A Geometric Characterization of Rational Groups

TL;DR

This paper provides a geometric criterion to identify finite rational groups by relating their structure to automorphisms of a finite geometry, offering a new perspective on their classification.

Contribution

It introduces a geometric characterization of finite rational groups based on their action on a finite geometry and flag fixation properties.

Findings

Finite rational groups can be characterized by their automorphism actions on finite geometries.

A group is rational iff it acts on a geometry with conjugate elements fixing the same number of flags.

Provides a new geometric perspective for understanding the structure of rational groups.

Abstract

We give a geometric characterization of finite rational groups. In particular, we prove that a finite group is rational if and only if there exists a finite geometry $\Gamma$ of type $I$ and action of $G$ on $\Gamma$ as a group of automorphisms such that if $g$ and $h$ are elements of $G$ fixing the same number of flags of type $J$ for all subsets $J$ of $I$, then $g$ and $h$ are conjugate in $G$.

Full text

1. Rational Groups

A finite group $G$ is said to be a rational group if every every ordinary (i.e., complex) irreducible character $\chi$ of G is a rational-valued. Since ordinary character values of a finite group are algebraic integers, and since any rational algebraic integer is an integer, it follows that the character table of a finite rational group consists entirely of integers. Examples of rational groups include the Coxeter groups of type $A_{n},$ (which are finite symmetric groups), $B_{n},D_{n},E_{6},E_{7},E_{8},F_{4},$ and $G_{2}$. Since each of these Coxeter groups acts in a natural way on the coset geometry of its maximal parabolic subgroups, it seems natural to seek geometric equivalents to rationality for finite groups.

Let $I$ be a nonempty set. An incidence geometry (or just a geometry) of type $I$ is a triple $\Gamma=(X,*,t)$ where $X$ is a non-empty set (the objects of the geometry), * is a symmetric and reflexive relation on $X$ (the incidence relation) and $t:X\rightarrow I$ is a function (the type function) such that if $t(x)=t(x^{\prime})$ and $x*x^{\prime}$ then $x=x^{\prime}$. The geometry is said to be finite if $X$ is finite. A flag of the geometry is a set $F$ of mutually incident elements of $X$, and the type of a flag $F$ is the set of types of the elements of $F$. An automorphism of $\Gamma$ is a bijection $\alpha:X\rightarrow X$ such that $\alpha$ and $\alpha^{-1}$ preserve incidence and types. The group of all automorphisms of $\Gamma$ will be denoted by $Aut(\Gamma)$, and we will say that $G$ acts on $\Gamma$ if there is a homomorphism $\theta:G\rightarrow Aut(\Gamma)$. When $G$ acts on $\Gamma$, for $g\in G$ and $J\subseteq I$, we will define $Fix_{J}(g)$ to be the set of flags of type $J$ fixed by $g$. The purpose of this note is to prove the following:

Theorem 1.1**.**

*Let $G$ be a finite group. Then the following are equivalent:

(1) $G$ is a rational group.

(2) There exists a finite geometry $\Gamma$ of type $I$ and action of $G$ on $\Gamma$ as a group

of automorphisms such that if $g$ and $h$ are elements of $G$ fixing the same

number of flags of type $J$ for all subsets $J$ of $I$, then $g$ and $h$ are conjugate in $G$.

Definition 1.2**.**

Let $G$ be a group, and let $I$ and $S$ be non-empty sets, and for each $i\in I$ let $f_{i}:G\rightarrow S$ be a class function on $G$ (that is, if $g$ and $h$ are conjugate in $G$, then $f_{i}(g)=f_{i}(h)$, so that $f_{i}$ is constant on conjugacy classes of $G$). We say that the set $\{f_{i}:i\in I\}$ of functions separates conjugacy classes of $G$ if whenever $g$ and $h$ are in distinct conjugacy classes of $G$, then there exists an $i$ such that $f_{i}(g)\neq f_{i}(h)$.

To prove Theorem 1.1, we will make use of the following theorem (see [3]):

Theorem 1.3**.**

*Let $G$ be a finite group, and let $S=\{g_{1},g_{2},g_{3},...,g_{k}\}$ be a complete set of representatives of the conjugacy classes of $G$. Then the following are equivalent:

(1) $G$ is a rational group.

(2) The permutation characters $\{1_{<g_{i}>}^{G}:i=1,2,...,k\}$ separate conjugacy classes

of $G$.

(3) $G$ has a finite collection $\mathcal{H}$ of subgroups such that the permutation characters

$\{1_{H}^{G}:H\in\mathcal{H}\}$ separate conjugacy classes of $G$.

(4) $G$ has a permutation character which separates the conjugacy classes of $G$.

(That is, G has a permutation representation such that any two elements of $G$

fixing the same number of letters are conjugate in $G$.)*

We can now give a proof of the main Theorem 1.1:

Proof.

(of Theorem 1.1) We will first show that (1) implies (2). Assume that $G$ is a rational group. Let $\{g_{1},g_{2},g_{3},...,g_{k}\}$ be a complete set of representatives of the conjugacy classes of $G$, and let $I=\{1,2,3,...,k\}$. For each integer $i\in I$, let $X_{i}$ be the set of left cosets of $\langle g_{i}\rangle$ in $G$. Let $X=\bigcup_{i\in I}X_{i}$. (Without loss of generality, we will assume that the sets $X_{i}$ are pairwise disjoint, for otherwise we could replace $X_{i}$ with $X_{i}\times\{i\}$.) Define the type of each element of $X_{i}$ to be $i$. Define the incidence relation * on $X$ by $x*x^{\prime}$ iff the cosets $x$ and $x^{\prime}$ are equal or have different types and non-empty intersection. This defines a finite geometry $\Gamma=(X,*,t)$. (In fact, $\Gamma$ is the coset geometry of the cyclic subgroups $\langle g_{i}\rangle$ for $i\in I$.) $G$ acts by left multiplication on the set $X$ of left cosets. So each $g\in G$ defines a function $\alpha_{g}:X\rightarrow X$ given by $\alpha_{g}(h\langle g_{i}\rangle)=gh\langle g_{i}\rangle$. This function $\alpha$ is a bijection with inverse $\alpha^{-1}=\alpha_{g^{-1}}$, and both $\alpha$ and $\alpha^{-1}$ preserve incidence and types. Therefore, $\alpha$ is an automorphism of $\Gamma$ and the mapping $g\mapsto\alpha_{g}$ defines a homomorphism $\alpha:G\rightarrow Aut(\Gamma)$; thus $G$ acts on $\Gamma$. Now assume that $g$ and $h$ are elements of $G$ fixing the same number of flags of type $J$ for all subsets $J$ of $I$ (and so in particular, for all $i\in I$, the number of flags of type $i$ fixed by $g$ equals the number of flags of type $i$ fixed by $h$). This means that for each $i\in I$, the number of cosets of $\langle g_{i}\rangle$ fixed by $g$ equals the number of cosets of $\langle g_{i}\rangle$ fixed by $h$. But $1_{\langle g_{i}\rangle}^{G}$ is the permutation character of $G$ acting on the cosets of $\langle g_{i}\rangle$, and so we have $1_{\langle g_{i}\rangle}^{G}(g)=1_{\langle g_{i}\rangle}^{G}(h)$ for all $i\in I$. But by Theorem 1.3 above, since $G$ is a rational group, the permutation characters $\{1_{\langle g_{i}\rangle}^{G}:i\in I\}$ separate conjugacy classes of $G$. So $g$ must be conjugate to $h$ in $G$. So we have shown that (1) implies (2).

Next, we will show that (2) implies (1). Assume that there exists a finite geometry $\Gamma$ of type $I$ and action of $G$ on $\Gamma$ such that if $g$ and $h$ are elements of $G$ fixing the same number of flags of type $J$ for all subsets $J$ of $I$, then $g$ and $h$ are conjugate in $G$. We want to show that $G$ is a rational group. Let $C=\{g_{1},g_{2},g_{3},...,g_{k}\}$ be a complete set of representatives of the conjugacy classes of $G$. Let $g_{i}$ and $g_{j}$ be distinct elements of $C$. Since $g_{i}$ and $g_{j}$ are not conjugate, there exists a subset $J$ of $I$ such that $g_{i}$ and $g_{j}$ fix a different number of flags of type $J$. It follows that there must exist a G-orbit $\mathcal{O}$ of flags of type $J$ on which $g_{i}$ and $g_{j}$ fix a different number of flags. Since $G$ is transitive on the G-orbit $\mathcal{O}$, the permutation character of $G$ on the G-orbit $\mathcal{O}$ is given by $1_{H}^{G}$, where $H$ is the stabilizer in $G$ of a flag of type $J$ in $\mathcal{O}$. So we have found a subgroup $H=H_{\{g_{i},g_{j}\}}$ (depending on the set $\{g_{i},g_{j}\}$) and a permutation character $1_{H}^{G}$ such that $1_{H}^{G}(g_{i})\neq 1_{H}^{G}(g_{j})$. Repeating this process for all sets $\{g_{i},g_{j}\}$ of distinct elements of $C$ gives us a finite collection $\mathcal{H}$ of subgroups such that the permutation characters $\{1_{H}^{G}:H\in\mathcal{H}\}$ separate conjugacy classes of $G$. By Theorem 1.3 above, it follows that $G$ is a rational group. Therefore (2) implies (1). This proves Theorem 1.1. ∎

Theorem 1.1 can be used to give a geometric proof of the well-known fact that the finite Coxeter groups of type $A_{n}$ (which are finite symmetric groups $Sym(\Omega)$) are rational groups. To do that, we will make use of the following lemma. Here, $Fix_{k}(g)$ represents the collection of subsets of $\Omega$ of cardinality $k$ fixed by the element $g$ of $Sym(\Omega)$.

Lemma 1.4**.**

Let $\Omega$ be a finite non-empty set, and let $g$ and $h$ be in $Sym(\Omega)$. If $|Fix_{k}(g)|=|Fix_{k}(h)|$ for every $k$, $(0\leq k\leq|\Omega|)$, then $g$ and $h$ are conjugate in $Sym(\Omega)$.

In other words, if two elements of $Sym(\Omega)$ fix the same number of subsets of $\Omega$ of cardinality $k$ for every $k$, $(0\leq k\leq|\Omega|)$, then they are conjugate in $Sym(\Omega)$.

Proof.

We will prove the equivalent contrapositive. Let $g$ and $h$ be in $Sym(\Omega)$ and assume that they are not conjugate in $Sym(\Omega)$. Then as permutations of $\Omega$, $g$ and $h$ do not have the same cycle structure. So there exists a least positive integer $m$ in $\{1,2,3,...,|\Omega|\}$ such that $g$ and $h$ have a different number of $m$-cycles. The number of subsets of $\Omega$ of cardinality $m$ fixed by a permutation of $\Omega$ equals the number of ways of forming a set of cardinality $m$ from cycles of that permutation, which is different for $g$ and $h$. Thus $|Fix_{m}(g)|\neq|Fix_{m}(h)|$. So it is not the case that $|Fix_{k}(g)|=|Fix_{k}(h)|$ for every $k$, $(0\leq k\leq|\Omega|)$. ∎

Example: Let $\Omega=\{1,2,3,4\}$, so $Sym(\Omega)$ acts on the 16 subsets of $\Omega$. The involution $(12)(34)$ fixes one subset of cardinality [math] (the empty set), no subsets of cardinality $1$, two subsets of cardinality $2$ (namely $\{1,2\}$ and $\{3,4\}$), no subsets of cardinality $3$, and one subset of cardinality $4$ (namely $\Omega$ itself). Also, the involution $(13)(24)$ fixes one subset of cardinality [math], no subsets of cardinality $1$, two subsets of cardinality $2$, no subsets of cardinality $3$, and one subset of cardinality $4$. So the lemma tells us that $(12)(34)$ and $(13)(24)$ are conjugate in $Sym(\Omega)$. On the other hand, the permutations $(12)(34)$ and $(1234)$ are not conjugate in $Sym(\Omega)$, and so by the lemma, they must fix different numbers of sets of cardinality $k$, for some $k$. Checking, we see that they both fix the same number of sets of cardinalities $0,1,3,$ and $4$, but while $(12)(34)$ fixes two subsets of cardinality 2, the permutation $(1234)$ fixes no subset of cardinality $2$. Note also that the elements $(12)(34)$ and $(123)$ each fix four subsets of $\Omega$, but are not conjugate; so it is not possible to weaken the hypotheses of Lemma 1.4 to the condition that $g$ and $h$ only fix the same total number of subsets of $\Omega$.

To give a geometric proof of the well-known fact that the finite Coxeter groups of type $A_{n}$ (which are finite symmetric groups $Sym(\Omega)$) are rational groups, we will construct a finite geometry $\Gamma$ on which $Sym(\Omega)$ acts, and then apply Theorem 1.1. Let $\Omega=\{1,2,...,n\}$ and let $X=\mathcal{P}(\Omega)$, the power set of $\Omega$. For each element $x$ of $X$ (i.e. for each subset $x$ of $\Omega$), define the type $t(x)$ of $x$ to be the cardinality of $x$. For elements $x$ and $x^{\prime}$ in $X$, define $x*x^{\prime}$ if and only if either $x\subseteq x^{\prime}$ or $x^{\prime}\subseteq x$. Then $\Gamma=(X,*,t)$ is a finite geometry of type $I$, where $I=\{0,1,2,...,n\}$. The usual action of $Sym(\Omega)$ on $\Omega$ induces a natural action of $Sym(\Omega)$ on the geometry $\Gamma$. We claim that this action satisfies condition (2) of the main Theorem 1.1: let $g$ and $h$ be arbitrary elements of $Sym(\Omega)$ which fix the same number of flags of type $J$ for all subsets $J$ of $I$; we want to show that $g$ and $h$ are conjugate. For each integer $i\in I=\{0,1,2,...,n\}$, the elements $g$ and $h$ fix the same number of flags of $\Gamma$ of type $i$, and therefore fix the same number of subsets of $\Omega$ of cardinality $i$. Then by Lemma 1.4, $g$ and $h$ are conjugate in $Sym(\Omega)$. So this action satisfies condition (2) of the main Theorem 1.1, and so $Sym(\Omega)$ is a rational group.

Concluding question: For each of the other previously mentioned finite Coxeter groups, $B_{n},D_{n},E_{6},E_{7},E_{8},F_{4},$ and $G_{2}$, can one - without assuming rationality - find a natural geometry (perhaps related to its geometry of maximal parabolic subgroups) which satisfies the geometric condition of Theorem 1.1? If so, this would provide a natural geometric proof of the rationality of these groups.