Monomial groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au
and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 20 May 2010

Monomial groups

A composition series of $G$ is a sequence $\{1\} = G_{0} \subseteq G_{1} \subseteq \dots \subseteq G_{n} = G$ with $G_{i - 1}$ normal in $G_{i} .$

A group is solvalbe if there exists a composition series of $G$ with $G_{i} / G_{i + 1}$ abelian.
A group is supersolvable if there exists a composition series with $G_{i}$ normal in $G$ and $G_{i} / G_{i - 1}$ cyclic.
A group is nilpotent if there exists a composition series of $G$ with $G_{i} / G_{i - 1} \subseteq Z (G / G_{i - 1}) .$ $nilpotent \Rightarrow supersolvable \Rightarrow solvable.$
A group is monomial if every irreducible representation of $G$ can be obtained via induction from a one dimensional representation of some subgroup.

Supersolvable groups are monomial.

Proof.

Assume that $G$ is supersolvable and let $M$ be a simple $G$ -module.

Case 1. Suppose that $K = \ker M \neq \{1\} .$ Then $M$ is a $G / K$ -module. $M = {Ind}_{K ⋊ {(G / K)}_{β}}^{K ⋊ {(G / K)}_{β}} (1_{K} \otimes M) = 1_{K} \otimes M .$ Since $|G / K| < |G|,$ by induction, $M = {Ind}_{H / K}^{G / K} (1_{ξ}),$ for some one dimensional representation $1_{ξ}$ of a subgroup $H / K .$

Case 2. If $K = \ker M = \{1\}$ then $M$ is a faithful representation of $G .$ Since $G$ is supersolvable, $G / Z (G)$ is supersolvable and has a composition series in which the first nontrivial term ${(G / Z (G))}_{1}$ is a cyclic subgroup of $G / Z (G) .$ The inverse image of ${(G / Z (G))}_{1}$ in $G$ is a normal abelian subgroup $A$ of $G$ which is not contained in the center of $G .$ Then $M = {Ind}_{A ⋊ {(I / A)}_{β}}^{A ⋊ {(G / A)}_{β}} (1_{ξ} \otimes M^{'}) = {Ind}_{I}^{G} (1_{ξ} \otimes M^{'}),$ where $I$ is the inertia group of $M$ and $1_{ξ} \otimes M$ is a simple $I$ module. Since $A$ is not contained in $Z (G)$ and $M$ is a faithful representation of $G$ , the group $A$ does not act on $M$ by scalars, but $A$ does act on $1_{ξ} \otimes M$ by scalars. So $1_{ξ} \otimes M^{'} \neq M$ and therefore $I \neq G .$ Thus, by induction, there is a one dimensional representation $1_{η}$ of a subgroup $H$ such that $1_{ξ} \otimes M^{'} = {Ind}_{H}^{I} (1_{η}) . So M = {Ind}_{I}^{G} ({Ind}_{H}^{I} (1_{η})) = {Ind}_{H}^{G} (1_{η}) . □$

Nilpotent groups are monomial.

Proof.

Let $M$ be an irreducible representation of $G .$ Let us assume that the theorem is proved for

groups of lower order (in the finite group case),
groups of lower dimension (in the Lie group case).

We need to show that

M

is obtained by induction from a one dimensional representation of a subgroup.

Let $ρ : G \to G L (M)$ be the homomorphism determined by $M .$ Let $K = \ker ρ .$ Then $G / K$ acts on $M$ and $M$ is an irreducible representation of $G / K .$ So, if,

$G / K$ has a lower order than $G$ (in the finite case),
or $G / K$ has lower dimension than $G$ (Lie case),

then

M ≅ {Ind}_{H / K}^{G / K} (1_{ξ}), as a G / K -module.

But then

M ≅ {Ind}_{H}^{G} (1_{ξ}),

since

{Ind}_{H}^{G} (1_{ξ})

and

{Ind}_{H / K}^{G / K} (1, ξ)

can be both viewed as functions on

G / H ≅ (G / K) / (H / K) .

So the theorem is proved for

K \neq \{1\} .

Now assume that $K = \{1\}$ , ie $ρ$ is injective. The last nontrivial term $𝒞^{k} (G)$ of the lower central series of $G$ is $Z (G) and 𝒞^{k - 1}$ is a subgroup of $G$ containing $Z (G) .$ Let $x \in 𝒞^{k - 1} (G)$ be such that $x \notin Z (G), x \in Z (𝒞^{k - 1} (G) / Z (G)) .$ Then let $\begin{array}{rcl} A & = & \{x^{k} z | k \in ℤ_{\geq 0}, z \in Z (G)\} & (finite case) \\ A & = & \{e^{t X} z | t \in ℝ, z \in Z (G)\}, & (Lie case) \end{array}$

so that

$A$ is cyclic (finite case)
$A$ is one dimensional (Lie case).

Then

A

is abelian since

x^{k_{1}} z_{1} x^{k_{2}} z_{2} = x^{k_{1} + k_{2}} z_{1} z_{2} = x^{k_{2}} z_{2} x^{k_{1}} z_{1}, for k_{i} \in ℤ_{\geq 0}, z_{i} \in Z (G) .

Let

L

be an irreducible

A

submodule of

M .

Then

\dim L = 1

(since

A

is abelian.) If

g \in G

then

g L

is a representation of

A

that looks just like

L

except

a (g l) = g (g^{- 1} a g l), for all l \in L .

Since

L

is a simple

A

module then so is

g L .

Now

\sum_{g \in G} g L is a G -submodule of M .

Since

M

is simple

M = \sum_{g \in G} g L .

Let

\tilde{A} = \{g \in G | g L ≅ L as A -modules\} .

Then

\tilde{L} = \sum_{\tilde{a} \in \tilde{A}} \tilde{a} L

is an

\tilde{A}

-module. Then

M = \sum_{g \in G} g L = \sum_{g_{i} \in G / \tilde{A}} g_{i} \tilde{L},

where

g_{i}

runs over a set of coset representatives of

G / \tilde{A} .

This is a decomposition of

M

as an

A

-module such that

every irreducible $A$ -submodule of $g_{i} \tilde{L}$ is isomorphic,
the irreducible $A$ -submodules of $g_{i} \tilde{L}$ and $g_{j} \tilde{L}$ are not isomorphic.

Since

A

acts on

M

by scalars,

A

acts on

\tilde{L}

by scalars. So

\tilde{L} \neq M

, since

ρ

is injective and

A \subseteq Z (G)

A \neq Z (G)

. So

\tilde{A} \neq G

and

M = {Ind}_{\tilde{A}}^{G} (\tilde{L}) .

Since

\tilde{A}

is smaller than

G

(since

G

is not abelian)

\tilde{L} ≅ {Ind}_{H}^{\tilde{A}} (1_{ξ}) .

M = {Ind}_{\tilde{A}}^{G} ({Ind}_{H}^{\tilde{A}} (1, ξ)) ≅ {Ind}_{H}^{G} (1, ξ) .

The theorem now follows from the fact that all representations of an abelian group are one dimensional.

□

$\{supersolvable groups\} \subseteq \{monomial groups\} \subseteq \{solvable groups\} .$

Exercise. Give examples to show that these inclusions are strict.

A Lie group is exponential is the map $𝔤 \overset{\exp}{\to} G$ is a diffeomorphism. $\{nilpotent Lie groups\} \subseteq \{exponential Lie groups\} \subseteq \{solvable Lie groups\} .$ Once again, the inclusions are strict.

A nilpotent Lie group is exponential.

		Proof.
	First note that if $X \in 𝔤$ then $e^{X} = 1 + X + \frac{X^{2}}{2!} + \dots$ is a finite sum, since $X^{n} = 0$ for sufficiently large $n .$ Furthermore, $X = \log (1 + (e^{X} - 1)) = (e^{X} - 1) - \frac{{(e^{X} - 1)}^{2}}{2} + \frac{{(e^{X} - 1)}^{3}}{3} - \dots$ is also finite, and so the map exp is invertible if $𝔤 \subseteq 𝔫 \subseteq 𝔤 𝔩_{n}$ for some $n . □$

Note that the proof of this theorem uses the fact that any nilpotent Lie algebera can be imbedded in the nilpotent Lie algebra $𝔫_{n}$ of strictly upper triangular matrices for some $n .$ This fact is proved in [CorGrn, Thm 1.1.11]. See also [Bou, I, 7] where this fact is called Ado's theorem. This statement is analogous to the statements that (a) for a finite group $G, G \subseteq S_{n}$ for some $n$ and (b) for an algebraic group $G, G \subseteq {GL}_{n}$ for some $n .$ The main idea in all of these proofs is to get $G$ to act on itself.

Every irreducible representation of an abelian group is one dimensional.

		Proof.
	Let $M$ be an irreducible $G$ -module. Let $ρ : G \to End (M)$ be the corresponding homomorphism. If $g \in G$ then $ρ (g) \in {End}_{G} (M)$ and so, by Schur's lemma, $ρ (g) = α {Id}_{M}$ for some $α \in ℂ .$ So every element $g \in G$ acts on $M$ by scalars. Thus, if $m \in M$ then $ℂ m \subseteq M$ is a submodule of $M .$ Since $M$ is irreducible, $ℂ m = M,$ and therefore $M$ is one dimensional. $□$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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