# Odd degree rational characters and the order of rational elements in finite
groups

Nicola Grittini The author is grateful to INPS - Istituto Nazionale di
Previdenza Sociale for supporting him financially during his research.

###### Abstract

We prove that, in a finite group, if every rational irreducible character has
odd degree, then all rational elements are $2$-elements, as it was originally
conjectured by Tiep and Tong-Viet.

##  1 Introduction

Let $G$ be a finite group. An irreducible character
$\chi\in\operatorname{Irr}(G)$ is said to be rational if
$\chi(x)\in\mathbb{Q}$ for each $x\in G$. On the other hand, an element $x\in
G$ is said to be rational in $G$ if it is $G$-conjugated to every generator of
the cyclic group $\langle x\rangle$, and the conjugacy class $x^{G}$ is said
to be rational if $x$ is rational.

It is known that all the irreducible characters of a group $G$ are rational if
and only if all the conjugacy class of $G$ are rational, while the principal
character is the only rational irreducible character of $G$ if and only if $G$
has odd order and, therefore, it does not have any rational elements other
then the identity. Moreover, it is proved in [5] that a finite group $G$ has
exactly two rational irreducible characters if and only if it has exactly two
rational conjugacy classes and, in this case, both rational irreducible
characters have odd degree. On the other hand, it is easy to see that, if $G$
has exactly two rational conjugacy classes, then the non-trivial class needs
to consist of elements of order $2$. Hence, it makes sense to conjecture, as
it was done by Tiep and Tong-Viet in [7], that, if every rational character of
a group $G$ has odd degree, then all rational elements of $G$ have order a
power of $2$. In this paper, we prove said conjecture.

######  Main Theorem.

Let $G$ be a finite group. If every rational irreducible character of $G$ has
odd degree, then all rational elements of $G$ are $2$-elements.

The conjecture was already proved in [7] for a fairly large number of cases.
In fact, it was proved that, if all the rational irreducible characters of a
finite group $G$ have odd degree and no simple group
$\operatorname{PSL}_{2}(3^{f})$, with $f$ odd, is involved in $G$, then $G$ is
solvable. Then, the conjecture was proved in this solvable context. In this
paper, we are able to generalize the approach adopted in [7] for solvable
groups, itself partially borrowed from [5], to the non-solvable case.

As it was already noted in [7], the converse of our Main Theorem does not
hold, since the dihedral group $D_{8}$ has a rational irreducible character of
degree $2$. On the other hand, however, the proof for the solvable case
presented in [7] implies that, if a group $G$ has a rational element of order
$p$, then it has a rational $B_{p}$-character, according to the definition
introduced by Isaacs in [4]. It should not be too difficult to prove that, in
this case, the converse actually holds and, if $G$ has a rational
$B_{p}$-character, then it has a rational $p$-element (and it is also fairly
easy to see that a rational $B_{p}$-character, for $p$ odd, must be of even
degree). Of course, the fact that $B_{p}$-characters are defined only for
$p$-solvable groups strongly limits the extent to which we could apply this
property. If, however, we can find a way to extend the definition of
$B_{p}$-characters also for non-solvable groups, as our Lemma 2.4 might
suggest, then we may find a new interesting relation between rational
characters and rational elements in finite groups.

##  2 Results

We first need to prove a result on the extension of characters in non-abelian
factor groups. We begin with a proposition on character triples isomorphisms
(see [3, Definition 11.23]).

######  Proposition 2.1.

Let $(G,N,\varphi)$ and $(\Gamma,A,\lambda)$ be isomorphic character triples
and suppose that $A\leq\Gamma^{\prime}$ and $\lambda$ is linear. Let $p$ be a
prime number such that $p\mid o(\lambda)$, then $p\mid\varphi(1)o(\varphi)$.

###### Proof.

Suppose $p\nmid\varphi(1)o(\varphi)$, let $P/N$ be a Sylow $p$-subgroup of
$G/N$ and let $Q/A$ be a Sylow $p$-subgroup of $\Gamma/A$ corresponding to
$P/N$ under the isomorphism of character triples. Since
$p\nmid\varphi(1)o(\varphi)$, we have by [3, Corollary 6.28] that $\varphi$
extends to $\hat{\varphi}\in\operatorname{Irr}(P\\!\mid\\!\varphi)$. Let
$\hat{\lambda}\in\operatorname{Irr}(Q\\!\mid\\!\lambda)$ corresponding to
$\hat{\varphi}$ under the isomorphism of character triples, then it follows
from [3, Lemma 11.24] that
$\hat{\lambda}(1)/\lambda(1)=\hat{\varphi}(1)/\varphi(1)=1$ and, thus,
$\lambda$ extends to $Q$. In particular, the $p$-part $\lambda_{p}\neq 1_{A}$
of $\lambda$ also extends to $Q$ and it follows from [3, Theorem 6.26] that
$\lambda_{p}$ extends to $\Gamma$, in contradiction with the assumption that
$A\leq\Gamma^{\prime}$. ∎

For a finite group $S$, we will write $M(S)$ to denote the Schur multiplier of
$S$. The reader can see [3, Definition 11.12] for a definition of Schur
multiplier.

######  Corollary 2.2.

Let $G$ be a finite group and let $N\lhd G$. Let $p$ be a prime number and let
$\varphi\in\operatorname{Irr}(N)$ be a $G$-invariant character such that
$\varphi(1)o(\varphi)$ is a power of $p$. Suppose that
$p\nmid\left|M(G/N)\right|$, then $\varphi$ extends to $G$.

Notice that the corollary holds also if we replace $p$ with a set of primes
$\pi$.

###### Proof.

Let $(\Gamma,A,\lambda)$ be a character triple isomorphic to $(G,N,\varphi)$
such that $A\leq\Gamma^{\prime}$ and $\left|A\right|=\left|M(G/N)\right|$ (see
[3, Theorem 11.17, Corollary 11.20 and Theorem 11.28]). Then it follows from
Proposition 2.1 that $\lambda=1_{A}$ and we have by [3, Theorem 11.7] that
$\varphi$ extends to $G$. ∎

######  Corollary 2.3.

Let $G$ be a finite group and let $N\lhd G$ such that $G/N\cong
S_{1}\times...\times S_{k}$ for some non-abelian simple groups $S_{i}$,
$i=1,...,k$. Let $p$ be a prime number and suppose that
$p\nmid\left|M(S_{i})\right|$ for each $i\in\\{1,...,k\\}$. Suppose that
$\varphi\in\operatorname{Irr}(N)$ is $G$-invariant and that
$\varphi(1)o(\varphi)$ is a power of $p$. Then, $\varphi$ has an irreducible
extension $\chi$ to $G$ such that $o(\chi)=o(\varphi)$.

###### Proof.

We prove it by induction on $k$. Let $N\leq S\lhd G$ such that $S/N\cong
S_{k}$, then it follows from Corollary 2.2 that $\varphi$ has an extension
$\hat{\varphi}\in\operatorname{Irr}(S)$. Since $S/N$ is perfect,
$\hat{\varphi}$ is the only extension of $\varphi$ to $S$ and it follows that
it is $G$-invariant. Moreover, $\det(\hat{\varphi})$ is the unique extension
to $S$ of $\det(\varphi)$ and it follows that $o(\hat{\varphi})=o(\varphi)$.
Hence, $\hat{\varphi}\in\operatorname{Irr}(S)$ is a $G$-invariant character
and $\hat{\varphi}(1)o(\hat{\varphi})=\varphi(1)o(\varphi)$ is a power of $p$.
Since $G/S\cong S_{1}\times...\times S_{k-1}$, it follows by induction that
$\hat{\varphi}$ has an extension $\chi\in\operatorname{Irr}(G)$ such that
$o(\chi)=o(\hat{\varphi})=o(\varphi)$, and
$\chi_{N}=\hat{\varphi}_{N}=\varphi$. ∎

The following lemma is, de facto, an extension of [5, Lemma 6.2] for (some)
non-solvable groups. Since we are not working with $p$-solvable groups, we
cannot just use the properties of Isaacs’ $B_{p}$-characters, as it is done in
[5]. However, we prove that some properties of those characters still hold in
our non-solvable context. It may be actually possible to extend the definition
of $B_{p}$-characters to non-solvable groups, by eventually imposing some
conditions, on the group structure, which are weaker then the $p$-solvability.
This, however, would go too far beyond the scope of our paper.

We should also mention that a weaker result then the following lemma may have
been sufficient for our purposes as well. However, our lemma could be of
interest also of its own, as an extension of [5, Lemma 6.2], and for this
reason the author decided to include it here.

######  Lemma 2.4.

Let $G$ be a finite group, let $p$ be an odd prime number, let $M\lhd G$ and
suppose that, for each subgroup $M\leq H\leq G$ and each non-abelian
composition factor $S$ of $H/M$, $p\nmid\left|M(S)\right|$. Let
$\varphi\in\operatorname{Irr}(M)$ and suppose there exists $U\leq M$ and
$\tau\in\operatorname{Irr}(U)$ such that $\varphi=\tau^{M}$ and
$\tau(1)o(\tau)$ is a power of $p$. Moreover, suppose that, for any
automorphism $b$ of $M$ fixing $\varphi$, $(U,\tau)$ and $(U,\tau)^{b}$ are
conjugated in $M$. Then, there exists a subgroup $U\leq W\leq G$ and a
character $\theta\in\operatorname{Irr}(W\\!\mid\\!\tau)$ such that
$\chi=\theta^{G}$ is irreducible and lies over $\varphi$, $\theta(1)o(\theta)$
is a power of $p$ and, for any automorphism $c$ of $G$ normalizing $M$ and
fixing $\chi$, $(W,\theta)$ and $(W,\theta)^{c}$ are conjugated in $G$.

Moreover, let
$\sigma\in\operatorname{Gal}(\mathbb{Q}_{\left|G\right|}/\mathbb{Q})$ of order
$p-1$ such that it fixes $p^{\prime}$th-roots of unity, and write
$\xi^{\sigma}=\xi^{k}$ when $\xi$ is any $p$th-root of unity. Suppose that $a$
acts as an automorphism on $G$ and it leaves both $M$ and $U$ invariant.
Assume that the order of $a$ on its action on $G/M$ is not divisible by $p$.
If $\tau^{a}=\tau^{\sigma}$, then $a$ normalizes $W$,
$\theta^{a}=\theta^{\sigma}$, and $\chi^{a}=\chi^{\sigma}$.

Finally, assume that there is no element $x\in G/E$ of order $p$ such that
$x^{a^{\sminus 1}}=x^{k}$ for every $a$-invariant $M\leq E\lhd G$. If $\tau$
has values in $\mathbb{Q}_{p}$, then also both $\theta$ and $\chi$ have values
in $\mathbb{Q}_{p}$.

###### Proof.

We prove the lemma by induction on $\left|G:M\right|$. Let $T=I_{G}(\varphi)$
be the inertia subgroup of $\varphi$ in $G$ and suppose first that $T<G$. By
induction, there exists $U\leq W\leq T$ and
$\theta\in\operatorname{Irr}(W\\!\mid\\!\tau)$ such that
$\theta^{T}=\psi\in\operatorname{Irr}(T\\!\mid\\!\varphi)$, then
$\chi=\psi^{G}=(\theta^{T})^{G}$ is irreducible and it lies over $\varphi$. In
particular, if $\tau$ has values in $\mathbb{Q}_{p}$, then by induction so has
$\theta$ and, therefore, so has $\chi$, too. Moreover, if $c$ acts as an
automorphism on $G$ preserving $M$ and it fixes $\chi$, then
$\varphi^{c}=\varphi^{g}$ for some $g\in G$. In particular, $x=cg^{\sminus 1}$
normalizes $M$ and fixes $\varphi$. It follows that $T^{x}=T$ and, since both
$\chi$ and $\varphi$ are fixed by $x$, then $x$ also fixes $\psi$. It follows
by induction that $(W,\theta)^{x}=(W,\theta)^{h}$ for some $h\in T$ and,
therefore, $(W,\theta)^{c}=(W,\theta)^{hg}$.

Finally, suppose that $\tau^{a}=\tau^{\sigma}$, then by induction also
$\theta^{a}=\theta^{\sigma}$. Notice that, for $g\in G$,

$(\theta^{a})^{G}(g)=\frac{1}{\left|W\right|}\sum_{x\in
G}\theta^{\circ}(x^{a^{\sminus 1}}g^{a^{\sminus 1}}(x^{a^{\sminus
1}})^{\sminus 1})=\frac{1}{\left|W\right|}\sum_{x\in
G}\theta^{\circ}(xg^{a^{\sminus 1}}x^{\sminus 1})=\theta^{G}(g^{a^{\sminus
1}})$

and, therefore, $(\theta^{a})^{G}=(\theta^{G})^{a}$. Since also
$(\theta^{\sigma})^{G}=(\theta^{G})^{\sigma}$, we have that
$\chi^{a}=\chi^{\sigma}$.

From now on, we can assume that $\varphi$ is $G$-invariant. Let
$\mathcal{H}\leq\mathrm{Aut}(G)$ consist of all the automorphisms of $G$
leaving $M$ invariant, so that $a\in\mathcal{H}$, and let $M<L\lhd G$ be
minimal such that $\mathcal{H}$ leaves $L$ invariant. Notice that $L/M$ is
either a elementary abelian group or it is the direct product of some non-
abelian simple groups.

Let $g\in L$, then $\varphi^{g}=\varphi$, because $\varphi$ is $G$-invariant,
and it follows from the hypotheses that $(U,\tau)^{g}=(U,\tau)^{h}$ for some
$h\in M$. Let $N=\operatorname{N}_{L}(U)$ and $T=I_{N}(\varphi)$, then we have
that $gh^{\sminus 1}\in T$ and, thus, $g\in TM$. It follows that $L=TM$.
Moreover, we have that $T\cap M=I_{\operatorname{N}_{M}(U)}(\tau)=U$, since
$\tau^{T\cap M}$ is irreducible, and it follows that $T/U\cong L/M$.

We claim that there exists $\gamma\in\operatorname{Irr}(T\\!\mid\\!\tau)$ such
that $\gamma(1)o(\gamma)$ is a power of $p$ and
$\gamma^{L}=\psi\in\operatorname{Irr}(L\\!\mid\\!\varphi)$. We also claim
that, if $d\in\mathcal{H}$ fixes $\psi$, then $(T,\gamma)$ and
$(T,\gamma)^{d}$ are conjugated in $L$. Moreover, if $U^{a}=U$ and
$\tau^{a}=\tau^{\sigma}$, we claim that also $T^{a}=T$ and
$\gamma^{a}=\gamma^{\sigma}$. Finally, if $\tau$ has values in
$\mathbb{Q}_{p}$ and there is no element $x\in L/M$ of order $p$ such that
$x^{a^{\sminus 1}}=x^{k}$, we claim that $\gamma$ has also values in
$\mathbb{Q}_{p}$.

If $T/U\cong L/M$ is either a $p^{\prime}$-group or the direct product of non-
abelian simple groups, then $\tau$ extends to $T$ (see Lemma 2.3 for the non-
abelian case) and we can call $\gamma$ the only extension of $\tau$ to $T$
such that $o(\gamma)=o(\tau)$. We have that
$[(\gamma^{L})_{M},\varphi]=[(\gamma_{U})^{M},\varphi]=[\tau^{M},\varphi]=1$,
thus, $\varphi$ is a constituent of $(\gamma^{L})_{M}$. However,
$\gamma^{L}(1)=\left|L:T\right|\gamma(1)=\left|M:U\right|\tau(1)=\varphi(1)$
and it follows that $\gamma^{L}=\psi\in\operatorname{Irr}(L)$ is an extension
of $\varphi$. If $d\in\mathcal{H}$ fixes $\psi$, then it also fixes $\varphi$
and, by hypothesis, $(U,\tau)^{d}=(U,\tau)^{g}$ for some $g\in G$; it follows
that $dg^{\sminus 1}$ normalizes $U$ and fixes $\tau$. In particular,
$dg^{\sminus 1}$ also normalizes $T=I_{N}(\tau)$ and fixes $\gamma$ (because
of the uniqueness properties of $\gamma$ as an extension of $\tau$). Hence, we
have that $(T,\gamma)^{d}=(T,\gamma)^{g}$. Finally, if $a$ normalizes $U$ and
$\tau^{a}=\tau^{\sigma}$, then $a$ also normalizes $T$, since
$I_{N}(\tau)=I_{N}(\tau^{\sigma})$. By uniqueness, if $\tau^{\sigma a^{\sminus
1}}=\tau$, then also $\gamma^{\sigma a^{\sminus 1}}=\gamma$ and, moreover, if
$\tau$ has values in $\mathbb{Q}_{p}$, then so does $\gamma$. Hence, we have
proved the claim in this case.

Assume now that $T/U\cong L/M$ is an abelian $p$-group and notice that, in
this case, both the order and the degree of every irreducible constituent of
$\tau^{T}$ are powers of $p$. By [8, Lemma 2.2], there exists a unique $M\leq
E\leq L$ maximal with respect to the property of $\varphi$ having a
$L$-invariant extension $\hat{\varphi}\in\operatorname{Irr}(E)$, and
$\hat{\varphi}$ is fully ramified in $L/E$. Since $L=TM$, there exists $U\leq
H_{1}\leq T$ such that $E=H_{1}M$. Moreover, if we apply [8, Lemma 2.2] to
$(T,U,\tau)$, we have that there exists a unique $U\leq H_{2}\leq T$ maximal
with respect to the property of $\tau$ having a $T$-invariant extension
$\hat{\tau}\in\operatorname{Irr}(H_{2})$, and $\hat{\tau}$ is fully ramified
in $T/U$. Now, by proceeding as in the previous paragraph we have that
$\hat{\tau}^{H_{2}M}$ lies over $\varphi$ and, since
$\hat{\tau}^{H_{2}M}(1)=\varphi(1)$, it follows that $\hat{\tau}^{H_{2}M}$
extends $\varphi$ to $H_{2}M$. Since $\hat{\tau}$ is $T$-invariant, then so is
$\hat{\tau}^{H_{2}M}$, and it follows from the maximality of $E=H_{1}M$ that
$H_{2}\leq H_{1}$. On the other hand, we may notice that

$[\hat{\varphi}_{U},\tau]=[\varphi_{U},\tau]=[\varphi,\tau^{M}]=1.$

In particular, we can choose $\hat{\tau}$ such that it is the only irreducible
constituent of $\hat{\varphi}_{H_{2}}$ lying over $\tau$, since
$[\hat{\varphi}_{H_{2}},\tau^{H_{2}}]=[\hat{\varphi}_{U},\tau]=1$, and we have
that there exists $\eta\in\operatorname{Irr}(H_{1})$ such that it is the only
irreducible constituent of both $\hat{\tau}^{H_{1}}$ and
$\hat{\varphi}_{H_{1}}$, since
$[\hat{\varphi}_{H_{1}},\tau^{H_{1}}]=[\hat{\varphi}_{H_{2}},\tau^{H_{2}}]=1$.
In particular, since $[\eta_{U},\tau]\leq[\hat{\varphi}_{U},\tau]=1$ and
$\tau$ is (by definition) $T$-invariant, we have that $\eta$ extends $\tau$.
Suppose that $\eta^{t}\neq\eta$ for some $t\in T$; since both $\hat{\varphi}$
and $\tau$ are $T$-invariant, both $\eta$ and $\eta^{t}$ are irreducible
constituents of $\hat{\varphi}_{H_{1}}$ lying over $\tau$, in contradiction
with the fact that $[\hat{\varphi}_{H_{1}},\tau^{H_{1}}]=1$. Hence, $\eta$ is
$T$-invariant and it follows that $H_{1}=H_{2}=H$, and
$\hat{\tau}^{HM}=\hat{\varphi}$ because they have the same degree. Now, since
$\hat{\tau}$ is fully ramified in $T/H$ and $\hat{\varphi}$ is fully ramified
in $L/HM$, we have that $\hat{\tau}^{T}=e\gamma$ and
$\hat{\varphi}^{L}=e\psi$, with the constant $e$ defined by
$e^{2}=\left|T:H\right|=\left|L:HM\right|$. In particular, we have that
$\gamma^{L}(1)=\left|L:T\right|e\tau(1)=e\left|M:U\right|\tau(1)=e\varphi(1)=\psi(1)$
and, since

$e[\gamma^{L},\psi]=[\gamma^{L},\hat{\varphi}^{L}]=[(\gamma^{L})_{HM},\hat{\varphi}]=e[\hat{\tau}^{HM},\hat{\varphi}]=e>0,$

it follows that $\gamma^{L}=\psi$.

Suppose that $d\in\mathcal{H}$ fixes $\psi$, then it also fixes both
$\hat{\varphi}$ and $\varphi$, since $\psi_{M}=e\varphi$, and it follows from
our hypothesis that $(U,\tau)^{d}=(U,\tau)^{g}$ for some $g\in M$. Hence,
$x=dg^{\sminus 1}$ stabilizes $(U,\tau)$ and, therefore, it normalizes both
$T$ and $H$. Since $\hat{\tau}$ is the only irreducible constituent of both
$\hat{\varphi}_{H}$ and $\tau^{H}$, and since $x$ fixes both $\hat{\varphi}$
and $\tau$, we have that $x$ also fixes $\hat{\tau}$ and, thus, it also fixes
$\gamma$. Hence, we have proved that $(T,\gamma)^{x}=(T,\gamma)$ and,
therefore, $(T,\gamma)^{d}=(T,\gamma)^{g}$.

Finally, if $U^{a}=U$ and $\tau^{a}=\tau^{\sigma}$, then $N^{a}=N$ and
$T^{a}=I_{N}(\tau^{a})=I_{N}(\tau^{\sigma})=T$. Moreover, since $H^{a}$ is
maximal in $T$ such that $\tau^{a}=\tau^{\sigma}$ has a $T$-invariant
extension to $H^{a}$, and $\tau^{\sigma}$ has a $T$-invariant extension if and
only if so has $\tau$, we have that $H^{a}=H$. Notice that $H/U$ is an
elementary abelian $p$-group. Define the function
$u\mapsto(u^{k})^{a}=u^{\rho}$ and notice that it is an automorphism of $H/U$
of order $s$ coprime with $p$. If there is no element $x\in H/U$ of order $p$
such that $x^{a^{\sminus 1}}=x^{k}$, then $\operatorname{C}_{H/U}(\rho)=1$.

Now, since for every $\lambda\in\operatorname{Irr}(H/U)$ and for every $x\in
H/U$ we have that $\lambda^{\sigma}(x)=\lambda(x^{k})$, we are justified in
our abusing the notation when we define the function $\rho=\sigma a^{\sminus
1}:\operatorname{Irr}(H/U)\mapsto\operatorname{Irr}(H/U)$, since
$\lambda^{\rho}(x)=\lambda(x^{\rho})$. In particular, we have that
$\tau^{\rho}=\tau$ and, thus, $\hat{\tau}^{\rho}=\lambda\hat{\tau}$ for some
$\lambda\in\operatorname{Irr}(H/U)$ by Gallagher’s theorem. Since, by coprime
action, we have

$H/U=\operatorname{C}_{H/U}(\rho)\times[H/U,\rho],$

then we can write

$\lambda=\mu(\nu^{\sminus 1})^{\rho}\nu,$

for some $\mu,\nu\in\operatorname{Irr}(H/U)$ with $\mu$ being fixed by $\rho$.
Hence, by replacing $\hat{\tau}$ with $\nu\hat{\tau}$ we have that
$\hat{\tau}^{\rho}=\mu\hat{\tau}$. It follows that
$\hat{\tau}^{\rho^{m}}=\mu^{m}\hat{\tau}$ for every integer $m$ and, in
particular, $\mu^{s}=1_{H/U}$; since $o(\mu)=p$ is coprime with $s$, it
follows that $\mu=1_{H/U}$ and $\hat{\tau}$ is fixed by $\rho$. Moreover, if
$\operatorname{C}_{H/U}(\rho)=1$, then $\hat{\tau}$ is the only irreducible
constituent of $\tau^{H}$ to be fixed by $\rho$ and, thus, $\tau$ and
$\hat{\tau}$ have the same field of values. Finally, since
$\hat{\tau}^{T}=a\gamma$, we have that $\gamma^{\rho}=\gamma$ and $\tau$ and
$\gamma$ have the same field of values. Now the claim is proved.

Now, it follows from our claim that the quadruple $(L,\psi,T,\gamma)$ verifies
the hypotheses of our lemma. Hence, by induction we have that there exists a
subgroup $T\leq W\leq G$ and a character
$\theta\in\operatorname{Irr}(W\\!\mid\\!\gamma)$ such that
$\theta(1)o(\theta)$ is a power of $p$ and
$\theta^{G}=\chi\in\operatorname{Irr}(G\\!\mid\\!\psi)$. In particular,
$\theta$ also lies over $\tau$ and $\chi$ over $\varphi$. Moreover, the
induction hypothesis guarantees that $(W,\theta)^{c}$ and $(W,\theta)$ are
conjugated in $G$ for every $c\in\mathcal{H}$ fixing $\chi$. If $U^{a}=U$ and
$\tau^{a}=\tau^{\sigma}$, then the claim guarantees that $T^{a}=T$ and
$\gamma^{a}=\gamma^{\sigma}$ and it follows by induction that also $W^{a}=W$,
$\theta^{a}=\theta^{\sigma}$ and $\chi^{a}=\chi^{\sigma}$. Finally, if $\tau$
has values in $\mathbb{Q}_{p}$ and no element $x\in G/E$ of order $p$ is such
that $x^{a^{\sminus 1}}=x^{k}$ for every $a$-invariant $M\leq E\lhd G$, then
$\gamma$ has values in $\mathbb{Q}_{p}$ and, by induction, the same is true
for $\theta$ and for $\chi$. Hence, the lemma is proved. ∎

We now list some properties of rational elements; most of them can be found
proved in [5], and they are reported also in [7, Lemma 2.4].

######  Lemma 2.5.

Let $G$ be a finite group and let $N\lhd G$.

  1. (a)

If $x\in G$ is rational in $G$, then $xN\in G/N$ is rational in $G/N$.

  2. (b)

If $g\in H\leq G$ is rational in $H$, then $g$ is also rational in $G$.

  3. (c)

Assume that $x\in G$ has prime order $p$. Then, $x$ is rational in $G$ if and
only if there exists a $p^{\prime}$-element $g\in G$ such that $x^{g}=x^{t}$,
where $t$ (mod $p$) is any generator of the multiplicative group
$\mathbb{Z}_{p}^{\times}$.

  4. (d)

If $x\in G$ is rational, then every power of $x$ is also rational; moreover,
its $\pi$-part $x_{\pi}$ is also rational for every set of primes $\pi$.

  5. (e)

If $x\in G$, $\operatorname{gdc}(o(x),\left|N\right|)=1$ and $xN$ is rational
in $G/N$, then $x$ is rational in $G$.

  6. (f)

If $G/N$ has a rational element of prime order $p$, then $G$ has also a
rational element of order $p$.

  7. (g)

If $H\leq G$ is such that $H=S_{1}\times...\times S_{n}$, for some
$S_{1},...,S_{n}\leq G$, and $x=x_{1}\times...\times x_{n}\in H$ is a rational
element of $G$ of prime power order, then $x_{i}$ is rational in $G$ for each
$i=1,...,n$. Moreover, if $H\lhd G$, then, for each $i=1,...,n$, $x_{i}$ is
rational in $\operatorname{N}_{G}(S_{i})$.

###### Proof.

All these statements except (g) can be found in [5, Lemmas 5.1 and 5.2]. To
prove (g), we may recall that, by definition, if $x$ is rational, then, for
every integer $m$ such that $x^{m}$ generates $\langle x\rangle$, there exists
$g\in G$ such that $x^{g}=x^{m}$. In particular, if $o(x)$ is a power of a
prime number $p$, then $x^{m}$ generates $\langle x\rangle$ if and only if
$p\nmid m$. Since, for any $i\in\\{1,...,n\\}$, $o(x_{i})$ is as well a power
of $p$, we have that, if $x_{i}\neq 1$, $x^{m}$ generates $\langle x\rangle$
if and only if ${x_{i}}^{m}$ generates $\langle x_{i}\rangle$. Hence, for
$i\in\\{1,...,n\\}$ such that $x_{i}\neq 1$ and for any positive integer $m$,
if ${x_{i}}^{m}$ generates $\langle x_{i}\rangle$, then there exists $g\in G$
such that $x^{g}=x^{m}$ and, since
$x^{g}={x_{1}}^{g}\times...\times{x_{n}}^{g}$, we have that
${x_{i}}^{g}={x_{i}}^{m}$ and, thus, $x_{i}$ is rational. Moreover, if $H\lhd
G$, then ${x_{i}}^{m}\in S_{i}\cap{S_{i}}^{g}$ and, since $S_{i}$ is a simple
group and $S_{i},{S_{i}}^{g}\lhd H$, it follows that $S_{i}={S_{i}}^{g}$ and,
thus, $g\in\operatorname{N}_{G}(S_{i})$. Hence, $x_{i}$ and ${x_{i}}^{m}$ are
conjugated in $\operatorname{N}_{G}(S_{i})$ and, since this holds for every
integer $p\nmid m$, it follows that $x_{i}$ is rational in
$\operatorname{N}_{G}(S_{i})$. ∎

It is proved in [7, Theorem C] that, if $G$ is an almost-simple group with
socle $S\ncong\operatorname{PSL}_{2}(3^{f})$, with $f$ odd, then $G$ has a
rational irreducible character of even degree. We now prove that it may happen
also when $S\cong\operatorname{PSL}_{2}(3^{f})$, provided that $G\neq S$.

######  Proposition 2.6.

Let $S\cong\operatorname{PSL}_{2}(3^{f})$, for some odd integer $f$, and let
$S<G\leq\operatorname{Aut}(S)$, where we identify $S$ with
$\operatorname{Inn}(S)$.

  1. (a)

$G$ has a rational element $x\in S$ of odd order if and only if $G$ has an
irreducible rational character of even degree not containing $S$ in its
kernel.

  2. (b)

If there exists $S<H\leq G$ such that $H\cong\operatorname{PGL}_{2}(3^{f})$,
then $G$ has an irreducible rational character of even degree not containing
$S$ in its kernel, and a rational element of odd order $x\in S$.

###### Proof.

Suppose first that there exists $S<H\leq G$ such that
$H\cong\operatorname{PGL}_{2}(3^{f})$, then we can see from [1, Theorem 38.1]
and from [6, Table III] that there exists $\varphi\in\operatorname{Irr}(S)$ of
degree $\varphi(1)=\frac{3^{f}-1}{2}$ such that
$\varphi^{H}=\eta\in\operatorname{Irr}(H)$ is a $G$-invariant rational
character of even degree $\eta(1)=3^{f}-1$. Since $G/H$ is a cyclic group of
odd order dividing $f$, it follows from a repeated application of [3, Theorem
6.30] that $\eta$ has a rational extension $\psi\in\operatorname{Irr}(G)$, of
even degree. Moreover, we can notice that there exists
$x=\begin{pmatrix}1&0\\\ 1&1\end{pmatrix}\in\operatorname{GL}_{2}(3^{f})$, of
order $3$, such that $x^{h}=x^{\sminus 1}$, where $h=\begin{pmatrix}1&0\\\
0&-1\end{pmatrix}$. Since
$x\notin\operatorname{Z}(\operatorname{GL}_{2}(3^{f}))$, the element
$\hat{x}\in\operatorname{PGL}_{2}(3^{f})$ corresponding to $x$ under the
canonical epimorphism is a rational element of order $3$. It follows that $H$
has a rational element of order $3$ and, thus, so has $G$.

We can now assume that $G/S$ is generated by a field automorphism $\tau$ of
odd order dividing $f$, and we let $\psi\in\operatorname{Irr}(S)$ be the
Steinberg character of $S$ and $\hat{\psi}$ be its unique rational extension
to $G$. We claim that, if $\chi\in\operatorname{Irr}(G)$ is rational such that
$\chi\notin\\{1_{G},\hat{\psi}\\}$, then $\chi(1)$ is even and does not
contains $S$ in its kernel. Suppose $\chi(1)$ is odd and let $\varphi$ be an
irreducible constituent of $\chi_{S}$, then $\varphi\neq\psi,1_{S}$ and
$\varphi(1)$ is odd, and it follows from [1, Theorem 38.1] that
$\varphi(1)=\frac{3^{f}-1}{2}$ and $\varphi$ is not rational. Since $\chi$ is
rational, we have that $\varphi$ is not the only irreducible constituent of
$\chi_{S}$ and, since $\varphi$ and $\bar{\varphi}$ are the only two
irreducible characters of $S$ of their degree, we have that
$\chi(1)=e(\varphi(1)+\bar{\varphi}(1))=2e\varphi(1)$, for some positive
integer $e$, is even.

We now claim that, if $x\in G$ is rational of prime power order and
$o(x)\notin\\{1,2\\}$, then $x\in S$ and $o(x)$ is odd. Suppose $o(x)$ is
even; since $G=\langle\tau\rangle\ltimes S$ and there are no rational elements
in $\langle\tau\rangle$, by Lemma 2.5(g) we have that $x\in S$. Now, let
$K\cong\operatorname{SL}_{2}(q)$, where $q=3^{f}$, so that
$K/\operatorname{Z}(K)\cong S$. We can see from [1, Theorem 38.1] that there
exists $a,b\in K$, with $o(a)=q-1$ and $o(b)=q+1$, such that
$\operatorname{Z}(K)\leq(\langle a\rangle\cap\langle b\rangle)$, and every
element of even order in $S\cong K/\operatorname{Z}(K)$ is conjugated to a
power of either $a\operatorname{Z}(K)$ or $b\operatorname{Z}(K)$. Since
$\left|\operatorname{Z}(K)\right|=2$, $4\nmid q-1$ and $8\nmid q+1$, we have
that $o(a\operatorname{Z}(K))$ is odd and $o(b\operatorname{Z}(K))_{2}=2$ and
it follows our claim.

Now, [5, Lemma 11.4] guarantees that $G$ has more than two rational
irreducible characters if and only if it has more than two rational conjugacy
classes. Hence, if $x\in S$ of odd order is rational in $G$, then there exists
$\chi\in\operatorname{Irr}(G)$ rational such that
$\chi\notin\\{1_{G},\hat{\psi}\\}$, and it follows from our first claim that
$\chi(1)$ is even. On the other had, if $G$ has a rational irreducible
character of even degree, then there exists $x\in G$ rational of prime power
order (by Lemma 2.5(d)) which does not belong to the (unique) conjugacy class
of elements of order $2$ in $G$, and it follows from our second claim that
$x\in S$ has odd order. ∎

The following proposition is a rather direct consequence of [7, Theorem C] and
of our Proposition 2.6, and it allows us to determine which non-abelian simple
groups can appear as composition factors in a group where all the rational
irreducible characters have odd degree.

######  Proposition 2.7.

Let $G$ be a finite group and let $M,L\lhd G$, with $L<M$, such that $M/L\cong
S_{1}\times...\times S_{n}$ is the direct product of some isomorphic non-
abelian simple groups $S_{i}\cong S$, for each $i=1,...,n$.

  1. (a)

If $S\ncong\operatorname{PSL}_{2}(3^{f})$ for any odd integer $f\geq 3$, then
$G$ has a rational irreducible character of even degree and a rational element
of odd order.

  2. (b)

If $S\cong\operatorname{PSL}_{2}(3^{f})$, for some odd integer $f\geq 3$, and
there exists $xL\in M/L$ of odd order which is rational in $G/L$, then $G$ has
a rational irreducible character of even degree.

###### Proof.

Working by induction on $\left|G\right|$, we can assume that $L=1$ and $M$ is
a minimal normal subgroup. Let $N=\operatorname{N}_{G}(S_{1})$ and
$C=\operatorname{C}_{G}(S_{1})\leq N$, then $N/C$ is isomorphic to a subgroup
of $\operatorname{Aut}(S_{1})$ containing $S_{1}$. If
$S\ncong\operatorname{PSL}_{2}(3^{f})$, as in case (a), then it follows from
[7, Theorem C] that there exists $\hat{\psi}\in\operatorname{Irr}(N/C)$
rational, of even degree and not containing $S_{1}C/C$ in its kernel. If
$S\cong\operatorname{PSL}_{2}(3^{f})$ and there exists $x\in M$ rational in
$G$ of odd order, then Lemma 2.5(g) allows us to assume, without loss of
generality, that $x\in S_{1}$, and that $x$ is rational in $N$. In particular,
$xC\in S_{1}C/C$ is rational in $N/C$ and it follows from Proposition 2.6 that
also in this case there exists $\hat{\psi}\in\operatorname{Irr}(N/C)$
rational, of even degree and not containing $S_{1}C/C$ in its kernel.

Hence, in both cases, there exists $\psi\in\operatorname{Irr}(N)$, containing
$C$ in its kernel but not $S_{1}$, such that $\psi$ is a rational character of
even degree. If $\delta\in\operatorname{Irr}(S_{1})$ is an irreducible
constituent of $\psi_{S_{1}}$, and $\varphi=\delta\times
1_{S_{2}}\times...\times 1_{S_{n}}\in\operatorname{Irr}(M)$, then
$T=I_{G}(\varphi)\leq N$ and there exists
$\theta\in\operatorname{Irr}(T\\!\mid\\!\varphi)$ such that $\theta^{T}=\psi$.
Hence, $\psi^{G}=(\theta^{T})^{G}$ is an irreducible rational character of
even degree.

Finally, it follows from [5, Theorem 11.1] that, if
$S\ncong\operatorname{PSL}_{2}(3^{f})$ for any odd integer $f\geq 3$, then
$M/L$ contains a rational element of odd prime order. It follows from Lemma
2.5(f) that also $M$ contains a rational element of odd prime order and, thus,
so does $G$. ∎

We are now ready to prove our main theorem. Our proof will use many of the
ideas of the proof of [7, Theorem A].

###### Proof of the Main Theorem.

Suppose, by contradiction, that $G$ is a counterexample of minimal order, and
let $M\lhd G$ be a minimal normal subgroup. Since
$\operatorname{Irr}(G/M)\subseteq\operatorname{Irr}(G)$, by minimality of
$\left|G\right|$ we have that $G/M$ does not contain any non-trivial rational
element of odd order. Let $x\in G$ be a rational element of odd order, then it
follows from Lemma 2.5(a) that $xM=M$ and thus, $x\in M$. Moreover, Lemma
2.5(d) allows us to assume that $x$ has $p$-power order, for some odd prime
number $p$.

Suppose that $M=S_{1}\times...\times S_{n}$ is the direct product of some non-
abelian simple groups, then it follows from Proposition 2.7(a) that, for each
$i=1,...,n$, $S_{i}\cong\operatorname{PSL}_{2}(3^{f})$ for some odd integer
$f$, and then it follows from Proposition 2.7(b) that $G$ has a rational
irreducible character of even degree, in contradiction with our hypothesis.

Assume now that $M$ is an elementary abelian $p$-group, then $x$ has order $p$
and, by Lemma 2.5(c), there exists a $p^{\prime}$-element $g\in G$ such that
$x^{g^{\sminus 1}}=x^{t}$, where $t$ is a generator for
$\mathbb{Z}_{p}^{\times}$. In particular, if $\lambda\in\operatorname{Irr}(M)$
is such that $\lambda(x)\neq 1$, then $\lambda^{g}=\lambda^{t}$. Let
$\sigma\in\operatorname{Gal}(\mathbb{Q}_{\left|G\right|}/\mathbb{Q})$ of order
$p-1$, fixing $p^{\prime}$th-roots of unity and such that
$\xi^{\sigma}=\xi^{t}$, where $\xi$ is a primitive $p$th-root of unity. We
have that $\lambda^{\sigma}=\lambda^{t}=\lambda^{g}$. Moreover, we have that,
for any $M\leq E\lhd G$ and for any $y\in G/E$ of order $p$, $y^{g^{\sminus
1}}\neq y^{t}$, since otherwise it would follow from Lemma 2.5(c) that $y$ is
a rational element of $G/E$ of odd order, in contradiction with the minimality
of $\left|G\right|$. We can also notice that, by Proposition 2.7(a), every
non-abelian simple group involved in $G$ is isomorphic to
$\operatorname{PSL}_{2}(3^{f})$, for some odd integer $f$. Now, every non-
solvable subgroup of $\operatorname{PSL}_{2}(3^{f})$ is isomorphic to
$\operatorname{PSL}_{2}(3^{e})$, for some integer $e\mid f$, by [2, II.8.27
Hauptsatz], and $\left|M(\operatorname{PSL}_{2}(q))\right|=2$ for every odd
prime power $q$. Hence, we have that $p\nmid\left|M(S)\right|$ for every non-
abelian composition factor $S$ of $H/M$, for every $M\leq H\leq G$, whenever
$p$ is an odd prime number. Hence, we can apply our Lemma 2.4, with
$(M,\lambda)$ playing the part of both $(M,\varphi)$ and $(U,\tau)$, and we
have that there exists $\chi\in\operatorname{Irr}(G\\!\mid\\!\lambda)$ having
values in $\mathbb{Q}_{p}$ and such that $\chi^{\sigma}=\chi^{g}=\chi$.

Since $\chi$ has values in $\mathbb{Q}_{p}$ and $\sigma$ is a generator for
$\operatorname{Gal}(\mathbb{Q}_{p}/\mathbb{Q})$, we have that $\chi$ is
rational and, by hypothesis, $\chi(1)$ is odd. It follows from [7, Lemma 2.1]
that $\lambda$ is a real-valued character and, since $p$ is odd, we have that
$\lambda=1_{M}$, in contradiction with our choice. ∎

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