# Rational elements in representations of simple algebraic groups, I

Alexandre Zalesski  School of Mathematics, University of East Anglia, Norwich,
UK

###### Key words and phrases:

Simple algebraic group, Modular representations, Rational elements, Eigenvalue
$1$

###### 2000 Mathematics Subject Classification:

20G05, 20G40

e-mail<EMAIL_ADDRESS>
Dedicated to Pham Huu Tiep on the occasion of his $60$-th birthday

Abstract An element $g$ of a group $G$ is called rational if $g$ is conjugate
to $g^{i}$ for every integer $i$ coprime to $|g|$. We determine all triples
$(G,g,\phi)$, where $G$ is a simple algebraic group of type $A_{n},B_{n}$ or
$C_{n}$ over an algebraically closed field of characteristic $p\geq 0$, $g\in
G$ is a rational odd order semisimple element and $\phi$ is an irreducible
representation of $G$ such that $\phi(g)$ has eigenvalue 1.

##  1\. Introduction

The primary motivation of this work lies in the study of eigenvalues of
semisimple elements of simple algebraic groups in their irreducible
representations. This problem is not treatable in full generality, however, a
number of special cases of it were discussed in literature. The earliest work
is probably by Wilson [13] who determined the irreducible representations of
simple algebraic groups in characteristic 2 in which some element of order 3
does not have eigenvalue 1. In [15] there appeared a number of useful
observations, and certain comments are available in surveys [14] and [16]. The
occurrence of the eigenvalue 1 deserves a particular attention, as this
carries some geometrical meaning, and is significant for applications. Robert
Guralnick and Pham Huu Tiep [6] determine the simple algebraic groups $G$
whose every element in every representation has eigenvalue 1. (If $\rho$ is a
representation of $G$ then every element of $G$ has weight 0 if and only if
$\rho$ has weight 0; the simple simply connected groups whose all
representations have weight 0 were listed in [15] and the groups of adjoint
type with this property were determined in [6]. In addition, in [6] the
authors determined all finite groups of Lie type with the above eigenvalue
property, some partial results appeared in [15].) In [17, 18] for groups
$G=SL_{n}(2)$ and $Sp_{2n}(2)$ I determine the irreducible 2-modular
representations of $G$ in which every element of $G$ has eigenvalue 1. It is
clear that one cannot expect to determine the elements $g\in G$ which have
eigenvalue 1 in every given irreducible representation. In this paper we
consider elements of ”highest symmetry”, specifically, the rational elements.
Recall that an element $g$ of a group $G$ is called rational if $g$ is
conjugate to $g^{i}$ for every integer $i$ coprime to $|g|$. In other words,
if $\langle g\rangle$ is the group generated by $g$ then all generators of
$\langle g\rangle$ are conjugate in $G$, equivalently, $N_{G}(\langle
g\rangle)/C_{G}(\langle g\rangle)\cong{\rm Aut}\,(\langle g\rangle)$. This
implies that for every representation $\phi$ of $G$ and every eigenvalue
$\lambda$ of $\phi(g)$ the powers $\lambda^{i}$ are eigenvalues of $\phi(g)$
for all $i$ coprime to $|g|$.

Let $G=G(F)$ be a simple simply connected algebraic group over an
algebraically closed field $F$ of characteristic $p\geq 0$. In Theorems 1.1,
1.3 and 1.4 below we determine all triples $(G,g,\phi)$, where $G$ is of type
$A_{n},B_{n}$, $C_{n}$, respectively, in $g\in G$ is a rational odd order
semisimple element and $\phi$ is an irreducible representation of $G$ such
that $\phi(g)$ has eigenvalue 1. Simple algebraic groups of the other types
will be considered in a subsequent paper. With additional efforts, one can
probably drop the restriction that $|g|$ is odd. Note that $\langle
g\rangle\cap Z(G)=1$ whenever $g$ is rational and $|g|$ is odd.

To state our results, we use a parametrization of the irreducible
representations of $G$ in terms of highest weights. If the rank of $G$ equals
$n$ then we denote by $\omega_{1},\ldots,\omega_{n}$ the fundamental weights
(ordered as in [1]). If $p=0$ then $p^{t}\omega_{i}$ is interpreted as
$\omega_{i}$. To describe the elements $g\in G$ in question, we use their
Jordan form at the natural representation of $G$, which we define as the
irreducible representation of $G$ with highest weight $\omega_{1}$. An
irreducible representation of $G$ with highest weight $\omega$ is denoted by
$\rho_{\omega}$ unless otherwise stated.

Given a natural number $m$, we denote by $R(m)$ and $\Phi(m)$ the set of all
$m$-roots of unity and all primitive $m$-roots of unity, respectively, and set
$\phi(m)=|\Phi(m)|$. The mapping $m\rightarrow\phi(m)$ for
$m\in\mathop{\mathbb{Z}}\nolimits^{+}$ is called the Euler function. For a
representation $\rho$ of $G$ denote by $E(\rho(g))$ the set of all eigenvalues
of $\rho(g)$ disregarding the multiplicities. If $g$ is of order $m$ then
$E(\rho(g))\subseteq R(m)$.

######  Theorem 1.1.

Let $G=SL_{n}(F),n>1,$ and let $g\in G$ be a semisimple odd order element. Let
$\rho$ be an irreducible representation of G with highest weight $\omega$.
Suppose that $g$ is rational. Then $1$ is an eigenvalue of $\rho(g)$ unless
one of the following holds for some integer $t\geq 0$:

$(1)$ $\dim\rho=n$, n even, and
$\omega\in\\{p^{t}\omega_{1},p^{t}\omega_{n-1}\\};$

$(2)$ $n=6$, $|g|=9$ and $\omega=p^{t}\omega_{3};$

$(3)$ $n=6$, $|g|=15$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$,
$|g_{1}|=5,|g_{2}|=3$ and $\omega=p^{t}\omega_{3};$

$(4)$ $n=8$, $|g|=15$ and $\omega\in\\{p^{t}\omega_{3},p^{t}\omega_{5}\\};$

$(5)$ $n=8$, $|g|=15$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2},g_{3})$,
$|g_{1}|=5,|g_{2}|=|g_{3}|=3$ and
$\omega\in\\{p^{t}\omega_{3},p^{t}\omega_{5}\\};$

$(6)$ $n=10$, $|g|=45$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$,
$|g_{1}|=5,|g_{2}|=9$ and
$\omega\in\\{p^{t}\omega_{3},p^{t}\omega_{7},p^{t}\omega_{5}\\};$

$(7)$ $n=14$, $|g|=45$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$,
$|g_{1}|=15,|g_{2}|=9$ and $\omega\in\\{p^{t}\omega_{3},p^{t}\omega_{11}\\}.$

Here $g_{i}$, $i=1,2,3$ is a diagonal matrix of size $\phi(|g_{i}|)$ with
diagonal entries $\Phi(|g_{i}|)$, and $g$ is identified with
$\rho_{\omega_{1}}(g)$.

As a by-product of our proof of Theorem 1.1, we have

######  Theorem 1.2.

In assumption of Theorem 1.1, every eigenvalue of $\rho_{\omega_{1}}(g)$ is an
eigenvalue of $\rho(g)$, unless possibly for the case where $n$ is even,
$\omega=\omega_{i}$ with $i$ even, $|g|=3l$ with $(3,l)=1$.

A more precise description of the exceptional case in Theorem 1.2 is given in
Lemma 3.4.

######  Theorem 1.3.

Let $G=Spin_{2n+1}(F),n>1$, $p\neq 2$, and let $g\in G$ be a semisimple
element. Let $\rho$ be an irreducible representation of G with highest weight
$\omega$. Suppose that $g$ is rational in G and $|g|$ is odd. Then $1$ is an
eigenvalue of $\rho(g)$ unless one of the following holds for some integer
$t\geq 0$:

$(A)$ $\omega=p^{t}\omega_{n}$ and one of the following holds:

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(1)$
$E(\rho(g))=\Phi(m)R(|g|/m)$ with $m\in\\{3,5,9\\}$ and $(m,|g|/m)=1;$ in
other words, $E(\rho(g))$
consists of all $|g|$-roots of unity whose orders are multiples of $m;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2)$
$E(\rho(g))=\Phi(5)(R(m)\setminus 1)R(|g|/5m)$ with $m\in\\{3,9\\}$ and
$(5m,|g|/5m)=1;$ in other words, $E(\rho(g))$ consists of all $|g|$-roots of
unity whose orders are multiples of $15;$

$(B)$ $\omega=p^{a}\omega_{1}+p^{b}\omega_{n}$,
$a,b\in\mathop{\mathbb{Z}}\nolimits^{+}$, and g is as in $(A)(2)$ above.

The Jordan normal form of $g$ on the natural $FG$-module can chosen to be
$D=\mathop{\rm diag}\nolimits(D_{1},\ldots,D_{r})$, where $D_{i}$,
$i=1,\ldots,r,$ is a diagonal matrix of size $\phi(m_{i})$ with non-zero
entries $\Phi(m_{i})$ (for some integer $m_{i}\geq 1$). Then $(2)$ holds if
and only if there is exactly one $m_{i}\in\\{3,5,9\\}$ and $(m_{i},m_{j})=1$
for $i,j\in\\{1,\ldots,r\\}$, $j\neq i$, and $(3)$ holds if and only if there
is exactly one $m_{i}\in\\{3,9\\}$, exactly one $j$ with $m_{j}=5$ and
$(15,m_{k})=1$ for $i,j,k\in\\{1,\ldots,r\\}$, for $k\neq i,j$.

######  Theorem 1.4.

Let $G=Sp_{2n}(F)$, and let $\rho$ be an irreducible representation of G with
highest weight $\omega$. Let $g\in G$ be a rational element of odd order.
Suppose that $\rho(g)$ does not have eigenvalue $1$. Then one of the following
holds:

$(A)$ $\omega=p^{t}\omega_{1};$

$(B)$ $\omega=p^{t}\omega_{3}$ and one of the following holds:

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(1)$ $n=3$,
$|g|=9;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2)$ $n=3$,
$|g|=15$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$, $|g_{1}|=5,|g_{2}|=3;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(3)$ $n=4$,
$|g|=15;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(4)$ $n=4$,
$|g|=15$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2},g_{3})$,
$|g_{1}|=5,|g_{2}|=|g_{3}|=3;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(5)$ $n=5$,
$|g|=45$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$, $|g_{1}|=5,|g_{2}|=9;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(6)$ $n=7$,
$|g|=45$, $g=\mathop{\rm diag}\nolimits(g_{1},g_{2})$, $|g_{1}|=15,|g_{2}|=9;$

$(C)$ $n=5$, $\omega=p^{t}\omega_{5}$, $g=\mathop{\rm
diag}\nolimits(g_{1},g_{2})$, $|g_{1}|=5$, $|g_{2}|=9;$

$(D)$ $p=2$, $\omega=2^{t}\omega_{n}$ and one of the following holds:

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(1)$
$g=\mathop{\rm diag}\nolimits(g_{1},y)$, where $g_{1}\in Sp_{2k}(F)$,
$k=1,2,3$, $|g_{1}|=2^{k}+1$ and $y\in Sp_{2n-2k}(F)$ is an arbitrary rational
element with $(|y|,|g_{1}|)=1;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2)$
$g=\mathop{\rm diag}\nolimits(g_{1},g_{2},y)$, where $g_{i}\in
Sp_{2k_{i}}(F)$, $(k_{1},k_{2})=(1,2)$ or $(2,3)$, $|g_{i}|=2^{k_{i}}+1$ for
$i=1,2$ and $y\in Sp_{2n-2(k_{1}+k_{2})}(F)$ is an arbitrary rational element
with $(|y|,|g_{1}g_{2}|)=1;$

$(E)$ $p=2$, $\omega=2^{s}\omega_{1}+2^{t}\omega_{n}$, $s,t\geq 0$,
$g=\mathop{\rm diag}\nolimits(g_{1},g_{2},g_{3})$, and one of the following
holds:

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(1)$ $g_{1}\in
Sp_{4}(F)$, $|g_{1}|=5$, $g_{2}\in Sp_{2}(F)$, $|g_{2}|=3$, $g_{3}\in
Sp_{2(n-3)}(F)$ is an arbitrary rational element of order coprime to $15;$

$\,\,\,\,\,\,\,\,\,\,~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}~{}(2)$ $g_{1}\in
Sp_{4}(F)$, $|g_{1}|=5$, $g_{2}\in Sp_{6}(F)$, $|g_{2}|=9$ and $g_{3}\in
Sp_{2(n-5)}(F)$ is an arbitrary rational element of order coprime to $15$.

If $p\neq 2$ and $\dim\rho>1$ then $E(\rho_{\omega_{1}}(g))\subset E(\rho(g))$
for $G=Sp_{2n}(F)$; see Lemma 4.3. As above, $g$ is identified here with
$\rho_{\omega_{1}}(g)$

Other results of the paper refine the above results for some special cases. In
particular, in some cases we observe that $E(\rho(g))=R(|g|)$. Note that with
some further analysis one can hope to determine all cases where
$E(\rho(g))=R(|g|)$.

Notation $\mathop{\mathbb{Z}}\nolimits$ is the ring of integers, and
$\mathop{\mathbb{Z}}\nolimits^{+}$ is the set of non-negative integers. For
$a,b\in\mathop{\mathbb{Z}}\nolimits$, $ab\neq 0$, we write $(a,b)$ for the
greatest common divisor of $a,b$, and $a|b$ means that $b$ is a multiple of
$a$. A diagonal $(n\times n)$-matrix with sequential entries
$x_{1},\ldots,x_{n}$ is denoted by $\mathop{\rm
diag}\nolimits(x_{1},\ldots,x_{n})$. By $\mathop{\mathbb{F}}\nolimits_{q}$ we
denote the finite field of $q$ elements. All representations below are over an
algebraically closed field $F$ of characteristic $p\geq 0$ (unless otherwise
stated). For a semigroup $H$ and subsets $X,Y\subset H$ we write $XY$ or
$X\cdot Y$ for the set $\\{xy:x\in X,y\in Y\\}$.

For $m\in\mathop{\mathbb{Z}}\nolimits^{+}$ we denote by $R(m)$ the set of all
$m$-roots of unity, and by $\Phi(m)$ the set of all primitive $m$-roots of
unity. So $|R(m)|=m$ and $|\Phi(m)|=\phi(m)$; we call $\phi$ the Euler
function (some authors use the term ”the Euler ’totient’ function”). Notations
$\Lambda_{i}(\Phi(m))$ and $\Delta(\Phi(m))$ are introduced in Section 2.

If $\phi$ is a representation of a group $X$ and $Y\subset X$ is a subgroup
then $\phi|_{Y}$ means the restriction of $\phi$ to $Y$.

Notation for algebraic groups and their weight systems will be introduced in
Section 2.

##  2\. Preliminaries

###  2.1. Cyclic groups and the Euler function

In this section we discuss some properties of the set $\Phi(m)$ with
$m\in\mathop{\mathbb{Z}}\nolimits^{+}$ which are useful in what follows.

######  Lemma 2.1.

Let $m$ be an odd natural number. Then $\Phi(m)\cdot\Phi(m)=R(m)$.

###### Proof.

Suppose first that $m$ is a $p$-power for some prime $p$. Then
$R(m)=R(m^{\prime})\cup\Phi(m)$, where $m^{\prime}=m/p$. Note that $R(m)$ is a
group and $R(m^{\prime})$ is a subgroup of index $p$. Let $\eta\in\Phi(m)$.
Then $\Phi(m)=\cup_{i=1}^{p-1}\eta^{i}R(m^{\prime})$ and
$\eta\cdot\eta^{i}R(m^{\prime})=\eta^{i+1}R(m^{\prime})$, whence
$R(m^{\prime})$ and $\eta^{j}R(m^{\prime})\subset\Phi(m)\cdot\Phi(m)$ for
$j\neq 1$. Also $\eta R(m^{\prime})=\eta^{2}\cdot\eta^{p-1}R(m^{\prime})$,
whence the result.

Let $m=m_{1}\cdots m_{k}$, where $k>1$ and $m_{1},\ldots,m_{k}$ are prime
powers coprime to each other. Then $\Phi(m)=\Phi(m_{1})\cdots\Phi(m_{k})$ and
hence
$\Phi(m)\cdot\Phi(m)=\Phi(m_{1})\cdot\Phi(m_{1})\cdots\Phi(m_{k})\cdot\Phi(m_{k})=R(m_{1})\cdots
R(m_{k})=R(m)$. ∎

######  Lemma 2.2.

Let $m_{1},m_{2}$ be odd natural numbers and $l=(m_{1},m_{2})>1$. Let
$M_{i}=R(m_{i})\setminus 1$ for $i=1,2$. Let $m_{3}=m_{1}m_{2}/l$ be the least
common multiple of $m_{1},m_{2}$. Then $M_{1}M_{2}=R(m_{3})$.

###### Proof.

By Lemma 2.1, the lemma holds if $m_{1}=m_{2}$, and, in addition,
$R(l)\subseteq M_{1}M_{2}$. Clearly, $M_{1}M_{2}$ contains 1 and
$R(m_{1})R(m_{2})\setminus(R(m_{1})\cup R(m_{2}))$. So it suffices to show
that $R(m_{1}),R(m_{2})\subseteq M_{1}M_{2}$. By assumption, $\Phi(l)\subseteq
M_{1}$, so there are distinct $\zeta_{1},\zeta_{2}\in\Phi(l)$. Then
$\zeta_{1}M_{2}\cup\zeta_{2}M_{2}=R(m_{2})\subseteq M_{1}M_{2}$. Similarly,
$R(m_{1})\subseteq M_{1}M_{2}.$∎

For $i\leq\phi(m)$ set $\Lambda_{i}(\Phi(m))=\eta_{1}\cdots\eta_{i}$, where
$\eta_{1},\ldots,\eta_{i}\in\Phi(m)$ are distinct. In other words,
$\Lambda_{i}(\Phi(m))$ is the set of all elements which are products of $i$
distinct elements in $\Phi(m)$.

######  Lemma 2.3.

For $\zeta\in R(m)$, m odd, there exists $\eta\in\Phi(m)$ such that
$\zeta\eta\in\Phi(m)$, and $\eta\neq\zeta$ if $m>3$.

###### Proof.

Note that $\Phi(m)=\Phi(m_{1})\cdots\Phi(m_{k})$, where $m_{1},\ldots,m_{k}$
are prime powers coprime to each other. Suppose first that $k=1$. If $m>3$ is
a prime then the claim is obvious. Otherwise, let $m=m^{\prime}p$ for $p$ a
prime. If $\zeta\in R(m^{\prime})$ then the claim is true for any
$\eta\in\Phi(m)$. Otherwise, $\zeta\in\Phi(m)$. If $p>3$ then take
$\eta=\zeta^{2}$. If $p=3$ then we can take $\eta=\zeta^{4}$ if $m>3$ and
$\eta=\zeta$ for $m=3$.

Let $k>1$. If $3\notin\\{m_{1},\ldots,m_{k}\\}$ then the result follows from
the above. Otherwise we can assume that $m_{1}=3$. Let $m^{\prime}=m/3$. Then
either $\zeta\in R(m^{\prime})$ and we take $\eta=\eta_{3}\eta^{\prime}$,
where $\eta_{3}\in\Phi(3)$, $\eta^{\prime}\in\Phi(m^{\prime})$ is such that
$\zeta\eta^{\prime}\in\Phi(m^{\prime})$, or $\zeta=\eta^{\pm
1}_{3}\zeta^{\prime}$ with $\zeta^{\prime}\in R(m^{\prime})$ and then we take
$\eta=\eta^{\pm 2}_{3}\eta^{\prime}$, where $\eta^{\prime}\in\Phi(m^{\prime})$
is such that $\zeta^{\prime}\eta^{\prime}\in\Phi(m^{\prime})$, which is
possible by the above. ∎

######  Lemma 2.4.

Let $m>3$ be an odd natural number. Then $\Lambda_{2}(\Phi(m))=R(m)$. In other
words, for every $\zeta\in R(m)$ there are distinct roots
$\eta_{1},\eta_{2}\in\Phi(m)$ such that $\eta_{1}\eta_{2}=\zeta$.

###### Proof.

By Lemma 2.3, $\zeta=\eta_{1}^{-1}\eta_{2}$ for some
$\eta_{1},\eta_{2}\in\Phi(m)$, and $\eta_{1}^{-1}\neq\eta_{2}$ whenever
$|\zeta|<m$. If $|\zeta|=m$ then choose
$\eta_{1}=\zeta^{2},\eta_{2}=\zeta^{-1}$. ∎

######  Lemma 2.5.

Let $m\neq 1,3,5,9,15$ be an odd natural number. Then there are six distinct
roots $\eta_{1}^{\pm 1},\eta_{2}^{\pm 1},\eta_{3}^{\pm 1}\in\Phi(m)$ such that
$\eta_{1}\eta_{2}\eta_{3}=1$. Equivalently, for every $\eta\in\Phi(m)$ there
exist $\eta_{1},\eta_{2}\in\Phi(m)$ such that $\eta=\eta_{1}\eta_{2}$ and
$\eta^{\pm 1},\eta_{1}^{\pm 1},\eta_{2}^{\pm 1}$ are distinct.

###### Proof.

Let $\eta\in\Phi(m)$. Note that $\eta^{2^{i}}\in\Phi$ for every
$i\in\mathop{\mathbb{Z}}\nolimits^{+}$.

(i) The lemma is true if $(m,3)=1$ and $m>5$.

Indeed, $\eta\cdot\eta^{2}=\eta^{3}\in\Phi(m)$ as $(m,3)=1$, and $m-3>3$ as
$m>5$. So $\eta^{\pm 1},\eta^{\pm 2},\eta^{\pm 3}$ are distinct.

(ii) The lemma is true for if $3|m$, $(m,5)=1$ and $m\geq 11$.

Indeed, $\eta\cdot\eta^{4}=\eta^{5}$ and $m-5>5$ for $m\geq 11$. So $\eta^{\pm
1},\eta^{\pm 4},\eta^{\pm 5}$ are distinct.

(iii) Let $m^{\prime}$ be the product of all primes dividing $m$. The lemma is
true if $m^{\prime}<m$.

Indeed, $m^{\prime}-1,m^{\prime}+1,2$ are coprime to $m$ so
$\eta^{m^{\prime}-1},\eta^{-m^{\prime}-1},\eta^{2}\in\Phi(m)$. As
$3m^{\prime}\leq m$, we have $m\geq 2m^{\prime}+m^{\prime}>2m^{\prime}+2$. If
$m^{\prime}>3$ then $\eta^{\pm(m^{\prime}+1)}$, $\eta^{\pm(m^{\prime}-1)}$,
$\eta^{\pm 2}$ are distinct, and
$\eta^{m^{\prime}-1}\eta^{-m^{\prime}-1}\eta^{2}=1$, as required. Let
$m^{\prime}=3$; then $m\geq 27$ by assumption. Then
$\eta^{8}\eta^{-10}\eta^{2}=1$ and $\eta^{\pm 8},\eta^{\pm 10},\eta^{\pm 2}$
are distinct.

(iv) The lemma is true if $m=m^{\prime}$.

In view of (i), (ii) and (iii) we can assume that $m>15$, $15|m$ and
$(15,m/15)=1$. Let $\eta_{1}=\zeta_{3}\zeta_{5}x\in\Phi(m)$, where
$\zeta_{i}\in\Phi(i)$ for $i=3,5$, and $x\in\Phi(m/15)$. Set
$\eta_{2}=\zeta_{3}\zeta_{5}x^{2}$ and
$\eta_{3}=\zeta_{3}\zeta_{5}^{3}x^{-3}$. Then $\eta_{2},\eta_{3}\in\Phi(m)$
and $\eta_{1}\eta_{2}\eta_{3}=1$ and $\eta_{1},\eta_{2},\eta_{3}$ are
distinct. ∎

######  Lemma 2.6.

Let $m>3$ be odd. Then either $\Lambda_{3}(\Phi(m))=R(m)$ or
$m\in\\{5,9,15\\}$ and $\Lambda_{3}(\Phi(m))=R(m)\setminus 1$.

###### Proof.

(i) Note that $\Phi(m)\subseteq\Lambda_{3}(\Phi(m))$. Indeed, if
$\zeta\in\Phi(m)$ then there is $\eta\in\Phi(m)$ such that
$\zeta\notin\\{\eta,\eta^{-1}\\}$, so $\eta\cdot\eta^{-1}\zeta=\zeta$. For
$m\in\\{5,9,15\\}$ the lemma follow by inspection. If $m\notin\\{5,9,15\\}$
then $1\in\Lambda_{3}(\Phi(m))$ by Lemma 2.5. This implies the lemma if $m$ is
a prime.

(ii) The lemma is true if $m$ is a prime power.

Indeed, let $m=p^{k}>9$, where $p$ is a prime, and $m^{\prime}=m/p$ if $p>3$
and $m/9$ if $p=3$. Let $\zeta\in R(m^{\prime})$. Suppose first that $p>5$.
Then, by Lemma 2.5, there are $\eta_{1},\eta_{2},\eta_{3}\in\Phi(m)$ such that
$\mu:=\eta_{1}\eta_{2}\eta_{3}\in R(m^{\prime})$ and
$\eta_{1},\eta_{2},\eta_{3}$ are in distinct cosets $rR(m^{\prime})$,
$r\in\Phi(m)$. Set $\nu=\mu^{-1}\zeta$. Then $\eta_{1},\eta_{2},\eta_{3}\nu$
are distinct and $\eta_{1}\eta_{2}\eta_{3}\nu=\mu\nu=\zeta$, as required.

Let $p=5$. For $\eta\in\Phi(m)$ set $x=\eta^{5}\in\Phi(m/5)$ and
$\nu=x^{-1}\zeta$. Then $\eta,\eta^{2},\eta^{2}\nu$ are distinct and
$\eta\eta^{2}\eta^{2}\nu=x\nu=\zeta$, as required.

Let $p=3$. For $\eta\in\Phi(m)$ set $x=\eta^{3}\in\Phi(m/3)$. Since $m>9$, the
elements $\eta,y\eta,y^{4}\eta$ are distinct for every $y\in\Phi(m/3)$ and
their product is $xy^{5}$. As $\\{y^{5}:y\in\Phi(m/3)\\}=\Phi(m/3)$, we
conclude that $x\Phi(m/3)\subset\Lambda_{3}(\Phi(m))$. By Lemma 2.3, this
implies $\Phi(m/3)\cap\Lambda_{3}(\Phi(m))\neq\emptyset$. Moreover,
$\Phi(m^{\prime})\cap\Lambda_{3}(\Phi(m))\neq\emptyset$ whenever
$m^{\prime}|m$ as $xu\in\Phi(m/3)$ for $u\in R(m/9)$. By replacing $\eta$ by
$\eta^{j}$ for integers $j$ coprime to $m$ we conclude that
$\Phi(m^{\prime})\subset\Lambda_{3}(\Phi(m))$ for every divisor $m^{\prime}$
of $m$, whence the result.

(iii) Let $m=m_{1}\cdots m_{k}$ be the primary decomposition of $m$, and
$\zeta=\zeta_{1}\cdots\zeta_{k}$, where $\zeta_{i}\in R(m_{i})$. If
$m_{i}\notin\\{3,5,9\\}$ for $i=1,\ldots,k$ then the result follows from
applying the above to every $\zeta_{i}$, $i=1,\ldots,k$.

(iv) Suppose that $m_{i}\in\\{3,5,9\\}$ for some $i\in\\{1,\ldots,k\\}$. Then
$m\in\\{3m^{\prime},5m^{\prime},9m^{\prime},15m^{\prime},45m^{\prime}\\}$,
where $(m^{\prime},m/m^{\prime})=1$. Then $\zeta=\alpha\beta$, where $\beta\in
R(m^{\prime})$ and $\alpha\in R(m/m^{\prime})$.

Suppose first that $m^{\prime}>1$. Then, by (iii),
$\beta=\sigma_{1}\sigma_{2}\sigma_{3}$, where
$\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(m^{\prime})$ are distinct. Set
$\eta_{j}=\theta_{j}\sigma_{j}$ for $j=1,2,3$ and
$\theta_{j}\in\Phi(m/m^{\prime})$. Then $\eta_{j}\in\Phi(m)$ and
$\eta_{1},\eta_{2},\eta_{3}$ are distinct as so are
$\sigma_{1},\sigma_{2},\sigma_{3}$. As $\theta_{j}\in\Phi(m/m^{\prime})$ are
arbitrary here, $\alpha=\theta_{1}\theta_{2}\theta_{3}$ for some
$\theta_{1},\theta_{2},\theta_{3}\in\Phi(m/m^{\prime})$, and hence
$\zeta=\eta_{1}\eta_{2}\eta_{3}$ with this choice of
$\theta_{1},\theta_{2},\theta_{3}$.

Suppose that $m^{\prime}=1$. Then $m=45$ as $m\neq 3,5,9,15$ by assumption. We
can again write $\zeta=\alpha\beta$, where $\alpha\in R(5)$, $\beta\in R(9)$.
As the case with $\zeta\in\Phi(45)\cup 1$ is settled in (i), we have
$\alpha=1,\beta\in R(9)\setminus 1$, or $\alpha\in\Phi(5)$, $\beta\in R(3)$.

Suppose that $\alpha=1,\beta\neq 1$. By (i),
$\beta=\sigma_{1}\sigma_{2}\sigma_{3}$, where
$\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(9)$ are distinct. Set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{2}\sigma_{2},\eta_{3}=\omega^{2}\sigma_{3}$
with $\omega\in\Phi(5)$. Then $\eta_{1},\eta_{2},\eta_{3}\in\Phi(45)$ are
distinct and $\beta=\eta_{1}\eta_{2}\eta_{3}$.

Suppose that $\alpha\neq 1$. Then $\alpha=\sigma_{1}\sigma_{2}\sigma_{3}$,
where $\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(5)$ are distinct. If $\beta=1$,
set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{7}\sigma_{2},\eta_{3}=\omega\sigma_{3}$,
where $\omega\in\Phi(9)$. If $\beta\in\Phi(3)$, set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{4}\sigma_{2},\eta_{3}=\omega^{7}\sigma_{3}$,
where $\omega\in\Phi(9)$ and $\omega^{12}=\beta$. Then
$\eta_{1},\eta_{2},\eta_{3}\in\Phi(45)$ are distinct and
$\alpha\beta=\eta_{1}\eta_{2}\eta_{3}$.∎

######  Lemma 2.7.

For $m>3$ odd set
$\Lambda^{*}_{3}\Phi(m)=\\{\eta_{1}\eta_{2}\eta_{3}^{-1}:\eta_{1},\eta_{2},\eta_{3}\in\Phi(m)$
are distinct$\\}$. Then $\Lambda^{*}_{3}\Phi(m)=R(m)$.

###### Proof.

(i) Note that $\Phi(m)\subseteq\Lambda^{*}_{3}(\Phi(m))$. Indeed, if
$\zeta\in\Phi(m)$ then there is $\eta\in\Phi(m)$ such that
$\zeta\notin\\{\eta,\eta^{-1}\\}$, so
$\eta\cdot\eta^{-1}(\zeta^{-1})^{-1}=\zeta$. For $m\in\\{5,9,15\\}$ the lemma
follow by inspection. If $m\notin\\{5,9,15\\}$ then
$1\in\Lambda^{*}_{3}(\Phi(m))$ by Lemma 2.5. This implies the lemma if $m$ is
a prime.

(ii) The lemma is true if $m$ is a prime power.

Indeed, let $m=p^{k}>9$, where $p$ is a prime, and $m^{\prime}=m/p$ if $p>3$
and $m/9$ if $p=3$. Let $\zeta\in R(m/p)$. Suppose first that $p>5$. Then, by
Lemma 2.5, there are $\eta_{1},\eta_{2},\eta_{3}\in\Phi(m)$ such that
$\mu:=\eta_{1}\eta_{2}\eta_{3}^{-1}\in R(m^{\prime})$ and $\eta^{\pm
1}_{1},\eta^{\pm 1}_{2},\eta^{\pm 1}_{3}$ are in distinct cosets
$rR(m^{\prime})$, $r\in\Phi(m)$. Set $\nu=\mu\zeta^{-1}$. Then
$\eta_{1},\eta_{2},\eta_{3}\nu$ are distinct and
$\eta_{1}\eta_{2}(\eta_{3}\nu)^{-1}=\mu\nu^{-1}=\zeta$, as required.

Let $p=5$. For $\eta\in\Phi(m)$ set $\nu=\zeta^{-1}$. Then
$\eta,\eta^{2},\eta^{3}\nu$ are distinct and
$\eta\eta^{2}(\eta^{3}\nu)^{-1}=\nu^{-1}=\zeta$, as required.

Let $p=3$. For $\eta\in\Phi(m)$ set $x=\eta^{3}\in\Phi(m/3)$. Since $m>9$, the
elements $\eta,y\eta,y^{2}\eta^{-1}$ are distinct for every $1\neq y\in
R(m/3)$ and $\eta\cdot\eta y\cdot(y^{2}\eta^{-1})^{-1}=xy^{-1}$. So we
conclude that $x(R(m/3)\setminus 1)=R(m/3)\setminus
x\subset\Lambda^{*}_{3}(\Phi(m))$. By replacing $\eta$ by $\eta^{2}$ we
conclude that $R(m/3)\subset\Lambda^{*}_{3}(\Phi(m))$. By (i),
$\Phi(m)\subseteq\Lambda^{*}_{3}(\Phi(m))$, whence the result.

(iii) Let $m=m_{1}\cdots m_{k}$ be the primary decomposition of $m$, and
$\zeta=\zeta_{1}\cdots\zeta_{k}$, where $\zeta_{i}\in R(m_{i})$. If
$m_{i}\notin\\{3,5,9\\}$ for $i=1,\ldots,k$ then the result follows from
applying the above to every $\zeta_{i}$, $i=1,\ldots,k$.

(iv) Suppose that $m_{i}\in\\{3,5,9\\}$ for some $i\in\\{1,\ldots,k\\}$. Then
$m\in\\{3m^{\prime},5m^{\prime},9m^{\prime},15m^{\prime},45m^{\prime}\\}$,
where $(m^{\prime},m/m^{\prime})=1$. We have $\zeta=\alpha\beta$, where
$\beta\in R(m^{\prime})$ and $\alpha\in R(m/m^{\prime})$.

Suppose first that $m^{\prime}>1$. Then, by (iii),
$\beta=\sigma_{1}\sigma_{2}\sigma_{3}^{-1}$, where
$\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(m^{\prime})$ are distinct. Set
$\eta_{j}=\theta_{j}\sigma_{j}$ for $j=1,2,3$ and
$\theta_{j}\in\Phi(m/m^{\prime})$. Then $\eta_{j}\in\Phi(m)$ and
$\eta_{1},\eta_{2},\eta_{3}$ are distinct as so are
$\sigma_{1},\sigma_{2},\sigma_{3}$. As $\theta_{j}\in\Phi(m/m^{\prime})$ are
arbitrary here, $\alpha=\theta_{1}\theta_{2}\theta_{3}^{-1}$ for some
$\theta_{1},\theta_{2},\theta_{3}\in\Phi(m/m^{\prime})$, and hence
$\zeta=\eta_{1}\eta_{2}\eta_{3}^{-1}$ with this choice of
$\theta_{1},\theta_{2},\theta_{3}$.

Suppose that $m^{\prime}=1$. Then $m=45$ as $m\neq 3,5,9,15$ by assumption. We
can again write $\zeta=\alpha\beta$, where $\alpha\in R(5)$, $\beta\in R(9)$.
As the case with $\zeta\in\Phi(45)\cup 1$ is settled in (i), we have
$\alpha=1,\beta\in R(9)\setminus 1$, or $\alpha\in\Phi(5)$, $\beta\in R(3)$.

Suppose that $\alpha=1,\beta\neq 1$. By (i),
$\beta=\sigma_{1}\sigma_{2}\sigma_{3}^{-1}$, where
$\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(9)$ are distinct. Set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{2}\sigma_{2},\eta_{3}=\omega^{3}\sigma_{3}$
with $\omega\in\Phi(5)$. Then $\eta_{1},\eta_{2},\eta_{3}\in\Phi(45)$ are
distinct and $\beta=\eta_{1}\eta_{2}\eta_{3}^{-1}$.

Suppose that $\alpha\neq 1$. Then
$\alpha=\sigma_{1}\sigma_{2}\sigma_{3}^{-1}$, where
$\sigma_{1},\sigma_{2},\sigma_{3}\in\Phi(5)$ are distinct. If $\beta=1$, set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{7}\sigma_{2},\eta_{3}=\omega^{8}\sigma_{3}$,
where $\omega\in\Phi(9)$. If $\beta\in\Phi(3)$, set
$\eta_{1}=\omega\sigma_{1},\eta_{2}=\omega^{-2}\sigma_{2},\eta_{3}=\sigma_{3}$,
where $\omega\in\Phi(9)$ with $\omega^{3}=\beta$. Then
$\eta_{1},\eta_{2},\eta_{3}\in\Phi(45)$ are distinct and
$\alpha\beta=\eta_{1}\eta_{2}\eta_{3}^{-1}$.∎

Recall that
$\Lambda_{i}(\Phi(m))=\\{\eta_{1}\cdots\eta_{i}:\eta_{1},\ldots,\eta_{i}\in\Phi(m)$
are distinct$\\}$.

######  Lemma 2.8.

If $m$ is odd and $\phi(m)>6$ then
$\Lambda_{3}(\Phi(m))\subseteq\Lambda_{i}(\Phi(m))$ for $3<i\leq\phi(m)-3$.

###### Proof.

Let $\zeta\in\Lambda_{3}(\Phi(m))$. Observe that
$\Lambda_{i}(\Phi(m))=(\Lambda_{\phi(m)-i}(\Phi(m)))^{-1}$. Indeed, the
product of all elements in $\Phi(m)$ equals 1, so if
$\eta_{1}\cdots\eta_{i}=\zeta$ for distinct elements
$\eta_{1},\ldots,\eta_{i}\in\Phi(m)$ then the product of $\phi(m)-i$ remaining
elements of $\Phi(m)$ equals $\zeta^{-1}$. So we can assume that
$i\leq\phi(m)/2$. Set $\Phi^{\prime}(m)=\Phi(m)\setminus\\{\eta_{1}^{\pm
1},\eta_{2}^{\pm 1},\eta_{3}^{\pm 1}\\}$. Then
$\Phi^{\prime}(m)^{-1}=\Phi^{\prime}(m)$. As $\phi(m)-6\geq 2(i-3)$, we can
choose $\eta_{4},\ldots,\eta_{i}\in\Phi^{\prime}(m)$ so that
$\eta_{j+1}=\eta_{j}^{-1}$ for $j\in\\{4,\ldots,i-1\\}$ even. Then
$\eta_{1}\cdots\eta_{i}=\zeta$ and $\eta_{1},\ldots,\eta_{i}$ are distinct. ∎

######  Lemma 2.9.

Let $m>3$ odd and $i\in\\{2,\ldots,\phi(m)-2\\}$. $(1)$ $R(m)\setminus
1\subseteq\Lambda_{i}(\Phi(m))$.

$(2)$ $\Lambda_{i}(\Phi(m))=R(m)$ unless $i$ is odd and $m\in\\{5,9,15\\}$.

###### Proof.

This follows from Lemmas 2.4, 2.6 and 2.8. ∎

###  2.2. The set $\Delta(m)$

For $m\in\mathop{\mathbb{Z}}\nolimits$, $m>1$ odd, denote by $\mathcal{P}(m)$
the set of all strings $\eta_{1},\ldots,\eta_{\phi(m)/2}\in\Phi(m)$ such that
$\eta_{1}^{\pm 1},\ldots,\eta_{\phi(m)/2}^{\pm 1}$ are distinct. Then set
$\Delta(m)=\\{\eta_{1}\cdots\eta_{\phi(m)/2}:\\{\eta_{1},\ldots,\eta_{\phi(m)/2}\\}\in\mathcal{P}(m)\\}$.
In other words, $x\in F$ lies in $\Delta(m)$ if and only if
$x=\eta_{1}\cdots\eta_{\phi(m)/2}$ for some
$\eta_{1},\ldots,\eta_{\phi(m)/2}\in\Phi(m)$ such that $\eta_{1}^{\pm
1},\ldots,\eta_{\phi(m)/2}^{\pm 1}$ are distinct.

Note that $x\in\Delta(m)$ implies $x^{i}\in\Delta(m)$ whenever $(i,m)=1$.

######  Lemma 2.10.

Let $m\not\in\\{1,3,5,9\\}$ be an odd natural number and $k=\phi(m)$. Then
$1\in\Delta(m)$. In other words, there are elements
$\eta_{1},\ldots,\eta_{k/2}\in\Phi$ such that $\eta_{1}^{\pm
1},\ldots,\eta_{k/2}^{\pm 1}$ are distinct and $\eta_{1}\cdots\eta_{k/2}=1$.

###### Proof.

(i) Suppose first that $m=p$ is a prime. Then $p>5$. Set
$b=1+2+\cdots+(p-1)/2$ and let $b=lp+t$, where $0\leq t<p$. As
$b=\frac{p^{2}-1}{8}$, we have $t\neq 0$. Suppose first that $t$ is even. Let
$b^{\prime}=b-t$. Then $b^{\prime}$ can be obtained from $b$ by replacing in
the sum $1+2+\cdots+(p-1)/2$ the term $t/2$ by $-t/2.$ Then $b^{\prime}$
satisfies the conclusion of the lemma. Suppose that $t$ is odd. Then we
replace the term $(p-1)/2$ by $-(p-1)/2$ to obtain $b+1$ as the new sum. Then
we replace the term $(t+1)/2$ by $-(t+1)/2$, which is possible unless
$(t+1)/2=(p-1)/2$. If the latter does not hold then the lemma follows. If
$(t+1)/2=(p-1)/2$ then $t=p-2$ and $\frac{p^{2}-1}{8}=lp+p-2$, whence
$p^{2}-(l+1)p=-15$ and $p\in\\{3,5\\}$.

(ii) Suppose that $m=p^{a}>p$. Let $m=pm^{\prime}$. Then the mapping
$h:R(m)\rightarrow R(p)$ defined by $\zeta\rightarrow\zeta^{m^{\prime}}$ for
$\zeta\in R(m)$ is a group homomorphism and $h(\Phi(m))=\Phi(p)$. Suppose
first that $p>5$. Then, by (i), there are
$\eta_{1},\ldots,\eta_{(p-1)/2}\in\Phi(m)$ such that $h(\eta_{1})^{\pm
1},\ldots,h(\eta_{(p-1)/2})^{\pm 1}$ are distinct, and $h(\eta_{1})\cdots
h(\eta_{(p-1)/2})=1$.

Set $Y_{i}=\\{\eta_{i}x:x\in R(m^{\prime})\\}$ (so
$h(Y_{i})=\eta_{i}^{m^{\prime}}$) and $Y=Y_{1}\cup\ldots\cup Y_{(p-1)/2}$.
Then $Y\subset\Phi(m)$. As $m$ is odd, $y\neq y^{-1}$ for $y\in Y_{i}$, and
obviously $y_{i}\neq y_{j}^{\pm 1}$ for $y_{i}\in Y_{i},y_{j}\in Y_{j}$ for
$i,j\in\\{1,\ldots,(p-1)/2\\}$, $i\neq j$. In addition, $\Pi_{y\in
Y_{i}}\,y=\Pi_{x\in R(m^{\prime})}\,\eta_{i}x=\eta_{i}^{m^{\prime}}$, so
$\Pi_{y\in Y}\,y=1$. This implies the result in this case.

Suppose that $p\in\\{3,5\\}$. If $m=27$ then $\phi(m)=18$. Set
$\\{\eta_{1},\ldots,\eta_{9}\\}=\\{\zeta,\zeta^{2},\zeta^{23},\zeta^{5},\zeta^{7},\zeta^{8},\zeta^{10},\\\
\zeta^{11},\zeta^{14}\\}$ for $\zeta\in\Phi(27)$. Then
$\Pi_{i}\eta_{i}=\zeta^{81}=1$ and $\eta_{i}\neq\eta_{i}^{-1}$,
$\eta_{i}\neq\eta_{j}^{\pm 1}$ for $i,j\in\\{1,\ldots,9\\}$, $i\neq j$, whence
the lemma in this case. If $27|m$, the result follows as above.

Suppose that $m=25$, so $k=20$. Set
$\\{\eta_{1},\ldots,\eta_{10}\\}=\\{\zeta,\zeta^{2},\zeta^{3},\zeta^{4},\zeta^{6},\zeta^{18},\zeta^{8},\zeta^{9},\zeta^{11},\zeta^{13}\\}$
for $\zeta\in\Phi(25)$. Then $\Pi_{i}\eta_{i}=\zeta^{75}=1$ and
$\eta_{i}\neq\eta_{i}^{-1}$, $\eta_{i}\neq\eta_{j}^{\pm 1}$ for
$i,j\in\\{1,\ldots,10\\}$, $i\neq j$. If $25|m$, the result follows as above.

(iii) Let $m=p^{a}m^{\prime}$ with $m^{\prime}>1$ and $(p,m^{\prime})=1$. Then
$\Phi(m)=\Phi(p^{a})\Phi(m^{\prime})$. Moreover, if
$\\{y_{1},\ldots,y_{\phi(p^{a})/2}\\}\in\mathcal{P}(p^{a})$ and
$Y=y_{1}\Phi(m^{\prime})\cup\cdots\cup y_{\phi(p^{a})/2}\Phi(m^{\prime})$ then
$Y\in\mathcal{P}(m)$ and the product of the elements in $Y$ equals
$y_{1}\cdots y_{\phi(p^{a})/2}$. If $p^{a}\neq 3,5,9$ then this product is
equal to 1 for some choice of the string $y_{1}\cdots y_{\phi(p^{a})/2}$. So
we are done in this case.

(iv) We can now assume $m$ to be coprime to any prime greater than 5, so
$m=3^{a}5^{b}$ for some $a,b\geq 0$, and $a\leq 2,b\leq 1$. Therefore, we are
left with the cases where $m\in\\{15,45\\}$, which can be varified by
inspection. For instance, if $m=15$ then $\phi(m)=8$; set
$\\{\eta_{1},\ldots,\eta_{4}\\}=\\{\zeta,\zeta^{2},\zeta^{4},\zeta^{8}\\}$ for
$\zeta\in\Phi(15)$. Then $\Pi_{i}\eta_{i}=\zeta^{15}=1$ and
$\eta_{i}\neq\eta_{j}^{\pm 1}$ for $i\neq j$, $i,j\in\\{1,2,3,4\\}$ and
$\eta_{i}\neq\eta_{i}^{-1}$. ∎

######  Corollary 2.11.

Let $m>1$ be an odd natural number and $k=\phi(m)$. Then
$\Phi(m)\subseteq\Delta(m)$.

###### Proof.

The cases with $m=3,5,9$ follow by inspection. If this does not hold, by Lemma
2.10 $\eta_{1}\cdots\eta_{k/2}=1$ for some
$\eta_{1},\ldots,\eta_{k/2}\in\Phi(m)$ such that
$\eta_{1}^{\pm},\ldots,\eta_{k/2}^{\pm}$ are distinct, that is,
$\\{\eta_{1},,\ldots,\eta_{k/2}\\}$ $\in\mathcal{P}(m)$. In this product
replace $\eta_{1}$ by $\eta_{1}^{-1}$. The set
$Y:=\\{\eta_{1}^{-1},\eta_{2},\ldots,\eta_{k/2}\\}$ lies in $\mathcal{P}(m)$
and the product of elements of $Y$ equals $\eta_{1}^{-2}\in\Phi(m)$. This
implies the result as $\Phi(m)=\\{x^{i}:(i,m)=1\\}$ for every fixed
$x\in\Phi(m)$. ∎

######  Lemma 2.12.

Let $m>1$ be an odd natural number and $k=\phi(m)$. Then $(R(m)\setminus
1)\subseteq\Delta(m)$. In other words, if $1\neq\zeta\in R(m)$ and $k=\phi(m)$
then there are elements $\eta_{1},\ldots,\eta_{k/2}\in\Phi(m)$ such that
$\eta_{1}\cdots\eta_{k/2}=\zeta$ and $\eta_{1}^{\pm 1},\ldots,\eta_{k/2}^{\pm
1}$ are distinct.

###### Proof.

By Corollary 2.11, $\Phi(m)\subset\Delta(m)$. In particular, the lemma is true
if $m$ is a prime.

(i) Suppose that $m=p^{a},a>1$ for some prime $p$. Set $m^{\prime}=m/p$. Then
$\Phi(m)=\cup_{i=1}^{p-1}\eta_{i}R(m^{\prime})$, where
$\eta_{1},\ldots,\eta_{p-1}$ are representatives of distinct cosets
$R(m)/R(m^{\prime})$.

Suppose first that $p>5$. By Lemma 2.10, there are
$\eta_{1},\ldots,\eta_{(p-1)/2}\in\Phi(m)$ such that $\eta_{1}^{\pm
m^{\prime}},\ldots,\\\ \eta_{(p-1)/2}^{\pm m^{\prime}}$ are distinct and
$\eta^{m^{\prime}}_{1}\cdots\eta^{m^{\prime}}_{(p-1)/2}=1$. Moreover,
$Y:=\eta_{1}R(m^{\prime})\cup\cdots\cup\eta_{(p-1)/2}R(m^{\prime})\subset\mathcal{P}(m)$
and the product of the elements in $Y$ equals 1. For every
$i=1,\ldots,(p-1)/2$ choose an arbitrary element
$\eta_{i}x_{i}\in\eta_{i}R(m^{\prime})$ and replace it in the product
$\Pi_{y\in Y}\,y$ by the inverse of it. Then we obtain the set
$Y^{\prime}:=(Y\setminus\\{\eta_{1}x_{1},\ldots,\eta_{(p-1)/2}x_{(p-1)/2}\\})\cup\\{\eta^{-1}_{1}x^{-1}_{1},\ldots,\eta_{(p-1)/2}^{-1}x_{(p-1)/2}^{-1}\\}$.
The product of the elements of $Y^{\prime}$ equals
$\eta^{-2}_{1}\cdots\eta^{-2}_{(p-1)/2}x_{1}^{-2}\cdots
x_{(p-1)/2}^{-2}=(\eta_{1}\cdots\eta_{(p-1)/2})^{-2}x_{1}^{-2}\cdots
x_{(p-1)/2}^{-2}$. As $\eta_{1}\cdots\eta_{(p-1)/2}\in R(m^{\prime})$ and
$x_{i}\in R(m^{\prime})$ are arbitrary, we conclude that $\Pi_{y\in
Y^{\prime}}\,y$ can be an arbitrary element of $R(m^{\prime})$ (for a suitable
choice of $x_{i}$’s). As $\Phi(m)\subset\Delta(m)$ by the above, the lemma in
this case follows.

Let $p\leq 5$. One can check that the lemma is true for $m=9,25,27$. In
addition, for $m\in\\{25,27\\}$, by Lemma 2.10, there are
$\eta_{1},\ldots,\eta_{\phi(m)/2}\in\Phi(m)$ such that $\eta_{1}^{\pm
1},\ldots,\eta_{\phi(m)/2}^{\pm 1}$ are distinct and
$\eta_{1}\cdots\eta_{\phi(m)/2}=1$. Then we repeat the above reasoning with
$m/m^{\prime}\in\\{25,27\\}$ in place of $p$ to obtain the result.

(ii) Let $m=p^{a}m^{\prime}$, where $m^{\prime}>1$ and $(p,m^{\prime})=1$.
Then $\Phi(m)=\Phi(p^{a})\Phi(m^{\prime})$. Let
$l=\phi(p^{a})/2=p^{a-1}(p-1)/2$.

Let $\sigma\in R(p^{a})$. By (i), for every $\sigma\neq 1$ there are
$\eta_{1},\ldots,\eta_{l}\in\Phi(p^{a})$ such that $\eta_{1}^{\pm
m^{\prime}},\ldots,\eta_{l}^{\pm m^{\prime}}$ are distinct and
$\eta^{m^{\prime}}_{1}\cdots\eta^{m^{\prime}}_{l}=\sigma$. If $\sigma=1$ and
$p^{a}>9$ then, by Lemma 2.10, there are
$\eta_{1},\ldots,\eta_{l}\in\Phi(p^{a})$ such that $\eta_{1}^{\pm
m^{\prime}},\ldots,\eta_{l}^{\pm m^{\prime}}$ are distinct and
$\eta^{m^{\prime}}_{1}\cdots\eta^{m^{\prime}}_{l}=1$. Then
$Y:=\eta_{1}\Phi(m^{\prime})\cup\cdots\cup\eta_{l}\Phi(m^{\prime})\subset\mathcal{P}(m)$.
Choose arbitrary elements $\eta_{i}x_{i}\in\eta_{i}\Phi(m^{\prime})$ for
$i=1,\ldots,l$ and replace them in the product $\Pi_{y\in Y}\,y$ by their
inverses. Then we obtain the set
$Y^{\prime}:=(Y\setminus\\{\eta_{1}x_{1},\ldots,\eta_{k}x_{l}\\})\cup\\{\eta^{-1}_{1}x^{-1}_{1},\ldots,\eta_{l}^{-1}x_{l}^{-1}\\}\subset\mathcal{P}(m)$.
The product of the elements of $Y^{\prime}$ equals
$\eta^{-2}_{1}\cdots\eta^{-2}_{l}x_{1}^{-2}\cdots
x_{l}^{-2}=(\eta_{1}\cdots\eta_{l})^{-2}x_{1}^{-2}\cdots
x_{l}^{-2}=\sigma^{-2}x_{1}^{-2}\cdots x_{l}^{-2}$. As
$x_{i}\in\Phi(m^{\prime})$ are arbitrary, by Lemma 2.1 we conclude that
$\sigma^{-2}R(m^{\prime})\subset\Delta(m)$. It follows that
$R(m)=R(p^{a})R(m^{\prime})\subset\Delta(m)$ unless $p^{a}\in\\{3,5,9\\}$,
$\sigma=1$ (and then $(R(p^{a})\setminus 1)R(m^{\prime})\subset\Delta(m)$).

So the lemma is true if $p|m$ for some prime $p>5$, and if $27|m$ or $25|m$,
hence we are left with the cases where $m=3^{a}5^{b}$ with $1\leq a\leq 2$ or
$b=1$, that is, with $m\in\\{15,45\\}$. In these cases the lemma can be
checked straightforwardly. ∎

Then, combining Lemmas 2.10 and 2.12, we have

######  Corollary 2.13.

$\Delta(m)=R(m)$ if $m\neq 3,5,9$, and $\Delta(m)=R(m)\setminus\\{1\\}$
otherwise.

###  2.3. Algebraic groups

Let $G$ be a simply connected simple algebraic group of rank $n$ defined over
an algebraically closed field $F$ of characteristic $p\geq 0$, and let $T$ be
a fixed maximal torus of $G$. Let $\Omega={\rm Hom}(T,F^{\times})$ be the
weight lattice of $G$. We also fix a Borel subgroup of $G$ containing $T$;
this defines the fundamental weights $\omega_{1},\ldots,\omega_{n}$ of $G$
ordered as in [1]. They form a $\mathop{\mathbb{Z}}\nolimits$-basis of
$\Omega$. The simple roots of $G$ are denoted by
$\alpha_{1},\ldots,\alpha_{n}$; these are elements of $\Omega$. A weight
$\omega\in\Omega$ is called radical if $\omega$ is a
$\mathop{\mathbb{Z}}\nolimits$-linear combination of simple roots. The weight
$\omega=\sum a_{i}\omega_{i}$ ($a_{i}\in\mathop{\mathbb{Z}}\nolimits$) is
called dominant if $a_{i}\geq 0$ for all $i$; the dominant weights form a
subset $\Omega^{+}$ of $\Omega$. Weight 0 is the one with $a_{i}=0$ for all
$i$. If $\omega,\omega^{\prime}\in\Omega$ then we write
$\omega^{\prime}\preceq\omega$ or $\omega\succeq\omega^{\prime}$ if the weight
$\omega-\omega^{\prime}=\sum b_{j}\alpha_{j}$ with $b_{j}\geq 0$. In [1]
$\Omega$ is embedded into a vector space over the rationals with basis
$\varepsilon_{1},\ldots,\varepsilon_{l}$, where $l=8$ if $G$ is of types
$E_{6},E_{7}$; $n+1$ if $G$ is of type $A_{n}$ and $n$ for the other types;
the expressions of the fundamental weights $\omega_{1},\ldots,\omega_{n}$ and
simple roots $\alpha_{1},\ldots,\alpha_{n}$ in terms of
$\varepsilon_{1},\ldots,\varepsilon_{l}$ are tabulated in [1, Planchees I -
III].

The irreducible representations of $G$ are parameterized by the elements of
$\Omega^{+}$, so we can identify them by writing $\rho_{\omega}$ for
$\omega\in\Omega^{+}$; here $\omega$ is called the highest weight of the
representation in question. The group $W=N_{G}(T)/T$ is called the Weyl group
of $G$. The conjugation action of $N_{G}(T)$ on $T$ induces the action of $W$
on $\Omega$. Note that $\rho(T)$ is always completely reducible so $V$, the
underlying $G$-module of $\rho$, is a direct sum of $T$-stable subspaces, on
each $T$ acts scalarly, and hence define a weight of $G$. The set of such
weights is denoted by $\Omega(V)$. Note that
$\varepsilon_{1},\ldots,\varepsilon_{l}$ can be interpreted as weights of
$\rho_{\omega_{1}}$ for groups $G$ of types $A_{n},B_{n},C_{n}$ and $D_{n}$.

The $G$-module $V$ with highest weight $\omega_{1}$ is called the natural
$G$-module for $G$ of classical types; there exists a basis of $V$ under which
the elements $t\in T$ acts on $V$ as diagonal matrices. This is used to
identify elements $g\in G$ in terms of their Jordan normal form. (If $G$ is of
types $B_{n}$ or $D_{n}$, then $t$ is determined by its matrix on $V$ up to a
multiple from $Z(G)$; this is immaterial for our purposes.) As every
semisimple element $g\in G$ is conjugate to an element of $T$, we can assume
that $g\in T$ so $\rho_{\omega_{1}}(g)$ is a diagonal matrix. If $|g|$ is odd
then $g,g^{\prime}\in T$ are conjugate in $G$ if
$\rho_{\omega_{1}}(g),\rho_{\omega_{1}}(g^{\prime})$ are conjugate in $GL(V)$.

The following lemma is often used without an explicit reference.

######  Lemma 2.14.

Let G be a semisimple algebraic group and $g\in G$ a semisimple element. Let V
be an irreducible G-module. If V has weight $0$ then g has eigenvalue $1$ on
V.

###### Proof.

This follows from the standard facts that the maximal tori of $G$ are
conjugate and every semisimple element is contained in a maximal torus. ∎

The following lemma is well known; for details see [12, Lemma 2.12 and Theorem
2.9]:

######  Lemma 2.15.

Let $G$ be a simple algebraic group of classical type in defining
characteristic $p\geq 0$, and $p\neq 2$ if G is of type $B,C,D$. Let
$\lambda,\mu\in\Omega^{+}$ with $\lambda$ $p$-restricted, and $V_{\lambda}$,
respectively, $V_{\mu}$ the associated irreducible $FG$-modules. Then the
following facts hold.

(1) If $\mu\prec\lambda$ then $\Omega(V_{\mu})\subseteq\Omega(V_{\lambda})$.

(2) If $\lambda+\mu$ is $p$-restricted then
$\Omega(V_{\lambda+\mu})=\Omega(V_{\lambda}\otimes
V_{\mu})=\Omega(V_{\lambda})+\Omega(V_{\mu})$.

(3) If $\lambda\neq 0$ is radical then some root is a weight of $V_{\lambda}$;
otherwise $\Omega(V_{\lambda})$ contains some minuscule weight.

(4) Let $\lambda$ be a $p$-restricted dominant weight. Then
$\Omega(V_{\lambda})=\\{w(\mu)\ |\ \mu\in\Omega^{+},\mu\preceq\lambda,w\in
W\\}$.

Note that minuscule weights are $\omega_{1},\ldots,\omega_{n}$ for type
$A_{n}$, $\omega_{n}$ for type $B_{n}$ and $\omega_{1}$ for $C_{n}$.

######  Lemma 2.16.

[18, Lemma 5.10(3)] Let $G=SL_{n}(F)$, $n>1$ odd, and $H=Sp_{n-1}(F)$. Let V
be an irreducible $G$-module. Then $V|_{H}$ has weight zero.

Recall that an element $g$ of a group $G$ is called real if $g$ is conjugate
to $g^{-1}$.

######  Corollary 2.17.

Let $G=SL_{n}(F)$, $n>1$ odd, and let $g\in G$ be a semisimple element. Let V
be a $FG$-module. Suppose that g is real. Then g has eigenvalue $1$ on $V$.

###### Proof.

By [18, Lemma 5.8], $g$ is contained in a subgroup isomorphic to
$H=Sp_{n-1}(F)$. Then the result follows from Lemma 2.16.∎

######  Lemma 2.18.

For $n\geq 4$ even let $g=\mathop{\rm diag}\nolimits(d_{1},\ldots,d_{n})\in
GL_{n}(V)$ be a diagonal matrix such that $g$ is odd and $g^{-1}$ is similar
to g. Suppose that $d_{i_{1}}d_{i_{2}}d_{i_{3}}=1$ for distinct
$i_{1},i_{2},i_{3}\in\\{1,\ldots,n\\}$. Then g has eigenvalue $1$ on the
$j$-th exterior power $V_{j}$ of V for $j=2,\ldots,n-2$.

###### Proof.

The eigenvalues of $g$ on $V_{j}$ are $d_{i_{1}},\ldots,d_{i_{j}}$ for
distinct $i_{1},\ldots,i_{j}\in\\{1,\ldots,n\\}$. By reordering of
$d_{1},\ldots,d_{n}$ we can assume that $d_{i+1}=d_{i}^{-1}$ for $i<n$ odd. If
$j$ is even then $d_{1}\cdots d_{j}=1$, so we assume $j$ to be odd. Then
$n\geq 6$, and $n=6$ is obvious as well as the cases where $j=3,n-3$ or
$1\in\\{d_{1},\ldots,d_{n}\\}$. Let $n>6,3<j<n-j$ and
$1\notin\\{d_{1},\ldots,d_{n}\\}$. Again we can assume that
$(i_{1},i_{2},i_{3})=(1,3,5)$. Then $d_{1}d_{3}d_{5}d_{7}d_{8}\cdots
d_{j+2}d_{j+3}=1$, whence the result. ∎

The following facts are obvious and we shall use it with no reference.

######  Lemma 2.19.

Let G be a group, and let $g\in G$ be a rational element. Let $\rho$ be a
representation of $G.$

$(1)$ Suppose $E(\rho(g))$ contains a primitive $m$-root of unity for some
integer $m$. Then $\Phi(m)\subseteq E(\rho(g))$.

$(2)$ Let $\rho=\rho_{1}\otimes\rho_{2}$, where both $\rho_{1},\rho_{2}$ are
representations of $G.$ Suppose that $m$ is odd and $\Phi(m)\subseteq
E(\rho_{i}(g))$ for $i=1,2$. Then $R(m)\subseteq E(\rho(g))$.

###### Proof.

(1) Let $\zeta\in E\,(\rho(g))$ be a primitive $m$-root of unity. As $g^{i}$
are conjugate whenever $(i,|g|)=1$, we have $E(\rho(g^{i}))=E(\rho(g))$ and
$\zeta^{i}\in E(\rho(g^{i}))$, we are left to observe that
$\Phi(m)=\\{\zeta^{i}:(i,|g|)=1\\}$.

(2) This follows from Lemma 2.1. ∎

######  Lemma 2.20.

Let G be a simple algebraic group defined over
$\mathop{\mathbb{F}}\nolimits_{p}$ and $g\in G$ a rational semisimple element.
Then g is conjugate to an element of $G(p)$.

###### Proof.

Let $Fr$ be the standard Frobenius map on $G$; then, by definition.
$G(p)=G^{Fr}=\\{g\in G:Fr(g)=g$. We first observe that the $G$-conjugacy class
of $g$ is $Fr$-stable. For this we use a criterion stated in [11, Ch.II, §3,
Corollary 3.3].

Let $T$ be a maximal torus of $G.$ Then $T$ can be chosen so that $Fr(T)=T$
and $Fr(t)=t^{p}$ for every $t\in T$. Let $\rho:G\rightarrow GL(V)$ be an
irreducible rational representation of $G$. Then $\rho(T)$ is contained in a
maximal torus of $GL(V)$. Identifying $GL(V)$ with group of matrices of size
$n=\dim V$ with non-zero determinant and replacing $\rho$ by a suitable
equivalent representation, we can assume that $\rho(T)$ consists of diagonal
matrices. Then $\rho(Fr(t))=\rho(t^{p})=\rho(t)^{p}$ for every $t\in T$. Since
$g$ is semisimple, we have $(p,|\rho(g)|)=1=(p,|g|)$. Since $\rho(x)$ is
rational in $GL(V)$, it follows that $\rho(t)$ and $\rho(Fr(t))=(\rho(t))^{p}$
are conjugate in $GL(V)$. As this holds for every $\rho$, by [11, Ch.II, §3,
Corollary 3.3] it follows that $Fr(t)$ and $t$ are conjugate in $G$. Finally,
by Lang’s theorem, we conclude that the conjugacy class of $t$ meets
$G(p)=G^{Fr}$, see [11, Ch.I, 2.7]. ∎

The following lemma is sometimes used without an explicit reference.

######  Lemma 2.21.

Let G be a simple algebraic group and let $g\in G$ be a rational semisimple
element. If $\rho$ is an irreducible representation of G with highest weight
$\omega$. Then $E(\rho(g))=E(\rho_{p\cdot\omega}(g))$.

###### Proof.

By Lemma 2.20, we can assume that $g\in G(p)=\\{x\in G:Fr(x)=x\\}$. In
addition, $\rho_{p\cdot\omega}(g)=\rho_{\omega}(Fr(g))=\rho_{\omega}(g)$,
whence the result. ∎

The following result is easy and well known.

######  Lemma 2.22.

$(1)$ Let $g\in H=SL_{2n}(F)$ be a real semisimple element. Then g is
conjugate to an element of $G_{1}=SO_{2n}(F)$ and $G_{2}=Sp_{2n}(F)$.

$(2)$ Let $g\in H=SL_{2n+1}(F)$ be a real semisimple element. Then g is
conjugate to an element of $G_{1}=SO_{2n+1}(F)$.

######  Lemma 2.23.

Let $G=Sp_{2n}(F)\subset H=SL_{2n}(F)$. Let
$\lambda_{1},\ldots,\lambda_{2n-1}$ be the fundamenttal weights of $H$, and
let $\tau_{i}$ be an irreducible representation of H with highest weight
$\lambda_{i}$, $i\leq n$.

$(1)$ $\rho_{\omega_{i}}$ is a composition factor of $\tau|_{G};$

$(2)$ if $(i,p)\neq(n,2)$ then every weight of $\tau|_{G}$ is a weight of
$\rho_{\omega_{i}}$.

$(3)$ Let $g\in G$ be a semisimple element. Then
$E(\rho_{\omega_{i}}(g)=E(\tau_{i}(g))$ unless possibly $(i,p)=(n,2)$.

###### Proof.

(1) By [18, Lemma 5.5], every composition factor of $\tau_{i}|_{G}$ is
isomorphic to $\rho_{\omega_{j}}$ for some $j\leq i$ with $i-j$ even. As a
Borel subgroup of $G$ is contained in a Borel subgroup of $H$, by [18, Lemma
5.5] we observe that $\rho_{\omega_{i}}$ is a composition factor
$\tau_{i}|_{G}$.

(2) By Lemma 2.15,
$\Omega(\rho_{\omega_{j}})\subset\Omega(\rho_{\omega_{i}})$, unless possibly
$p=2$. If $i<n$ and $p=2$ then this remains true by [16, Theorem 15]. Whence
the claim.

(3) follows from (2).∎

######  Lemma 2.24.

[12, Lemma 5.2] Let $\Omega$ be the root systems of types $A_{n}$, $n>1$,
$B_{n}$, $n>2$, or $C_{n}$, $n>2$, and let $\omega\in\Omega$ be a dominant
weight. Let $U_{1},U_{2}$ be the subsets of $\Omega$ defined in Table below.

$(1)$ $\omega\succeq\nu$ for some $\nu\in U_{1};$

$(2)$ if $\omega\notin U_{1}$ then $\omega\succeq\nu$ for some $\nu\in U_{2};$

$(3)$ if $\Omega$ is of type $B_{n}$, $\omega\succ 0$ and
$\omega\neq\omega_{1}$ then $\omega\succeq\omega_{2}$.

$\begin{array}[]{|l|c|c|}\hline\cr\Phi&U_{1}&U_{2}\\\ \hline\cr A_{n},n\geq
1&0,\omega_{1},\dots,\omega_{n}&2\omega_{1},2\omega_{n},\omega_{1}+\omega_{n},\omega_{1}+\omega_{i},\omega_{i}+\omega_{n},i=2,\dots,n-1\\\
\hline\cr B_{n},n\geq 2&0,\omega_{n}&\omega_{1},\omega_{1}+\omega_{n}\\\
\hline\cr C_{n},n>2&0,\omega_{1}&\omega_{2},\omega_{3}\\\
\hline\cr\end{array}$

######  Lemma 2.25.

Let $\Omega$ be a root system of type $B_{n}$ and $\omega\in\Omega$ a dominant
weight. Suppose that $\omega\succ\omega_{n}$ and
$\omega\neq\omega_{1}+\omega_{n}$. Then $\omega\succeq\omega_{2}+\omega_{n}$.

###### Proof.

By Lemma 2.24, we have $\omega\succ\omega_{1}+\omega_{n}$. Let $\omega=\sum
a_{i}\omega_{i}$, where $a_{i}$ are non-negative integers. Note that
$0\prec\omega_{1}\prec\cdots\prec\omega_{n-1}\prec 2\omega_{n}$ are radical
weights. As $\omega\succ\omega_{n}$, this is not a radical weight and hence
$a_{n}$ is odd. Set $\omega^{\prime}=\omega-\omega_{n}$. Then
$\omega^{\prime}\succ 0$. By Lemma 2.24, either $\omega^{\prime}=\omega_{1}$
(which is excluded by assumption) or $\omega^{\prime}\succeq\omega_{2}$. It
follows that $\omega\succeq\omega_{2}+\omega_{n}$, as claimed. ∎

##  3\. Rational elements in representations of $SL_{n}(F)$

In this section $G=SL_{n}(F),n>1,$ and we identify a semisimple element $g\in
G$ with $\rho_{\omega_{1}}(g)$. So we can assume that $g=\mathop{\rm
diag}\nolimits(d_{1},\ldots,d_{n})$ is a diagonal matrix of determinant 1. We
view $d_{1},\ldots,d_{n}$ as elements of $F^{\times}$. Set
$D=\\{d_{1},\ldots,d_{n}\\}$. We assume that $|g|$ is odd and $g$ is rational,
equivalently, $\Phi(|d_{i}|)\subseteq D$ for every $i=1,\ldots,n$.

######  Lemma 3.1.

Let $G=SL_{n}(F),n>4,$ and let $g\in G$ be a semisimple odd order element.
Suppose that $g$ is rational. Then $1$ is not an eigenvalue of
$\rho_{\omega_{3}}(g)$ if and only if one of the following holds:

$(1)$ $n=6$, $D=\Phi(9)$ and $|g|=9;$

$(2)$ $n=6$, $D=\Phi(5)\cup\Phi(3)$ and $|g|=15;$

$(3)$ $n=8$, $D=\Phi(15)$ and $|g|=15;$

$(4)$ $n=8$, $D=\Phi(5)\cup\Phi(3)\cup\Phi(3)$ and $|g|=15;$

$(5)$ $n=10$, $D=\Phi(5)\cup\Phi(9)$ and $|g|=45;$

$(6)$ $n=14$, $D=\Phi(15)\cup\Phi(9)$ and $|g|=45$.

In addition, $E(\rho_{\omega_{1}}(g))\subseteq E(\rho_{\omega_{3}}(g))$.

###### Proof.

We can assume that $g=\mathop{\rm diag}\nolimits(d_{1},\ldots,d_{n})$ is a
diagonal matrix, and let $D=\\{d_{1},\ldots,d_{n}\\}$ be the diagonal entries
of $g$. The eigenvalues of $\rho_{\omega_{3}}(g)$ are the products
$d_{i_{1}}d_{i_{2}}d_{i_{3}}$ for distinct elements
$i_{1},i_{2},i_{3}\in\\{1,\ldots,n\\}$.

The additional claim is equivalent to saying that every primitive
$|d_{i}|$-root of unity is an eigenvalue of $\rho_{\omega_{3}}(g)$. This is
almost obvious as for every $i$ there are $j,k$ distint from $i$ such that
$d_{j}d_{k}=1$ and then $d_{i}=d_{i}d_{j}d_{k}\in E(\rho(g))$. In particular,
if $d_{i}=1$ for some $i$ then $1\in E(\rho_{\omega_{3}}(g))$. So we can
assume that $d_{i}\neq 1$ for every $i$.

Furthermore, $1\in\Lambda_{3}(|d_{i}|)\subset\rho_{\omega_{3}}(g)$ unless
$|g_{i}|\in\\{3,5,9,15\\}$. So the lemma is true unless possibly $D$ is a
union of subsets $\Phi(3)$, $\Phi(5)$, $\Phi(9)$, $\Phi(15)$, a priory with
repetitions. In fact, if a repetition occurs, say, $\Phi(m)$ occurs twice then
we can try to choose $d_{i_{1}},d_{i_{2}}$ in one copy of $\Phi(m)$ and
$d_{i_{3}}$ in another copy of it with
$d_{i_{3}}^{-1}=d_{i_{1}}d_{i_{2}}\in\Phi(m)$; this is possible by Lemma 2.4
if $|m|>3$. So $\Phi(m)$ with $m>3$ occurs at most once. In addition,
$\Phi(3)$ occurs at most twice as otherwise we choose
$d_{i_{1}}=d_{i_{2}}=d_{i_{3}}$ in distinct copies of $\Phi(3)$, and then
$d_{i_{1}}d_{i_{2}}d_{i_{3}}=1$.

Next, we rule out the case where $d_{j}$ is a multiple of $d_{i}$ and
$|d_{i}|<|d_{j}|$ for some $i,j\in\\{1,\ldots,n\\}$. Indeed, in this case
either $|d_{i}|=3,|d_{j}|\in\\{9,15\\}$ or $|d_{i}|=5,|d_{j}|=15$ for somr
$d_{i},d_{j}\in D$. Then there are $d_{k},d_{l}\in\Phi(|d_{j}|)$, $k\neq l$,
such that $d_{t}:=d_{k}d_{l}\in\Phi(|d_{i}|)$. Then
$d_{t}^{-1}\in\Phi(|d_{i}|)$ and $d_{t}^{-1}d_{k}d_{l}=1$.

This excludes all possibilities for $D$ except those listed in the lemma. The
fact that $1\notin E(\rho(g))$ for $g$ listed in items $(1)-(6)$ is
straightforward. ∎

######  Remark 3.2.

$(1)$ As the representations $\rho_{\omega_{3}}$ and $\rho_{\omega_{n-3}}$ are
dual to each other, Lemma 3.1 implies a similar result for $i=n-3$.

$(2)$ For arbitrary rational element $g\in G$ primitive $|g|$-roots of unity
do not always occur as eigenvalues of $\rho_{\omega_{3}}(g)$. For instance,
this happens if $D=\cup_{i}\Phi(d_{i})$ with $i=1,2,3,4$ and
$d_{1},d_{2},d_{3},d_{4}$ are coprime to each other.

######  Lemma 3.3.

Suppose that $g\in SL_{n}(F)$ is real. Then $E(\rho_{\omega_{i}}(g))\subseteq
E(\rho_{\omega_{i+2}}(g))$ for $i\leq j\leq n/2$.

###### Proof.

Let $\rho_{\omega_{1}}(g)=\mathop{\rm diag}\nolimits(d_{1},\ldots,d_{n})$.
Then $E(\rho_{\omega_{i}}(g))=\\{d_{j_{1}}\cdots d_{j_{i}}:1\leq
j_{1}<\cdots<j_{i}\leq n\\}$. Let $\zeta\in\Phi(m)$, so $\zeta=d_{j_{1}}\cdots
d_{j_{i}}$ for some distinct $j_{1},\ldots,j_{i}$. Let $D^{\prime}\subseteq D$
be a minimal set such that $d_{j_{1}},\ldots,d_{j_{i}}\in D^{\prime}$ and
$D^{\prime}=(D^{\prime})^{-1}$. Then $|D\setminus D^{\prime}|\geq n-2i$ and
$D\setminus D^{\prime}$ coincides with its inverse. So we can pick two
elements $x,y\in D\setminus D^{\prime}$ with $xy=1$. Then
$\zeta=d_{j_{1}}\cdots d_{j_{i}}=d_{j_{1}}\cdots d_{j_{i}}xy\in
E(\rho_{\omega_{i+2}}(g))$, as required. ∎

######  Lemma 3.4.

Let $G=SL_{n}(F)$, n even, let $g\in G$ be a rational odd order element, and
$\rho$ a non-trivial irreducible representation of G. Then
$E(\rho_{\omega_{1}}(g))\subset E(\rho(g))$, unless n is even,
$\omega=p^{a}\omega_{j}$ with $j$ even, and $\rho_{\omega_{1}}(g)$ has exactly
two eigenvalues of order multiple to $3$.

###### Proof.

Let $\rho_{\omega_{1}}(g)=\mathop{\rm diag}\nolimits(d_{1},\ldots,d_{n})$ and
$D=\\{d_{1},\ldots,d_{n}\\}$. Set $m=|d_{i}|$ for $i\in\\{1,\ldots,n\\}$. We
show that $\Phi(m)\subset\rho(g)$ for $m\neq 3$. Observe that the lemma is
true for $m=1$. Indeed, if $n$ is odd then the result is contained in
Corollary 2.17. If $n$ is even then $g$ is contained in a subgroup $H$ of $G$
isomorphic to $SL_{n-1}(F)$. So the result for $m=1$ follows by applying the
result to an irreducible constituent of $\rho|_{H}$.

From now on we assume that $m>1$. Let $\omega$ be the highest weight of
$\rho$. We first consider some special cases. Note that the lemma is trivial
for $\omega=\omega_{1}$, and the case with $\omega=\omega_{n-1}$ follows by
duality reason.

(i) The lemma is true if $\omega=\omega_{j}$ for $j$ even. In addition, if
$\Phi(3)\subset E(\rho_{\omega_{1}}(g))$ then $xy\in E(\rho(g))$ for all
$x\in\Phi(3)$, $y\in\Phi(l)$ whenever $\Phi(l)\subseteq D$.

Suppose first that $\omega=\omega_{2}$. Then the eigenvalues of $\rho(g)$ are
$d_{i}d_{j}$ for $i,j\in\\{1,\ldots,n\\}$, $i\neq j$. If $m>3$ then the claim
follows from Lemma 2.4. If $m=3$ and $d_{k}\in\\{1,d_{i}\\}$ for some
$k\in\\{1,\ldots,n\\}$ then the inclusion $\Phi(3)\subset\rho(g)$ is obvious;
this holds if $n$ is odd as $d_{j}=1$ for some $j\in\\{1,\ldots,n\\}$. If
there is $m^{\prime}$ with $3|m^{\prime}$ and $\Phi(m^{\prime})\subseteq
D\setminus\Phi(3)$ then the claim is again true by Lemma 2.4. If $\Phi(3)$
occurs once and $\Phi(m^{\prime})\notin D$ whenever $3|m^{\prime}$ then
$E(\rho_{\omega_{2}}(g))=\\{1\\}\cup\\{xy\\}\cup Y$ for some $Y\subseteq
R(|g|/3)$ and all $x\in\Phi(3),y\in D\setminus\Phi(3),$ so $\Phi(3)$ is not
contained in $E(\rho_{\omega_{2}}(g))$. This also implies the additional
claim. (It is trivial that $1\in E(\rho_{\omega_{2}}(g))$.)

For $j$ even with $2<j\leq n/2$ the result follows by Lemma 3.3. For
$\omega=\omega_{n-j}$ the lemma is true by duality reason.

(iii) The lemma is true if $\omega=\omega_{j}$ with $j$ is odd.

For $j=3$ the result follows from Lemma 2.6, and from Lemma 3.3 for $j$ with
$3<j\leq n/2$.

(iv) The lemma is true if $\omega=\omega_{1}+\omega_{n-1}$.

Indeed, the eigenvalues of $\rho(g)$ in this case are $d_{i}d_{j}^{-1}$ for
$i\neq j$, $i,j\in\\{1,\ldots,n\\}$. So $\Phi(m)\Phi(m)\subset\rho(g)$ for
$m=|d_{i}|$. So $R(m)\subseteq E(\rho(g))$ in this case by Lemma 2.1.

(v) The lemma is true if $\omega$ is $p$-restricted.

By Lemma 2.15(3), $\omega\succeq\mu$ for some
$\mu\in\\{\omega_{1},\ldots,\omega_{n-1},0\\}$, and if $0\neq\omega\succ 0$
then $\omega\succeq\omega_{1}+\omega_{n-1}$. In addition, all weights of
$\rho_{\mu}$ are weights of $\rho$, and if $\mu=0$ then all weights of
$\rho_{\omega_{1}+\omega_{n-1}}$ are weights of $\rho$. So the result follows
from the above, unless possibly $\omega\succ\omega_{j}$ with $j$ even,
$\Phi(3)\subset D$ and $\rho_{\omega_{1}}(g)$ has only two eigenvalues of
order 3. If $\omega\neq\omega_{j}$ here then
$\omega\succeq\omega_{1}+\omega_{j-1}$ or $\omega_{j+1}+\omega_{n}$ by Lemma
2.24. Then
$E(\rho_{\omega_{1}+\omega_{j-1}}(g))=E(\rho_{\omega_{1}}(g))E(\rho_{\omega_{j-1}}(g))$
and
$E(\rho_{\omega_{j+1}+\omega_{n}}(g))=E(\rho_{\omega_{j+1}}(g))E(\rho_{\omega_{n}}(g))$.
Here $j-1,j+1$ are odd, so, by (iii) above, $E(\rho_{\omega_{1}}(g))\subseteq
E(\rho_{\omega_{j-1}}(g))\cap E(\rho_{\omega_{j+1}}(g))$. Therefore,
$R(m)=\Phi(m)\Phi(m)\subseteq E(\rho_{\omega_{1}+\omega_{j-1}}(g))\cap
E(\rho_{\omega_{n}+\omega_{j+1}}(g))\subset E(\rho(g))$, as requred.

(vi) The lemma is true if $\omega$ is not $p$-restricted.

By Lemma 2.21, we can assume that $\rho_{\omega}$ decomposes as a tensor
product of at least two non-trivial factors. Suppose first that
$\rho=\rho_{\lambda}\otimes\rho_{\mu}$, where $\rho_{\lambda},\rho_{\mu}$ are
tensor-indecomposable. Then $\lambda=p^{a}\lambda^{\prime}$,
$\mu=p^{b}\mu^{\prime}$, where $\lambda^{\prime},\mu^{\prime}$ are
$p$-restricted and $a,b\in\mathop{\mathbb{Z}}\nolimits^{+}$. As the
eigenvalues of $V_{\lambda}$ and $V_{\lambda^{\prime}}$ are the same (Lemma
2.21), we can deal with
$\rho^{\prime}:=\rho_{\lambda^{\prime}}\otimes\rho_{\mu^{\prime}}$. If
$\Phi(m)\subseteq D$ then, by (v), $\Phi(m)\subseteq
E(\rho_{\lambda^{\prime}}(g))$ unless, possibly, $m=3$. If $\Phi(3)\subset
E(\rho_{\lambda^{\prime}}(g))$ and $\Phi(3)\subset E(\rho_{\mu^{\prime}}(g))$
then $\Phi(m)\subset E(\rho_{\lambda^{\prime}}(g))\cdot
E(\rho_{\mu^{\prime}}(g))$. Otherwise, we can assume that $\Phi(3)$ is not
contained in $E(\rho_{\lambda^{\prime}}(g))$. Then by (i), $xy\in
E(\rho_{\lambda^{\prime}}(g))$ for every $x\in\Phi(3)$, $y\in\Phi(l)$ for
every $l\neq 3$. By the above, $y^{-1}\in E(\rho_{\mu^{\prime}}(g))$. Then
$x\in E(\rho_{\lambda^{\prime}}(g))\cdot E(\rho_{\mu^{\prime}}(g))$, whence
the result in this case. If $\rho_{\omega}$ is a tensor product of more than
two non-trivial factors then we conclude similarly. ∎

###### Proof of Theorem 1.2.

This is a special case of Lemma 3.4. ∎

######  Lemma 3.5.

Let $G=SL_{n}(F)$, $n\geq 10,$ and let $g=\mathop{\rm
diag}\nolimits(d_{1},\ldots,d_{n})\in G$ be an odd order element. Let $\rho$
be an irreducible representation of G with highest weight $\omega_{i}$, $5\leq
i\leq n-5$. Suppose that $g$ is rational. Then $1$ is an eigenvalue of
$\rho(g)$ unless $n=10$, $D=\Phi(5)\cup\Phi(9)$, $|g|=45$ and $i=5$.

###### Proof.

Set $m=|g|$. We can assume that $i\leq n/2$ as the representations
$\rho_{\omega_{i}}$ and $\rho_{\omega_{n-i}}$ are dual. Note that the
eigenvalues of $\rho(g)$ are $d_{j_{1}}\cdots d_{j_{i}}$, where
$j_{1},\ldots,j_{i}\in\\{1,\ldots,n\\}$ are distinct. The case with $i$ even
is obvious so we assume $i$ to be odd. In addition, by Lemma 3.3, $1\in
E(\rho_{\omega_{3}})(g)$ implies $1\in E(\rho_{\omega_{i}})(g)$, so we examine
the cases in Lemma 3.1. As $n\geq 10,$ we are left with the cases in items
(5),(6) of Lemma 3.1.

Suppose that (5) holds. We show that $1$ is not an eigenvalue of
$\rho_{\omega_{5}}(g)$. The eigenvalues of $\rho_{\omega_{5}}(g)$ are
$d_{i_{1}}d_{i_{2}}d_{i_{3}}d_{i_{4}}d_{i_{5}}$ for distinct
$d_{1},d_{i_{2}},d_{i_{3}},d_{i_{4}},d_{i_{5}}\in\Phi(5)\cup\Phi(9)$. Suppose
that $d_{i_{1}}d_{i_{2}}d_{i_{3}}d_{i_{4}}d_{i_{5}}=1$. Let $s,t$ be the
number of terms $d_{i_{1}},\ldots,d_{i_{5}}$ of the left hand side that lie in
$\Phi(5)$,$\Phi(9)$, respectively. By Lemma 2.6, $s,t>0$. Obviously, $st\neq
1$. As $|\Phi(9)|=6$ and $\Pi_{\zeta^{k}\in\Phi(9)}\zeta^{k}=1$, it follows
that $t\neq 5$. Similarly, $s\neq 4$. Then the product of all terms of the
product that are in $\Phi(9)$ equals 1 as well as the product of those in
$\Phi(5)$. Therefore, $(s,t)\in\\{(2,3),(3,2)\\}$. So we have
$d_{i_{1}}d_{i_{2}}d_{i_{3}}=1$ where the terms are either in $\Phi(5)$ or in
$\Phi(9)$. By Lemma 2.6, this is false.

If (6) holds then let $\zeta\in\Phi(15)$, $\mu\in\Phi(9)$ be such that
$\zeta^{5}=\mu^{3}$. Then $\zeta^{-5}=\zeta^{-1}\zeta^{-4}$ and
$\mu.\mu^{4}.\mu^{7}=\mu^{3}$, so
$\mu\cdot\mu^{4}\cdot\mu^{7}\cdot\zeta^{-1}\cdot\zeta^{-4}=1$. Hence the case
with $i=5$ is ruled out. The case with $i=7$ is ruled out by Lemma 3.3. ∎

###### Proof of Theorem 1.1.

The case $\omega=0$ is trivial, and will be ignored. We can assume that the
matrix $\rho_{\omega_{1}}(g)$ on $V_{\omega_{1}}$ is diagonal and let
$D=\\{d_{1},\ldots,d_{n}\\}$ be the diagonal entries of it regarding the
multiplicities.

Set $m={\rm max}\\{|d_{1}|,\ldots,|d_{n}|\\}$. We can assume that
$\\{d_{1},\ldots,d_{k}\\}=\Phi(m)$, where $k=\phi(m)$. In addition, we can
reorder $d_{1},\ldots,d_{n}$ and assume that $d_{i+1}=d_{i}^{-1}$ for $i<n$
odd and $d_{n}=1$ if $n$ is odd.

i) The theorem is true if $n=2$.

In this case $|g|=3$ and $\omega=a\omega_{1}$ for some integer $a>1$. The
eigenvalues of $\rho(g)$ are $\zeta^{a},\zeta^{a-2},\ldots,\zeta^{-a}$, where
$\zeta\in\Phi(3)$. As $\zeta^{3}=1$, the result easily follows in this case.
So we assume $n>2$ until the end of the proof.

(ii) The theorem is true if $n>3$ and $\omega=\omega_{i}$ for
$i\in\\{2,\ldots,n-2\\}$.

This follows from Lemmas 3.1 and 3.5 for $i$ is odd and from Lemma 2.1 for $i$
even.

(iii) The theorem is true if $\omega=\omega_{2}+\omega_{n-1}$.

We have
$\omega_{2}+\omega_{n-1}=2\varepsilon_{1}+2\varepsilon_{2}+\varepsilon_{3}+\cdots+\varepsilon_{n-1}=\varepsilon_{1}+\cdots+\varepsilon_{n}+\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{n}$.
Then
$(\omega_{2}+\omega_{n-1})(g)=(\varepsilon_{1}+\varepsilon_{2}-\varepsilon_{n})(g)=d_{1}d_{2}d_{n}^{-1}$
as $(\varepsilon_{1}+\cdots+\varepsilon_{n})(g)=1$. By reordering
$d_{1},\ldots,d_{n}$ we can assume that $d_{1},d_{2},d_{n}\in\Phi(m)$, and,
moreover, by Lemma 2.7, we can assume that $d_{1}d_{2}d_{n}^{-1}=1$ (if $m=3$
then this is straghtforward).

(iv) The theorem is true if $\omega$ is $p$-restricted.

By Lemma 2.15, $\omega\succ\mu$ implies
$\Omega(\rho_{\mu})\subset\Omega(\rho_{\omega})$, where
$\Omega(\rho_{\mu}),\Omega(\rho_{\omega})$ stand for the set of weights of the
respective representations. If $\mu=0$ then $\rho$ has weight zero, whence the
result in this case. If $\mu\in\\{\omega_{2},\ldots,\omega_{n-2}\\}$ and $g,n$
are as in the exceptional cases of Lemmas 3.1 and 3.5 then $\nu(g)=1$ for some
weight $\nu$ of $\rho_{\mu}$ by Lemmas 3.1 and 3.5; as $\nu$ is also a weight
of $\rho$, the result follows in this cases.

Suppose that $\mu\in\\{\omega_{1},\omega_{n-1}\\}$ or $g,n$ are as in one of
the exceptional cases of Lemmas 3.1 and 3.5. Let $\sigma$ be a minimal
dominant weight such that $\mu\prec\sigma\preceq\omega$. By Lemma 2.24, if
$\mu\in\\{\omega_{1},\ldots,\omega_{n-1}\\}$ then
$\sigma\in\\{\omega_{1}+\omega_{i-1},\omega_{i+1}+\omega_{n-1}\\}$ for
$1<i<n-1$, and $\sigma=\omega_{2}+\omega_{n-1}$ and
$\sigma=\omega_{1}+\omega_{n-2}$ for $i=1,n-1$, respectively. So we have to
inspect the following cases:

(a) $\sigma=\omega_{2}+\omega_{n-1}$ or $\omega_{1}+\omega_{n-2}$ (the latter
can be ignored as these two weights are the highest weights of dual modules);

(b) $n>4$, $\sigma=\omega_{1}+\omega_{2}$ or $\omega_{4}+\omega_{n-1}$ and one
the exceptional cases of Lemmas 3.1 holds;

(c) $\sigma=\omega_{1}+\omega_{4}$ or $\omega_{6}+\omega_{9}$ (the latter can
be ignored by the duality reason) and the exceptional case of Lemmas 3.5
holds.

Case (a) is considered in (iii), so in this case the theorem is true. Consider
(b). Suppose first that $\sigma=\omega_{1}+\omega_{2}$ and $g$ is as in items
$(1)-(6)$ of Lemma 2.5. As
$\omega_{1}+\omega_{2}=2\varepsilon_{1}+\varepsilon_{2}$, by reordering of
$d_{1},\ldots,d_{n}$ we can assume that $d_{1}=\zeta$, $d_{2}=\zeta^{-2}$,
where $\zeta\in\Phi(5)$ in cases $(1)-(5)$, and $\zeta\in\Phi(15)$ in case
$(6)$. Then $(\omega_{1}+\omega_{2})(g)=1$, as desired.

Suppose that
$\sigma=\omega_{4}+\omega_{n-1}=2\varepsilon_{1}+2\varepsilon_{2}+2\varepsilon_{3}+2\varepsilon_{4}+\varepsilon_{5}+\cdots+\varepsilon_{n-1}=\varepsilon_{1}+\cdots+\varepsilon_{n}+\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}+\varepsilon_{4}-\varepsilon_{n}$,
whence
$\mu(g)=(\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}+\varepsilon_{4}-\varepsilon_{n})(g)$.
If $n=6$ then this case can be ignored by the duality reason. So we assume
that $n\geq 8$ and (5) or (6) of Lemma 3.1 holds. Then $\Phi(9)\subset D$; let
$\zeta\in\Phi(9)$. Then, by reordering of $d_{1},\ldots,d_{n}$, we can assume
that $d_{1}=\zeta,$ $d_{2}=\zeta^{2}$, $d_{3}=\zeta^{4}$, $d_{4}=\zeta^{7}$,
$d_{n}=\zeta^{5}$. Then $\sigma(g)=\zeta^{1+2+4+7-5}=1$, as desired.

Consider (c). Here that $n=10$, $g$ is in Lemma 3.5 and we can assume that
$\sigma=\omega_{1}+\omega_{4}$. We have
$\omega=\omega_{1}+\omega_{4}=2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}+\varepsilon_{4}$.
Then we can assume that $d_{1}=\zeta$, $d_{2}=\zeta^{7}$, $d_{3}=\zeta^{4}$,
$d_{4}=\zeta^{5}$, where $\zeta\in\Phi(9)$. Then
$(\omega_{1}+\omega_{4})(g)=(2\varepsilon_{1}+\varepsilon_{2}+\varepsilon_{3}+\varepsilon_{4})(g)=1$.

(v) The theorem is true if $\omega$ is not $p$-restricted.

As $1\in\Phi(m)\Phi(m)$, the reasoning in item (vi) of Lemma 3.4 shows that
the result is true if $|g|>3$. The case with $|g|=3$ is easy and is left to a
reader.∎

##  4\. The symplectic groups

Throughout this section $G=Sp_{2n}(F)$ and $H=SL_{2n}(F)$. Let
$\omega_{1},\ldots,\omega_{n}$ be the fundamental weights of $G$ and
$\lambda_{1},\ldots,\lambda_{2n-1}$ the fundamental weights of $H$. Both
$\omega_{1},\ldots,\omega_{n}$ and $\lambda_{1},\ldots,\lambda_{2n-1}$ can be
viewed as elements of a lattice with basis
$\varepsilon_{1},\ldots,\varepsilon_{2n}$ introduced in [1] and the
expressions of $\omega_{i}$ for $i=1,\ldots,n$ and $\lambda_{j}$ for
$j=1,\ldots,2n-1$ are given in [1, Planches I, III]. In fact,
$\varepsilon_{1},\ldots,\varepsilon_{2n}$ are functions on the set of diagonal
matrices in $H$ defined by $\varepsilon_{i}(\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{2n}))=t_{i}$ for $i=1,\ldots,2n$. Note that
$\varepsilon_{1},\ldots,\varepsilon_{n}$ is a basis of the lattice
$\Omega(G)$. We view $G,H$ as the images of the irreducible representations of
the respective algebraic groups with highest weight $\omega_{1},\lambda_{1}$,
respectively.

######  Lemma 4.1.

$(1)$ Let $\tau$ be an irreducible representation of $Sp_{2n}(F)$ with highest
weight $\omega_{3}$, where $(n,g)$ are as in items $(1)-(6)$ of Lemma 2.5.
Then $1$ is not an eigenvalue of $\tau(g)$.

$(2)$ Let $\tau$ be an irreducible representation of $Sp_{10}(F)$ with highest
weight $\omega_{5}$ and g as in item $(5)$ of Lemma 2.5. Then $1$ is not an
eigenvalue of $\tau(g)$.

###### Proof.

This follows from Lemma 2.22.∎

Let $g\in G$ be a semisimple element. We can assume that $g=\mathop{\rm
diag}\nolimits(d_{1},\ldots,d_{2n})$, where $d_{2n-i+1}=d_{i}^{-1}$. Then $g$
preserves the non-degenerate bilinear form with Gram matrix
$\begin{pmatrix}0&\mathop{\rm Id}\nolimits_{n}\\\ -\mathop{\rm
Id}\nolimits_{n}&0\end{pmatrix}$, and hence we can assume that $g\in G\subset
H$.

######  Lemma 4.2.

Let $G=Sp_{2n}(F)$, and let $\rho$ be an irreducible representation of G with
highest weight $\omega_{i}$, where $1\leq i\leq n$ and $(i,p)\neq(n,2)$. Let
$g\in G$ be a rational semisimple element of odd order. Then
$E(\rho_{\omega_{1}}(g))\subset E(\rho(g))$ unless $i$ is even and $g$ is as
in the exceptional case of Lemma 3.4. In addition, $1$ is not an eigenvalue of
$\rho(g)$ if and only if one of the following holds:

$(1)$ $i=3$ and $2n,g$ are as in items $(1)-(6)$ of Lemma 3.1;

$(2)$ $2n=10$, $i=5$ and $g$ is as in Lemma 3.5.

###### Proof.

Let $\tau_{i}$ be an irreducible representation of $H=SL_{2n}(F)$ with highest
weight $\lambda_{i}$. By Lemma 2.23, $E(\rho_{\omega_{i}}(g))\subseteq
E(\tau_{i}(g))$, where the equality holds unless $(i,p)=(n,2)$. So the result
follows in this case from Lemma 3.4. The additional claim follows from Theorem
1.1.∎

######  Lemma 4.3.

Let $p\neq 2$, $G=Sp_{2n}(F)$, and let $\rho$ be a irreducible representation
of G with highest weight $\omega\neq 0,p^{t}\omega_{1}$. Let $g\in G$ be a
rational semisimple element of odd order.

$(1)$ $1\in E(\rho(g))$ unless $\omega=p^{a}\omega^{\prime}$ for some integer
$a\geq 0$ and $(g,2n,\omega^{\prime})$ are as in Lemmas 3.1 or 3.5.

$(2)$ $E(\rho_{\omega_{1}}(g))\subset E(\rho(g))$ unless
$\omega=p^{a}\omega_{i}$ with $i$ even and g is as in the exceptional case of
Lemma 3.4.

###### Proof.

If $\omega\in\\{p^{a}\omega_{1},\ldots,p^{a}\omega_{n}\\}$ then the lemma
follows from Lemma 4.2.

Suppose first that $\omega$ is $p$-restricted. Then $\omega\succ\mu$, where
$\mu\in\\{\omega_{1},\omega_{2}\\}$ and $E(\rho_{\mu})\subseteq E(\rho)$
(Lemma 2.15).

(a) Suppose that $\mu=\omega_{1}$. Then
$\Omega(\rho_{\omega_{1}})\subset\Omega(\rho_{\omega})$, whence (2) in this
case. By Lemma 4.2, we can assume that
$\omega\notin\\{\omega_{1},\ldots,\omega_{n}\\}$, and then
$\omega\succeq\omega_{1}+\omega_{2}$ by [18, Lemma 2.2, item 5]. The weights
of $\rho_{\omega_{1}+\omega_{2}}$ are exactly the same as those of
$\rho_{\omega_{1}}\otimes\rho_{\omega_{2}}$ (Lemma 2.15(2)). So
$E(\rho_{\omega_{1}+\omega_{2}}(g))=E(\rho_{\omega_{1}}(g))\cdot
E(\rho_{\omega_{2}}(g))$. If $E(\rho_{\omega_{1}}(g))\subset
E(\rho_{\omega_{2}}(g))$, then (1) follows as $E(\rho_{\omega_{1}}(g))$ is a
union of $\Phi(m)$ for certain divisors $m$ of $|g|$, and
$\Phi(m)\Phi(m)=R(m)$ for every $m$ by Lemma 2.1. Suppose that
$E(\rho_{\omega_{1}}(g))$ is not contained in $E(\rho_{\omega_{2}}(g))$. Then
$\Phi(3)\cup\Phi(m)\subseteq E(\rho_{\omega_{1}}(g))$ for some $m$ with
$(3,m)=1$. By Lemma 2.9(2), we have $R(m)\subseteq E(\rho_{\omega_{2}}(g))$,
and $1\in\Phi(m)\cdot R(m)\subseteq E(\rho_{\omega_{1}}(g))\cdot
E(\rho_{\omega_{2}}(g))\subset E(\rho(g))$, as desired.

(b) Suppose that $\mu=\omega_{2}$. Then $1\in E(\rho_{\omega_{2}}(g))$. As
$\omega\succ\omega_{2}$ and $\omega\notin\\{\omega_{1},\ldots,\omega_{n}\\}$,
we have $\omega\succ 2\omega_{1}$ by [18, Lemma 2.2(5)]. The weights of
$\rho_{2\omega_{1}}$ are exactly the same as those of
$\rho_{\omega_{1}}\otimes\rho_{\omega_{1}}$ (Lemma 2.15(2)). So
$E(\rho_{2\omega_{1}}(g))=E(\rho_{\omega_{1}}(g))\cdot
E(\rho_{\omega_{1}}(g))$. As above, it follows that
$E(\rho_{\omega_{1}}(g))\subseteq E(\rho_{2\omega_{1}}(g))\subseteq
E(\rho(g))$, completing the proof for this case and for arbitrary
$p$-restricted $\omega$.

Suppose that $\omega$ is not $p$-restricted. Then $\omega=\sum_{i\geq
0}p_{i}\lambda_{i}$, where $\lambda_{i}$ are $p$-restricted. By Lemma 2.21, we
can assume that the sum here have at least 2 non-zero terms. As in item (vi)
of the proof of Lemma 3.4, we observe that
$E(\rho(g))=\otimes_{i}E(\rho_{\lambda_{i}}(g))$, and this product consists of
at least two terms. As above, $E(\rho_{\lambda_{i}}(g))$ contains
$E(\rho_{\omega_{1}}(g))$ or $E(\rho_{\omega_{2}}(g))$. We have shown above
that $1\cup E(\rho_{\omega_{1}}(g))\subseteq E(\rho_{\omega_{i}}(g))\cdot
E(\rho_{\omega_{j}}(g))$ for $(i,j)\in(1,1),(1,2)$. It is easy to observe that
this also true for $(i,j)=(2,2)$. ∎

Lemma 4.3 implies Theorem 1.4 for $p\neq 2$. So we assume until the end of
this section that $p=2$.

######  Lemma 4.4.

Let $G=Sp_{2n}(F)$, $p=2$, and let $g\in G$ be a rational element of odd
finite order. Let $\rho$ be an irreducible representation of G of highest
weight $\omega=a_{1}\omega_{1}+\cdots+a_{n}\omega_{n}$. Suppose that $a_{n}=0$
and $\omega\neq 0,2^{l}\omega_{1}$ with $l\geq 0$ an integer. Then
$1\notin\rho(g)$ if and only if cases $(1)$ or $(2)$ of Lemma 4.3 holds.

###### Proof.

By Lemma 2.20, we can assume that $g\in X=Sp_{2n}(2)$. By [18, Theorem 1.3],
if $g$ is real then $\rho(g)$ has eigenvalue 1 unless possibly
$\lambda=2^{t}\omega_{i}$ for some $i$. For $g$ rational this case is examined
in Lemma 4.2 for $i<n$ which yields the result in this case. ∎

Next we consider the case where $p=2$ and $\rho=\rho_{\omega_{n}}$. By Lemma
2.20, every rational element in $G$ is contained in a subgroup $X$ isomorphic
to $Sp_{2n}(2)$. Let $V$ be the natural
$\mathop{\mathbb{F}}\nolimits_{2}X$-module, and $V=\oplus_{i=1}^{m}V_{i}$,
where $V_{1},\ldots,V_{m}$ are minimal non-degenerate $g$-stable subspaces of
$V$. Then $2d_{i}:=\dim V_{i}$ is even, and in some basis of $V$ compatible
with the above decomposition one can write $g=\mathop{\rm
diag}\nolimits(g_{1},\ldots,g_{m})$, where $g_{i}\in
Sp_{2d_{i}}(2)=Sp(V_{i})$. If $g_{i}$ is irreducible in $Sp(V_{i})$ then
$|g_{i}|$ divides $2^{d_{i}}+1$, otherwise $|g_{i}|$ divides $2^{d_{i}}-1$.
Following [18, Theorem 4.5], an index $i$ is called singular if
$|g_{i}|=2^{d_{i}}+1$ and $(|g_{i}|,|g_{j}|)=1$ for $j\in\\{1,\ldots,m\\}$
with $j\neq i$, and the number of singular indices is denoted by $Si(g)$.

######  Lemma 4.5.

Let $g\in X$ be a rational semisimple element. Then $Si(g)\leq 2$. More
precisely, if $i$ is a singular index for $g$ then
$(d_{i},|g_{i}|)\in\\{(1,3),(2,5),(3,9)\\}$. In addition, if $i\neq j$ are
singular indices then $5\in\\{|g_{i}|,|g_{j}|\\}$.

###### Proof.

Let $i$ be a singular index for $g$. Then $|g_{i}|=2^{d_{i}}+1$ and
$(|g_{i}|,|g_{j}|)=1$ for $j\neq i$. This implies $g_{i}$ to be rational in
$Sp_{2d_{i}}(2)$. Indeed, if $xgx^{-1}=g^{i}$ for some $x\in Sp_{2n}(2)$ then
$xV_{i}=V_{i}$, whence the restriction of $x$ to $V_{i}$ lies in
$Sp_{2d_{i}}(2)$. (See also [4, Lemma 3.6]). Therefore,
$2d_{i}\geq\phi(2^{d_{i}}+1)$. We show that $g_{i}$ rational implies
$d_{i}\in\\{1,2,3\\}$ with $|g_{i}|=3,5,9$, respectively.

Suppose first that $6\neq 2d_{i}\geq 4$. By Zsigmondy’s theorem [10, Theorem
5.2.14], there is a prime $p$ such that $p|(2^{2d_{i}}-1)$ and coprime to
$2^{t}-1$ for $t<2d_{i}$. This implies $Sp_{2d_{i}}(2)$ to have an irreducible
element $h$ of order $p$, and $p|(2^{d_{i}}+1)$. Therefore, $p>2d_{i}$. So
$p>2d_{i}\geq\phi(2^{d_{i}}+1)\geq\phi(p)=p-1$. If the latter inequality is
strict then $p>\phi(2^{d_{i}}+1)\geq 2(p-1)$ as $\phi(2^{d_{i}}+1)$ is a
multiple of $\phi(p)$, a contradiction. So
$2d_{i}\geq\phi(2^{d_{i}}+1)=\phi(p)=p-1$, whence $2^{d_{i}}+1=p$ and
$2d_{i}\geq 2^{d_{i}}=p-1$, which implies $d_{i}\in\\{1,2\\}$. If $d_{i}=1$
then $|g_{i}|=3$ and if $d_{i}=2$ then $|g_{i}|=5$. If $2d_{i}=6$ then
$|g_{i}|=9$, as claimed. Whence the first statement of the lemma.

As $(|g_{i}|,|g_{k}|)=1$ whenever $i$ is a singular index and $k\neq i$, the
additional statement follows. ∎

######  Lemma 4.6.

Let $p=2$, $G=Sp_{2n}(F)$, and let $g\in G$ be a rational element of odd
finite order. Let $\rho=\rho_{\omega_{n}}$ be an irreducible representation of
G of highest weight $\omega_{n}$. Then either $\rho(g)$ has eigenvalue $1$ or
one of the following holds:

$(1)$ $g=\mathop{\rm diag}\nolimits(g_{1},y)$, where $g_{1}\in Sp_{2k}(F)$,
$k=1,2,3$, $|g_{1}|=2^{k}+1$ and $y\in Sp_{2n-2k}(F)$ is an arbitrary rational
element with $(|y|,|g_{1}|)=1;$

$(2)$ $g=\mathop{\rm diag}\nolimits(g_{1},g_{2},y)$, where $g_{i}\in
Sp_{2k_{i}}(F)$, $(k_{1},k_{2})=(1,2)$ or $(2,3)$, $|g_{i}|=2^{k_{i}}+1$ for
$i=1,2$ and $y\in Sp_{2n-2(k_{1}+k_{2})}(F)$ is an arbitrary rational element
with $(|y|,15)=1$.

###### Proof.

This follows from Lemma 4.5 and [18, Lemma 3.10], where a similar problem is
solved for arbitrary semisimple elements in $Sp_{2n}(2)$. To justify, observe
that every rational element in $G$ is contained in a subgroup $X$ isomorphic
to $Sp_{2n}(2)$ (Lemma 2.20) and $\rho|_{X}$ is irreducible.

If $Si(g)=1$ then we have (1). If $g$ has 2 singular indices $i,j$, say, then
$|g_{i}|,|g_{j}|$ (up to reordering) are $3,5$ or $5,9$. This is recorded in
(2). ∎

Let $\omega=\sum a_{i}\omega_{i}\neq\omega_{n}$. In view of Lemma 4.6, we can
assume that $a_{n}\neq 0$. We use [18, Proposition 4.9], which is stated in
terms of $Si(g)$. To be precise, set
$\omega^{\prime}=a_{1}\omega_{1}+\cdots+a_{n-1}\omega_{n-1}$; then we can
write $\omega^{\prime}=\sum 2^{k}\nu_{k}$, where $\nu_{k}$’s are 2-restricted
weights, and set $\nu=\sum\nu_{k}$. Set
$\nu=b_{1}\omega_{1}+\cdots+b_{n-1}\omega_{n-1}$ and $\delta(\nu)=\sum
b_{j}j$. Then, by [18, Proposition 4.9], $\rho_{\omega}(g)$ does not have
eigenvalue 1 if and only if $\delta(\nu)<Si(g)$. Here $\delta(\nu)\geq 1$ as
$\nu\neq 0$, so $1\notin E(\rho(g))$ implies $Si(g)>1$ and also $Si(g)\leq 2$
by Lemma 4.5. Whence $Si(g)=2$ and $\delta(\nu)=1$. The letter yields
$\nu=\omega_{1}$ and $\omega^{\prime}=2^{a}\omega_{1}$ for some $a\geq 0$,
whence $\omega=2^{k}\omega_{1}+2^{t}\omega_{n}$.

Combining this with previous information, we have:

######  Theorem 4.7.

Let $G=Sp_{2n}(F)$, $p=2$, and let $\rho$ be an irreducible representation of
G of highest weight $\omega=a_{1}\omega_{1}+\cdots+a_{n}\omega_{n}$, where
$a_{n}\neq 0$ and $\omega\neq p^{t}\omega_{n}$. Let $g\in G$ be a rational
element of odd order. Suppose that $1$ is not an eigenvalue of $\rho(g)$. Then
$\omega=p^{s}\omega_{1}+p^{t}\omega_{n}$, $s,t\geq 0$. Moreover, if V is the
natural G-module then there are g-stable non-degenerate subspaces
$V_{1},V_{2},V_{3}$ of $V$ such that $V=V_{1}\oplus V_{2}\oplus V_{3}$, the
restriction $g_{i}$ of g on $V_{i}$ for $i=1,2,3$ are rational elements of
$Sp(V_{i})$ and one of the following holds:

$(1)$ $\dim V_{1}=4$, $|g_{1}|=5$, $\dim V_{2}=2$, $|g_{2}|=3$ and
$(|g_{3}|,15)=1;$

$(2)$ $\dim V_{1}=4$, $|g_{1}|=5$, $\dim V_{2}=6$, $|g_{2}|=9$ and
$(|g_{3}|,15)=1.$

###### Proof.

The first claim of the theorem has already been established in the reasoning
prior to the statement. By Lemma 2.20 we can assume that every rational
element in $G$ is contained in a subgroup $g\in X\cong Sp_{2n}(2)$. Let $U$ be
the natural $\mathop{\mathbb{F}}\nolimits_{2}X$-module. We have also seen
above that $Si(g)=2$, so there are exactly two singular indices; we can assume
these to be $1,2$ and then $|g_{1}|=5$ and $|g_{2}|\in\\{3,9\\}$ by Lemma 4.5.
Therefore, we can write $g=\mathop{\rm diag}\nolimits(g_{1},g_{2},g_{3})$,
where $(|g_{3}|,15)=1$. It follows that there is a $g$-stable orthogonal
decomposition $V=V_{1}\oplus V_{2}\oplus V_{3}$, such that $\dim U_{1}=4$ and
$\dim U_{2}\in\\{2,6\\}$ and $g_{i}$ acts trivially on $V_{j}$ for
$i,j\in\\{1,2,3\\},i\neq j$. So the lemma follows. ∎

###### Proof of Theorem 1.4.

The theorem follows from Theorem 4.3 for $p\neq 2$, and from Lemmas 4.4, 4.6
and Theorem 4.7 for $p=2$. ∎

##  5\. The group of type $B_{n}$

As $B_{2}\cong C_{2}$, we can assume here that $n>2$ and $p\neq 2$. Then
universal (simply connected) version of a simple algebraic group $G$ of type
$B_{n}$ is known to be $Spin_{2n+1}(F)$, and $\rho_{\omega_{1}}(G)\cong
SO_{2n+1}(F)$.

The simple roots of $B_{n}$ are $\alpha_{i}=\varepsilon_{i}-\varepsilon_{i+1}$
for $i=1,\ldots,n-1$ and $\alpha_{n}=\varepsilon_{n}$. In addition,
$\omega_{i}=\varepsilon_{1}+...+\varepsilon_{i}$ for $i<n$ are radical weights
and $\omega_{n}=\frac{1}{2}(\varepsilon_{1}+...+\varepsilon_{n})$ is not
radical. Let $T$ be a maximal torus of $G$ and $t\in T$. Then we can assume
that $\rho_{\omega_{1}}(t)=\mathop{\rm
diag}\nolimits(t_{1},\ldots,t_{n},1,t_{n}^{-1},\ldots,t_{1}^{-1})$ with
$t_{1},\ldots,t_{n}\in F^{\times}$, and $\varepsilon_{i}(t)=t_{i}$.

For a diagonal matrix $D$ we denote by $[D]$ the set of diagonal entries of
$D$. Recall that $\Delta(m)=\\{x\in F:x=\eta_{1}\cdots\eta_{\phi(m)/2}\\}$,
where $\eta_{1},\ldots,\eta_{\phi(m)/2}\in\Phi(m)$ are such that
$\eta_{1}^{\pm},\ldots,\eta_{\phi(m)/2}^{\pm}$ are distinct.

######  Lemma 5.1.

Let $g\in G=Spin_{2n+1}(F)$ be a rational semisimple element and
$D=\rho_{\omega_{1}}(g)=\mathop{\rm
diag}\nolimits(D_{1},\ldots,D_{r},\mathop{\rm Id}\nolimits)\in SO_{2n+1}(F)$,
where $D_{i}$ is a diagonal $(k_{i}\times k_{i})$-matrix $(i=1,\ldots,r)$ such
that $k_{i}=\phi(m_{i})>1$ for some $m_{i}>1$ odd, and $[D_{i}]=\Phi(m_{i})$.
Then $E(\rho_{\omega_{n}}(g))=\Delta(m_{1})\cdots\Delta(m_{r})$.

###### Proof.

We can reorder the diagonal entries of $D$ so that the new matrix
$D^{\prime}$, say, be of shape $\mathop{\rm
diag}\nolimits(M_{1},\ldots,M_{r},\mathop{\rm
Id}\nolimits,M_{r}^{-1},,\ldots,,M_{1}^{-1})$, where $D_{i}=\mathop{\rm
diag}\nolimits(M_{i},M_{i}^{\prime})$ for $i=1,\ldots,r$. Let
$d_{1},\ldots,d_{n}$ be the top diagonal entries of $D^{\prime}$. Then
$\varepsilon_{j}(g)=d_{j}$ for $j=1,\ldots,n$. The weights of
$\rho_{\omega_{n}}$ are
$\frac{1}{2}(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})$. As $|g|$ is odd
and $g$ is rational in $G$, the eigenvalues of $\rho(g)$ and of $\rho(g^{2})$
are the same, and hence the eigenvalues of
$(\frac{1}{2}(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}))(g)$ and of
$(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(g)$ are the same. It follows
that
$E(\rho_{\omega_{n}}(g))=(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(D)=d_{1}^{\pm
1}\cdots d_{n}^{\pm 1}=\Delta(m_{1})\cdots\Delta(m_{r})$. ∎

######  Lemma 5.2.

In notation of Lemma 5.1 we have $E(\rho_{\omega_{n}}(g))=R(|g|)$, unless
there is exactly one $i$ with $m_{i}\in\\{3,5,9\\}$, $(m_{i},m_{j})=1$ for
$j\neq i$ and then $E(\rho_{\omega_{n}}(g))=\Delta(m_{i})R(|g|/m_{i})$, or
there is exactly one $i$ with $m_{i}=5$ and exactly one $j$ with
$m_{j}\in\\{3,9\\}$, $(m_{s},15)=1$ for $s\in\\{1,\ldots,r\\}$, $s\neq i,j$
and then $E(\rho_{\omega_{n}}(g))=\Delta(5)\Delta(m_{j})R(|g|/5m_{j})$.

###### Proof.

By Corollary 2.13, $\Delta(m_{i})=R(m_{i})$ unless $m_{i}\in\\{3,5,9\\}$. One
observes that $\Delta(3)\Delta(9)=R(9)$ and $\Delta(m)\Delta(m)=R(m)$ for
$m\in\\{3,5,9\\}$. In addition, $R(m_{i})\cdot R(m_{j})=R(c)$, and if
$(m_{i},m_{j})>1$ then $(R(m_{i})\setminus 1)\cdot(R(m_{j})\setminus 1)=R(c)$,
where $c$ is the least common multiple of $m_{i},m_{j}$ (see Lemma 2.2). Note
that $|g|$ is the least common multiple of $m_{1},\ldots,m_{r}$. It follows
from Lemma 5.1 that $E(\rho_{\omega_{n}}(g))=R(|g|)$, unless there is exactly
one $i\in\\{1,\ldots,r\\}$ with $m_{i}\in\\{3,9\\}$ or exactly one $i$ with
$m_{i}=5$ and exactly one $j\in\\{1,\ldots,r\\}$ with $m_{j}\in\\{3,9\\}$. In
the former case $E(\rho_{\omega_{n}}(g))=\Delta(m_{i})R(|g|/m_{i})$ with
$m_{i}\in\\{3,5,9\\}$, in the latter one
$E(\rho_{\omega_{n}}(g))=\Delta(5)\Delta(m_{j})R(|g|/5m_{j})$ with
$m_{i}\in\\{3,9\\}$. ∎

######  Lemma 5.3.

Let $G=Spin_{2n+1}(F)$, let $\rho$ be an irreducible representation of G of
highest weight $\omega_{n}$ and $g\in G$ an odd order element. Suppose that g
is rational. Then one of the following holds:

$(1)$ $E(\rho(g))=R(|g|);$

$(2)$ $E(\rho(g))=(R(m)\setminus 1)R(|g|/m)=(R(|g|)\setminus R(|g|/m)$ with
$m\in\\{3,5,9\\}$ and $(m,|g|/m)=1;$

$(3)$ $E(\rho(g))=(R(m)\setminus 1)\Phi(5)R(|g|/5m)$ with $m\in\\{3,9\\}$ and
$(5m,|g|/5m)=1;$ in other words, $E(\rho(g))$ consists of all elements of
$R(|g|)$ whose orders are multiples of $15.$

In addition, in notation of Lemma 5.1 let $D=\mathop{\rm
diag}\nolimits(D_{1},\ldots,D_{r},\mathop{\rm
Id}\nolimits)=\rho_{\omega_{1}}(g)$ and let $m_{i}$ be as in Lemma 5.2. Then
$(2)$ holds if and only if there is exactly one $m_{i}\in\\{3,5,9\\}$ with
$[D_{i}]=\Phi(m_{i})$ and $(m_{i},m_{j})=1$ for $j\neq i$
$(i,j\in\\{1,\ldots,r\\})$, and $(3)$ holds if and only if there is exactly
one $m_{i}\in\\{3,9\\}$ with $[D_{i}]=\Phi(m_{i})$, exactly one $j$ with
$[D_{j}]=\Phi(5)$, $i,j\in\\{1,\ldots,r\\}$, and $(5m_{i},m_{k})=1$ for $k\neq
i,j$.

###### Proof.

By Corollary 2.13, $\Delta(m_{i})=R(m_{i})$ unless $m_{i}\in\\{3,5,9\\}$ and
$\Delta(m)=(R(m)\setminus 1)$ for $m\in\\{3,5,9\\}$. One observes that
$\Delta(3)\Delta(9)=R(9)$ and $\Delta(m)\Delta(m)=R(m)$ for $m\in\\{3,5,9\\}$.
In addition, $R(m_{i})\cdot R(m_{j})=R(c)$, and if $(m_{i},m_{j})>1$ then
$(R(m_{i})\setminus 1)\cdot(R(m_{j})\setminus 1)=R(c)$, where $c$ is the least
common multiple of $m_{i},m_{j}$ (see Lemma 2.2). Note that $|g|$ is the least
common multiple of $m_{1},\ldots,m_{r}$. It follows from Lemma 5.1 that
$E(\rho_{\omega_{n}}(g))=R(|g|)$, unless there is exactly one
$i\in\\{1,\ldots,r\\}$ with $m_{i}\in\\{3,5,9\\}$ or exactly one $i$ with
$m_{i}=5$ and exactly one $j\in\\{1,\ldots,r\\}$ with $m_{j}\in\\{3,9\\}$. In
the former case $E(\rho_{\omega_{n}}(g))=\Delta(m_{i})R(|g|/m_{i})$ with
$m_{i}\in\\{3,5,9\\}$, in the latter one
$E(\rho_{\omega_{n}}(g))=\Delta(5)\Delta(m_{j})R(|g|/5m_{j})$ with
$m_{i}\in\\{3,9\\}$.

For more details, let $V$ be the underlying space for $\rho_{\omega_{1}}$;
then for every semisimple element of $G$ the matrix of $\rho_{\omega_{1}}(g)$
is conjugate to $\mathop{\rm
diag}\nolimits(d_{1},\ldots,d_{n},1,d_{n}^{-1},\ldots,d_{1}^{-1})$ under a
suitable basis of $V$. The set of diagonal matrices (under this basis) forms a
maximal torus of $G$ and $\varepsilon_{i}(g)=d_{i}$ for $i=1,\ldots,n$. As $g$
is rational, $E(\rho(g))=E(\rho(g^{2}))$ for arbitrary representation $\rho$
of $G$.

By Lemma 5.1, the values of
$(\frac{1}{2}(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n}))(g^{2})$ and of
$(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(g)$ are the same. So we
determine $(\pm\varepsilon_{1}\pm\cdots\pm\varepsilon_{n})(g)$, which is are
$d_{1}^{\pm 1}\cdots d_{n}^{\pm 1}$, where $d_{i}\neq d_{j}^{\pm 1}$ for
$i\neq j$, $i,j\in\\{1,\ldots,n\\}$. ∎

######  Theorem 5.4.

Let $G=Spin_{2n+1}(F)$, $p\neq 2$, let $\rho$ be an irreducible representation
of G of highest weight $\omega$ and $g\in G$ an element of finite order.
Suppose that g is rational and $(2,|g|)=1$. Then one of the following holds:

$(1)$ $\rho(g)$ has eigenvalue $1;$

$(2)$ $\omega=p^{j}\omega_{n}$ $(j\geq 0)$, and $(2)$ or $(3)$ of Lemma 5.3
holds;

$(3)$ $\omega=p^{i}\omega_{1}+p^{j}\omega_{n}$ $(i,j\geq 0)$, and $(3)$ of
Lemma 5.3 holds.

###### Proof.

Suppose that (1) does not hold. If $\omega=p^{j}\omega_{n}$ then the result
follows from Lemma 5.3. Suppose that (1) holds and $\omega\neq
p^{j}\omega_{n}$.

$(i)$ Suppose that $\omega$ is $p$-restricted. Then either $\omega\succ 0$ or
$\omega\succ\omega_{n}$. In the former case, by Lemma 2.15, $\rho$ has weight
zero, and the result follows by Lemma 2.14.

Suppose that $\omega\succ\omega_{n}$. Then, by Lemma 5.3,
$\omega_{n}\prec\lambda\preceq\omega$, where $\lambda=\omega_{1}+\omega_{n}$.

(a) Suppose first that $\omega=\lambda$. Note that
$E(\rho(g))=E(\rho_{\omega_{1}}(g))\cdot E(\rho_{\omega_{n}}(g))$ by Lemma
2.15. We use Lemma 5.3. If $(1)$ of Lemma 5.3 holds then $E(\rho(g))=R(|g|)$
and the claim is obvious. Suppose that (2) of Lemma 5.3 holds. We show that
$E(\rho(g))=R(|g|)$. Indeed, we have $\Phi(m)\subset E(\rho_{\omega_{1}}(g))$
and $E(\rho_{\omega_{n}}(g))$ coincides with $\Phi(m)\cdot
R(|g|/m)=R(|g|)\setminus R(|g|/m)$, where $m\in\\{3,5,9\\}$ and $(m,|g|/m)=1$.
As $\Phi(m)\cdot\Phi(m)=R(m)$, it follows that $R(m)\cdot R(|g|/m)\subset
E(\rho(g))$. As $(m,|g|)=1$, we have $R(m)\cdot R(|g|/m)=R(|g|)$.

Suppose that (3) of Lemma 5.3 holds. Then
$E(\rho_{\omega_{n}}(g))=\Phi(5)\cdot(R(m)\setminus 1)\cdot R(|g|/5m)$, where
$m\in\\{3,9\\}$ and $(15,|g|/5m)=1$. In other words, these are all elements of
$R(|g|)$ whose order are multiples of $15$. The spectrum of
$\rho_{\omega_{1}}(g)$ coincides with $\Phi(5)\cup\Phi(m)\cup X$, where
$X\subseteq R(|g|/5m)$. We have

$(\Phi(5)\cup\Phi(m))\cdot\Phi(5)\cdot(R(m)\setminus
1)=(R(5)\cup\Phi(5m))\cdot(R(m)\setminus 1))=$ $=(R(5m)\setminus
R(5))\cup(\Phi(5m)\cdot(R(m)\setminus 1))=(R(5m)\setminus
R(5))\cup(\Phi(5)\cdot R(m)\setminus\Phi(5m))=R(5m)\setminus 1$

since the elements of order 5 occurs in $\Phi(5)\cdot R(m)\setminus\Phi(5m)$.
So $E(\rho(g))=R(|g|)\setminus R(|g|/5m)$. In particular, $1\notin E(\rho(g))$
and $\Phi(5)\cdot R(|g|/5m)\subset E(\rho(g))$.

(b) Suppose that $\omega\succ\lambda$. Then $E(\rho(g))=|g|$.

Indeed, by Lemma 2.25, $\omega\succeq\omega_{2}+\omega_{n}$. Then
$E(\rho_{\omega_{2}}(g))\cdot E(\rho_{\omega_{n}}(g))\subseteq E(\rho(g))$ by
Lemma 2.15. If $E(\rho_{\omega_{n}}(g))=R(|g|)$ then $E(\rho(g))=R(|g|)$, and
we have (1). Suppose that $E(\rho_{\omega_{n}}(g))\neq R(|g|)$; then $(2)$ or
$(3)$ of Lemma 5.3 holds.

As $n>2$, the weights of $\rho_{\omega_{2}}$ are
$\pm\varepsilon_{1}\pm\varepsilon_{2}\cup 0$. Suppose that (2) of Lemma 5.3
holds. By Lemma 5.2, $E(\rho_{\omega_{n}}(g))=\Phi(m)R(|g|/m)$,
$m\in\\{3,5,9\\}$, and then $E(\rho_{\omega_{2}}(g))$ contains the set
$\Phi(m)X$, where $X$ is the set of the eigenvalues of $\rho_{\omega_{1}}(g)$
whose orders are comprime to $m$. As $X\subset R(|g|/m)$, by Lemma 2.1 we have
$E(\rho_{\omega_{2}}(g))\cdot E(\rho_{\omega_{n}}(g))=R(m)\cdot
R(|g|/m)=R(|g|)$ as here $(m,|g|/m)=1$.

Suppose that (3) of Lemma 5.3 holds. By Lemma 5.2,
$E(\rho_{\omega_{n}}(g))=(R(m)\setminus 1)\Phi(5)R(|g|/m)$, $m\in\\{3,9\\}$.
As $\Phi(5m)=\Phi(5)\Phi(m)\subset E(\rho_{\omega_{2}}(g))$, we have

$R(5m)=\Phi(5m)\Phi(5m)\subseteq\Phi(5m)(R(m)\setminus
1)\Phi(5)=\Phi(5)\Phi(m)(R(m)\setminus 1)\Phi(5)$

by Lemma 2.1, and then $E(\rho_{\omega_{2}}(g))\cdot E(\rho_{\omega_{n}}(g))$
contains $R(|g|)$ as desired.

$(ii)$ Suppose that $\omega$ is not $p$-restricted, say, $\omega=\sum_{i\geq
0}p^{i}\mu_{i}$, where $\mu_{i}$’s are $p$-restricted. In view of Lemma 2.21
we have $E(\rho(g))=E(\otimes_{i}\rho_{i}(g))$, where each $\rho_{i}$ is a
$p$-restricted irreducible representation of $G$ with highest weight
$\mu_{i}$. If $1\in E(\rho_{i}(g))$ for every $i$, then $1\in E(\rho(g))$.
Suppose that $1\notin E(\rho(g_{i}))$ for some $i$; we can assume that $i=0$
here. For our purpose we can assume that $\mu_{i}\neq 0$ for every $i$. By
Lemma 2.24, either $\mu_{i}\succeq\omega_{n}$ or $\mu_{i}\succeq\omega_{1}$;
by the above we can assume that $\mu_{0}\succeq\omega_{n}$.

Suppose first that $\mu_{j}\succeq\omega_{n}$ for some $j>0$, and we can
assume that $j=1$. Then we show that $E(\rho_{0}(g))E(\rho_{1}(g))=R(|g|)$. By
Lemma 2.15, $E(\rho_{\omega_{n}}(g))\subseteq E(\rho_{0}(g))\cap
E(\rho_{1}(g))$, so it suffices to observe that
$E(\rho_{\omega_{n}}(g))E(\rho_{\omega_{n}}(g))=R(m)$. By Lemma 5.3, we have
to inspect the cases (2),(3) of that lemma. Observe that $g$ is fixed here, so
one can immediately conclude from the expressions for
$E(\rho_{\omega_{n}}(g))$ in Lemma 5.3 that the required equality holds.

So we are left with the case where $\mu_{j}\succeq\omega_{1}$ for all $j>0$.
If $\mu_{i}\neq\omega_{1}$ then, by (i)(b) above,
$E(\rho_{\mu_{1}}(g))=R(|g|)$, whence $E(\rho(g))=|g|$.

Suppose that $\mu_{i}=\omega_{1}$ for $i>0$. We show that $i\leq 1$.

Indeed, suppose that $\mu_{2}=\omega_{1}.$ As $p>2$ here, we have
$E(\rho_{\omega_{1}}(g))\cdot
E(\rho_{\omega_{1}}(g))=E(\rho_{2\omega_{1}}(g))$ and
$E(\rho_{\omega_{n}}(g))\cdot
E(\rho_{2\omega_{1}}(g))=E(\rho_{2\omega_{1}+\omega_{n}}(g))$ by Lemma 2.15.
Then $1\in E(\rho_{2\omega_{1}+\omega_{n}}(g))=R(|g|)$ by (i)(b), whence the
result. ∎

###### Proof of Theorem 1.3.

This is a self-contained form of Theorem 5.4. ∎

Acknowledgement This work has been completed during the author’s visit to the
Isaac Newton Institute for Mathematical Sciences, Cambridge (June 26th - July
29th, 2022), for his participation in the research programme ‘Groups,
representations and applications: new perspectives’. I would like to thank the
Isaac Newton Institute for hospitality during my visit and acknowledge
financial support, EPSRC grant no. EP/R014604/1, very encouraging scientific
activity at the Institute including seminars and lectures on the latest
progress in the area of the programme.

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