived series is characteristic, hence normal, in $G$ by [F2] and [F3]; so $g G^{(k)} g^{- 1} = G^{(k)} \leq g (\ker λ) g^{- 1}$ for every $g \in G$ by step 5.1. Hence $G^{(k)} \leq ⋂_{g \in G} g (\ker λ) g^{- 1} = \ker χ$ by the kernel formula [F7] applied to $χ = {Ind}_{H}^{G} λ$ , which completes the induction step.

F3F7step 5.1

7.1

By step 1.1 and step 6.1 the assertion $G^{(k)} \leq \ker χ$ holds for every $k \in {1, \dots, r}$ and every irreducible $χ$ with $χ (1) = f_{k}$ . Taking $k = r$ and using that the series is descending [F2], $G^{(r)} \leq \ker χ$ for every irreducible character $χ$ of $G$ , so $G^{(r)} \leq ⋂_{χ \in Irr (G)} \ker χ = 1$ by [F8]. Thus $G^{(r)} = 1$ , $G$ is solvable with $d l (G) \leq r = ∣ c d (G) ∣$ . ∎

F2F8step 1.1step 6.1discharge-induction: step 1.2

Depends on

Used by

Implications and limits for M-groups Remark

Dependency tree · two levels

48 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

SLMath, Character Theory of Finite Groups, Chapter 9 — slides 391–404 (Taketa's theorem, ordered character-degree induction) (standard reference, not scraped)