# Finite groups whose real irreducible representations have unique dimensions

Thomas Breuer  T. Breuer: Lehrstuhl für Algebra und Zahlentheorie, RWTH Aachen
Universty, Germany<EMAIL_ADDRESS>,  Frank Calegari  F. Calegari:
Department of Mathematics, University of Chicago, Chicago, IL 60637, USA
<EMAIL_ADDRESS>,  Silvio Dolfi  S. Dolfi: Dipartimento di Matematica,
Università di Firenze, 50134 Firenze, Italy<EMAIL_ADDRESS>,  Gabriel
Navarro  G. Navarro: Departament de Matemàtiques, Universitat de València,
46100 Burjassot, València, Spain<EMAIL_ADDRESS>and  Pham Huu Tiep  P. H.
Tiep: Department of Mathematics, Rutgers University, Piscataway, NJ 08854, USA
<EMAIL_ADDRESS>
###### Abstract.

We determine the finite groups whose real irreducible representations have
different degrees.

###### Key words and phrases:

Characters, Fields of Values, Degrees

###### 2010 Mathematics Subject Classification:

Primary 20D20; Secondary 20C15

T. B. is supported by the German Research Foundation (DFG) – Project-ID
286237555 – within the SFB-TRR 195 Symbolic Tools in Mathematics and their
Applications. F.C. was supported in part by NSF Grant DMS-2001097. The
research of G. N. is supported by Ministerio de Ciencia e Innovación
PID2022-137612NB-I00 funded by MCIN/AEI/ 10.13039/501100011033 and ERDF “A way
of making Europe.” P. H. T. gratefully acknowledges the support of the NSF
(grants DMS-1840702 and DMS-2200850), the Simons Foundation, and the Joshua
Barlaz Chair in Mathematics.

##  1\. Introduction

In this paper, we answer the following question: what are the finite groups
whose real irreducible representations all have different degrees? In fact, we
work under a slightly more general hypothesis. If $G$ is a finite group, let
${\rm Irr}(G)$ be the set of the irreducible complex characters of $G$, and
recall that the Frobenius–Schur indicator of $\chi\in{\rm Irr}(G)$,
$\nu(\chi)$, is 1 if $\chi$ can be afforded by a representation over the real
numbers; it is $-1$, if it is real-valued but cannot be afforded by a real
representation; and it is $0$ if it is not real-valued.

######  THEOREM A.

Let $G$ be a finite group. Suppose that whenever $\alpha,\beta\in{\rm Irr}(G)$
have the same degree and Frobenius–Schur indicator, then $\alpha$ and $\beta$
are complex-conjugate.

  1. (a)

If $G$ is solvable, then $G$ is a factor group of either $(C_{2}\times
C_{2}\times C_{2})\rtimes(C_{7}\rtimes C_{3})$, or $(C_{p}\times
C_{p})\rtimes{\rm SL}_{2}(3)$ for $p\in\\{3,5\\}$.

  2. (b)

If $G$ is non-solvable, then $G$ is almost simple, and isomorphic to one of
the following groups in the list

(1.1) $\mathcal{L}=\\{{\sf A}_{8},{\rm SL}_{3}(2),{\rm M}_{11},{\rm
M}_{22},{\rm M}_{23},{\rm M}_{24},{\rm SU}_{3}(3),{\rm McL},{\rm Th},{\rm
SL}_{2}(8).3,{\rm O}_{8}^{+}(2).3\\}.$

Of these groups, ${\rm McL}$ is the unique group which admits real
representations $\alpha$, $\beta$ of the same dimension ($3520$) but with
distinct Frobenius–Schur indicator. Furthermore, the groups in part (a) of
Theorem A are isomorphic to $C_{3}$, $C_{7}\rtimes C_{3}$, ${\sf A}_{4}$,
${\rm SL}_{2}(3)$, $(C_{3}\times C_{3})\rtimes{\rm SL}_{2}(3)$, $(C_{2}\times
C_{2}\times C_{2})\rtimes(C_{7}\rtimes C_{3})$, and $(C_{5}\times
C_{5})\rtimes{\rm SL}_{2}(3)$.

If $G$ is a finite group, then recall that the degrees of the irreducible real
representations are $\chi(1)$ if $\nu(\chi)=1$, $2\chi(1)$ if $\nu(\chi)=-1$,
and $2\chi(1)$ for the representatives of the non-real irreducible characters
of $G$. (See, for instance, page 108 of [Ser77].) Using this, we easily derive
the following.

######  COROLLARY B.

Let $G$ be a finite group. Suppose that the degrees of the real irreducible
representations of $G$ are all different.

  1. (a)

If $G$ is solvable and non-trivial, then $G$ is either $C_{3}$, ${\sf A}_{4}$,
$C_{7}\rtimes C_{3}$, or $(C_{2}\times C_{2}\times C_{2})\rtimes(C_{7}\rtimes
C_{3})$.

  2. (b)

If $G$ is non-solvable, then $G$ is either ${\sf A}_{8}$, ${\rm M}_{11}$,
${\rm M}_{22}$, ${\rm M}_{23}$, ${\rm M}_{24}$, ${\rm McL}$, ${\rm Th}$ or
${\rm SL}_{2}(8).3$.

######  Remark.

If one asks instead that the degrees of the _complex_ irreducible
representations $G$ are all different, then there are no non-trivial examples;
a minimal example $G$ would be a non-solvable simple group all of whose
representations are rational, and thus by [FS89, Corollary B.1] either
$G=\operatorname{Sp}_{6}(2)$ or ${\rm O}_{8}^{+}(2)$. But neither of these
groups has the required property by inspection. Alternatively, the result
follows from [BCH92, Lemma 1].

##  2\. Solvable groups

For convenience, we say that a finite group $G$ satisfies Hypothesis C if any
two $\alpha,\beta\in{\rm Irr}(G)$ with the same Frobenius–Schur indicator and
degree are complex-conjugate.

We remark that if $G$ satisfies Hypothesis C, then every factor group of $G$
also satisfies Hypothesis C, and moreover for every positive integer $d$,
${\rm Irr}(G)$ contains at most four characters of degree $d$. We also have
${\bf O}^{2}(G)=G$, since $G$ does not contain a non-trivial linear character
of order $2$. (Recall that ${\bf O}^{p}(G)$ is the smallest normal subgroup
$N$ of the finite group $G$ such that $G/N$ is a $p$-group.)

######  Lemma 2.1.

If $G>1$ is a nilpotent group that satisfies Hypothesis C, then $G$ is a
cyclic group of order $3$.

###### Proof.

Since ${\bf O}^{2}(G)=G$, we have that $G$ has odd order. Hence,
$G/G^{\prime}$ has odd order and every non-principal character $\lambda\in{\rm
Irr}(G/G^{\prime})$ is linear and non-real. So ${\rm
Irr}(G/G^{\prime})=\\{1,\lambda,\overline{\lambda}\\}$ and
$G/G^{\prime}=C_{3}$. (In this paper, $G^{\prime}$ denotes the derived
subgroup of $G$, that is, the smallest normal subgroup with abelian factor
group.) In particular, if $\Phi(G)$ is the Frattini subgroup of $G$, then we
have that $G/\Phi(G)$ is cyclic, using that since $G$ is nilpotent, then
$G^{\prime}\subseteq\Phi(G)$. Therefore $G$ itself must be cyclic, and hence
$G=C_{3}$. ∎

In this paper, we use the notation for characters in [Isa76]. Thus, if $H\leq
G$ and $\tau\in{\rm Irr}(H)$, then $\tau^{G}$ denotes the induced character
from $H$ to $G$. Also recall that irreducible characters of groups of odd
order are non-real (by a theorem of Burnside, see Problem 3.16 of [Isa76]).

######  Lemma 2.2.

Assume that $G$ satisfies Hypothesis C. If $N\trianglelefteq\,G$, $G/N$ has
odd order and cyclic Sylow subgroups, and $\tau\in{\rm Irr}(N)$ is non-real,
then $\tau^{G}\in{\rm Irr}(G)$.

###### Proof.

Let $G_{\tau}$ be the inertia subgroup of $\tau$ in $G$. (That is, $G_{\tau}$
is the set of elements $g\in G$ such that $\tau^{g}=\tau$, where $\tau^{g}$ is
the character of $N$ defined by $\tau^{g}(n)=\tau(gng^{-1})$.) The assumption
that the Sylow subgroups of $G/N$ are cyclic implies that $\tau$ has an
extension $\gamma\in{\rm Irr}(G_{\tau})$ (by Corollary 11.22 of [Isa76].). If
$G_{\tau}>N$ then, as $G_{\tau}/N\leq G/N$ is solvable, there exists a linear
non-principal $\lambda\in{\rm Irr}(G_{\tau}/N)$. Hence, by the Clifford
correspondence (Theorem 6.11 of [Isa76]), Gallagher’s theorem (Corollary 6.17
of [Isa76]), and [IN12, Lemma 2.1], the induced characters $\gamma^{G}$ and
$(\lambda\gamma)^{G}$ of $G$ are irreducible of the same Frobenius–Schur
indicator and degree. Thus (since $G$ satisfies Hypothesis C) they are
complex-conjugate. This implies that $\tau$ and its complex conjugate
$\overline{\tau}$ are by some element $g\in G$. Then $\tau^{g^{2}}=\tau$.
Since $G/N$ has odd order, $\tau$ is real, and this is a contradiction. ∎

######  Lemma 2.3.

If $G$ is a non-trivial group of odd order that satisfies Hypothesis C, then
$G$ is either cyclic of order $3$ or a Frobenius group of order $21$.

###### Proof.

Let $N=G^{\prime}$, $K=N^{\prime}$ and $L=K^{\prime}$. We have that
$G/N=C_{3}$. Let $1\neq\lambda,\nu\in{\rm Irr}(N/K)$. Since $G/N$ has prime
order, then either $\lambda$ extends to $G$ or $G_{\lambda}=N$ and therefore
$\lambda^{G}$ is irreducible (by the Clifford correspondence). Since linear
characters have $N$ in their kernel, we then have that $\lambda^{G}$ and
$\nu^{G}\in{\rm Irr}(G)$ should be irreducible. They also have the same
Frobenius–Schur indicator 0, because groups of odd order do not have non-
trivial real-valued irreducible characters. So $\lambda^{G}$ and $\nu^{G}$ are
either equal or complex-conjugate. Thus $\lambda$ is $G$-conjugate to $\nu$ or
to $\bar{\nu}$, and it follows that the action of $G/N$ on ${\rm
Irr}(N/K)-\\{1_{N}\\}$ has two orbits of size 3. (Notice that $\nu$ and
$\bar{\nu}$ cannot be $G$-conjugate, since $G$ has odd order.) Thus $N/K$ has
order 7, and $G/K=(C_{7}:C_{3})$. Suppose that $K>L$. Let
$1\neq\lambda,\nu\in{\rm Irr}(K/L)$. By Lemma 2.2, we have that
$\lambda^{G},\nu^{G}\in{\rm Irr}(G)$. Since these characters have the same
Frobenius–Schur indicator and degree, we conclude that they are equal or
complex-conjugate. Thus, again, $\lambda$ is $G$-conjugate to $\nu$ or to
$\bar{\nu}$. Since $G_{\lambda}=K$ (using that $\lambda^{G}$ is irreducible
and Problem 6.1 of [Isa76]), we have that the $G$-orbit of $\lambda$ has size
21. Therefore $K/L$ has order 43. Thus $K/L$ is cyclic of order 43. But then
$(G/L)/{\bf C}_{G/L}(K/L)$ is abelian, because it is isomorphic to a subgroup
of the automorphism group of a cyclic group. Then $N/L\subseteq{\bf
C}_{G/L}(K/L)$, and therefore $K/L$ is in the center of $N/L$. This is
impossible since there are characters of $K/L$ that induce irreducibly to
$G/L$. ∎

We write ${\rm Irr_{+}}(G)=\\{\chi\in{\rm Irr}(G)\mid\nu(\chi)=1\\}$, and
denote by $A\Gamma(2^{3})$ the affine group $C_{2}^{3}\rtimes(C_{7}\rtimes
C_{3})$.

######  Proposition 2.4.

If $G$ satisfies Hypothesis C and has a normal Sylow $2$-subgroup, then it is
isomorphic to a factor group of either ${\rm SL}_{2}(3)=Q_{8}\rtimes C_{3}$ or
$A\Gamma(2^{3})$.

###### Proof.

Let $N$ be the normal Sylow $2$-subgroup of $G$. By Lemma 2.3, $G/N\cong
C_{3}$ or $G/N\cong C_{7}\rtimes C_{3}$. We can assume $N\neq 1$.

Assume first $G/N\cong C_{3}$. Then $N=G^{\prime}$, since ${\bf O}^{2}(G)=G$.
If $\lambda\in{\rm Irr}(N)$ is linear of order $2$, then $\lambda^{G}\in{\rm
Irr_{+}}(G)$, so by Hypothesis C and Clifford’s Theorem there is a unique
$G$-orbit of such characters. Hence, $N/\Phi(N)\cong C_{2}\times C_{2}$ is an
irreducible $T$-module, where $T\cong C_{3}$ is a Sylow $3$-subgroup of $G$.
This implies that $N/N^{\prime}$ is an indecomposable $T$-module, so it is
homocyclic. For $\lambda\in{\rm Irr}(N/N^{\prime})$ with $o(\lambda)>2$, it
follows from Lemma 2.2 that $\lambda^{G}$ is an irreducible character of $G$
and it is non-real (by [IN12, Lemma 2.1]), so there is at most a $T$-orbit of
characters of order $4$ in ${\rm Irr}(N/N^{\prime})$. Since $N/N^{\prime}$ is
homocyclic and $C_{2^{n}}\times C_{2^{n}}$ has $2^{2n}-2^{2}>3$ linear
characters of order $>2$ when $n>1$, this implies $N/N^{\prime}\cong
C_{2}\times C_{2}$ and $G/N^{\prime}\cong{\sf A}_{4}$. If $N^{\prime}\neq 1$,
then $N$ is either dihedral, semidihedral or a generalized quaternion group
and, observing that ${\rm Aut}(N)$ is not a $2$-group, we conclude that
$N\cong Q_{8}$ and $G\cong{\rm SL}_{2}(3)$.

Assume now that $G/N\cong C_{7}\rtimes C_{3}$ and let $H=TR$ be a complement
of $N$ in $G$, where $T\cong C_{7}$ and $R\cong C_{3}$ are Sylow subgroups of
$G$. By considering a suitable factor group of $G$, we may assume first that
$N$ is minimal normal in $G$. We observe that $T$ does not centralize $N$, as
otherwise (recalling that ${\bf O}^{2}(G)=G$) $N$ is a faithful irreducible
$R$-module, so $N\cong C_{2}\times C_{2}$ and $G$ would be a Frobenius group
with kernel $N\times T$, so $G$ would have $9$ irreducible characters of
degree $3$, a contradiction. By [Isa76, Theorem 15.16] (with $G=H$, $V=N$,
$N=T$, $H=R$, and $H_{0}=1$), we have $\dim N=\dim{\bf C}_{N}(1)=3\dim{\bf
C}_{N}(R)$, and thus there exists a non-trivial element $x\in{\bf C}_{N}(R)$,
so by minimality $N={\langle x\rangle}^{G}={\langle x\rangle}^{T}\cong
C_{2}\times C_{2}\times C_{2}$ and $G\cong A\Gamma(2^{3})$. (Alternatively,
the faithful irreducible representations of $G/N$ over $\mathbb{F}_{2}$ all
have dimension $3$ and are unique up to automorphisms of $G/N$.)

In order to conclude the proof, we show that $N$ is minimal normal in $G$. We
first observe that $N/\Phi(N)$ is an irreducible $H$-module: otherwise, since
it is a completely reducible $H$-module, by the previous paragraph there exist
distinct normal subgroups $K_{1}$ and $K_{2}$ of $G$ such that $G/K_{i}\cong
A\Gamma(2^{3})$, for $i=1,2$, so, as $A\Gamma(2^{3})$ has two irreducible non-
real characters of degree $7$, there would be at least four of them in $G$,
against Hypothesis C.

As $N/\Phi(N)$ is irreducible, then $N/N^{\prime}$ is an indecomposable
$H$-module and, using Lemma 2.2 as in the second paragraph of this proof,
Hypothesis C implies that $N^{\prime}=\Phi(N)$. We now show that
$K=N^{\prime}$ is trivial. Working by contradiction, we may assume (by taking
a suitable quotient of $G$) that $K$ is minimal normal in $G$, so $K={\bf
Z}(N)$. For $1\neq\mu\in{\rm Irr}(K)$, by [Wol78, Lemma 2.2] there is a unique
subgroup $U_{\mu}$ with $K\leq U_{\mu}\leq N$ such that $\mu$ extends to
$\lambda\in{\rm Irr}(U_{\mu})$ and $\lambda$ is fully ramified in $N/U_{\mu}$.
As $\mu$ does not extend to $N$ and $|N/U_{\mu}|$ is a square (using [Isa76,
Theorem 6.18(b)]), it follows $|U_{\mu}/K|=2$. So, ${\rm
Irr}(N|\mu)=\\{\theta_{1},\theta_{2}\\}$ has order $2$, $\theta_{i}(1)=2$ and
$G_{\theta_{i}}=G_{\mu}$ for $i=1,2$. Note that $K$ is a faithful $H$-module,
otherwise $H_{\mu}$ has order divisible by $7$ for some $1\neq\mu\in{\rm
Irr}(K)$, and this gives more than $4$ irreducible characters of $G$ with the
same degree. In particular, we see that $T\in{\rm Syl}_{7}(H)$ acts fixed
point freely on $K$. An application of [Isa76, Theorem 15.16] to the action of
$H$ on the dual group $\widehat{K}$ (as ${\bf C}_{\widehat{K}}(T)={\bf
C}_{K}(T)=1$) yields the existence of a non-principal irreducible character
$\mu$ of $K$ such that $R\leq H_{\mu}$. So, $G_{\theta_{i}}=NR$ for
$\theta_{i}\in{\rm Irr}(N|\mu)$, $i=1,2$, and hence $G$ has at least six
irreducible characters of degree $14$, a contradiction. Hence $K=1$ and and
$G\cong A\Gamma(2^{3})$. ∎

######  Theorem 2.5.

Let $G$ be a solvable group that satisfies Hypothesis C. Then $G$ is
isomorphic to a factor group of either $A\Gamma(2^{3})$ or $(C_{p}\times
C_{p})\rtimes{\rm SL}_{2}(3)$ for $p\in\\{3,5\\}$.

###### Proof.

Let $N={\bf O}^{2^{\prime}}(G)$ and $K={\bf O}^{2}(N)$. If $K=N$, then $N$ has
no quotients of order $2$ or of order prime to $2$; since $N$ is assumed to be
solvable, this implies that $N$ is trivial and we are done by Lemma 2.3; hence
we can assume $K<N$. By Proposition 2.4, we deduce that $G/K$ is isomorphic to
either $A\Gamma(2^{3})$, ${\sf A}_{4}$ or ${\rm SL}_{2}(3)$, and hence we can
assume that $K\neq 1$.

(I.) Suppose first $G/K\cong A\Gamma(2^{3})$ and that $K$ is a minimal normal
subgroup of $G$. Let $|K|=p^{n}$, for an odd prime $p$. We claim that $K$ is
the Fitting subgroup ${\bf F}(G)$. Otherwise, $N\leq{\bf F}(G)$, because $N/K$
is the only minimal normal subgroup of $G/K$. But then $N$ would be nilpotent
and thus isomorphic to a direct sum of $K$ and ${\bf O}_{2}(N)\cong N/K$, and
$G/{\bf O}_{2}(N)$ would be a group of odd order that satisfies Hypothesis C,
contradicting Lemma 2.3. Let $M\in{\rm Syl}_{2}(N)$ and $H={\bf N}_{G}(M)$.
Then $H$ is a complement of $K$ in $G$, as $G=KH$ by the Frattini argument and
$K\cap H={\bf C}_{K}(M)={\bf Z}(N)\cap K\trianglelefteq\,G$, so $K\cap H=1$ as
$K$ is minimal normal in $G$. Let $T\cong C_{7}$ be a Sylow $7$-subgroup of
$H$. Then $MT$ is a Frobenius group and hence ${\bf C}_{K}(T)\neq 1$ by
[Isa76, Theorem 15.16]. Thus, there exists a non-principal $\mu\in{\rm
Irr}(K)$ such that $T\leq H_{\mu}$. Note that the inertia subgroup $N_{\mu}$
is a proper subgroup of $N$, since $\mu$ extends to $N_{\mu}$ by coprimality,
and hence $N_{\mu}=K$ because $N_{\mu}$ is $T$-invariant and $N/K$ is an
irreducible $T$-module. Thus $\theta=\mu^{N}\in{\rm Irr}(N)$. We show that
$\theta$ is non-real. In fact, otherwise there exists an element $x\in M$ (as
$M$ is a Sylow $2$-subgroup of $G$) such that
$\mu^{x}=\overline{\mu}=\mu^{-1}$. So, considering the action of $H$ on
$\widehat{K}$, ${\bf N}_{M}(\langle\mu\rangle)$ is a non-trivial and
$T$-invariant subgroup of $M$. Hence ${\bf N}_{M}(\langle\mu\rangle)=M$, since
$M$ is an irreducible $T$-module. But then ${\bf C}_{M}(\langle\mu\rangle)$ is
a $T$-invariant subgroup of index $2$ in $M$, a contradiction. As
$\theta\in{\rm Irr}(N)$ is non-real, then $\theta^{G}\in{\rm Irr}(G)$ by Lemma
2.2, which is a contradiction as $T\leq G_{\mu}\leq I_{G}(\theta)$. Therefore,
$K=1$ if $G/K\cong A\Gamma(2^{3})$.

(II.) We now assume $G/K\cong{\sf A}_{4}$ and, again, that $K$ is a minimal
normal subgroup of $G$, $|K|$ odd. As above, $K={\bf F}(G)$, so ${\bf
C}_{K}(N)=1$ and by the Frattini argument $K$ has a complement $H=MR\cong{\sf
A}_{4}$ in $G$, where $M\in{\rm Syl}_{2}(G)$ and $R\in{\rm Syl}_{3}(G)$.
Again, by [Isa76, Theorem 15.16] there exists a non-principal and
$R$-invariant $\mu\in{\rm Irr}(K)$ and, as $M_{\mu}$ is an $R$-invariant and
proper (as ${\bf C}_{\widehat{K}}(M)={\bf C}_{K}(M)=1$) subgroup of $M$, we
deduce that $M_{\mu}=1$ and hence $\theta=\mu^{N}\in{\rm Irr}(N)$. As in the
previous paragraph, one shows that $\theta$ is non-real, giving a
contradiction by Lemma 2.2.

(III.) Finally, we assume $G/K\cong{\rm SL}_{2}(3)$.

(III.a) To start with, we assume that $K$ is a $p$-group for some prime $p$.
By Proposition 2.4, $p>2$. Let $Q\cong Q_{8}$ be a Sylow $2$-subgroup of $G$
and $Z={\bf Z}(Q)$.

If $K$ is abelian, then $K={\bf C}_{K}(Z)\times[K,Z]$ by coprimality. Since
$KZ\trianglelefteq\,G$, $[K,Z]=[K,KZ]\trianglelefteq\,G$ and then
$L=[K,Z]Z\trianglelefteq\,G$, since $L/[K,Z]$ is the Sylow $2$-subgroup of the
abelian normal subgroup $KZ/[K,Z]$ of $G/[K,Z]$. So $\overline{G}=G/L$
satisfies Hypothesis C and $\overline{G}/\overline{K}\cong{\sf A}_{4}$, with
$\overline{K}={\bf O}^{2}({\bf O}^{2^{\prime}}(\overline{G}))$, and hence by
case (II) $\overline{K}=1$, which gives ${\bf C}_{K}(Z)=1$. So, the involution
$z\in Z$ acts by inverting all elements of $K$ and $KQ$ is a Frobenius group.

We now assume that $K$ is nilpotent with $\Phi(K)=1$, but not necessarily a
$p$-group. Since ${\bf C}_{K}(Q)=1$, by the Frattini argument $K={\bf F}(G)$
has a complement $H=QR\cong{\rm SL}_{2}(3)$ in $G$, where $|R|=3$. If
$\mu\in{\rm Irr}(K)$ is such that $H_{\mu}=1$ (so $G_{\mu}=K$), then
$\chi=\mu^{G}\in{\rm Irr}(G)$ and $\chi$ is a real character as
$\overline{\mu}=\mu^{-1}=\mu^{z}$. By [Isa76, Lemma 10.8], the Schur index
$m_{\mathbb{R}}(\chi)$ divides $\mu(1)=1$ and hence $\chi\in{\rm Irr_{+}}(G)$.
Therefore, by Hypothesis C there is at most one regular orbit of $G/K$ on
$V=\operatorname{Irr}(K)$.

Assuming now that $K$ is a $p$-group with $\Phi(K)=1$ and writing $|V|=p^{n}$,
we show that then

$(p,n)\in\\{(3,2),(5,2),(7,2)\\}.$

Let ${\rm Syl}_{3}(H)=\\{R_{i}|1\leq i\leq 4\\}$ and observe that ${\bf
C}_{V}(R_{i})\cap{\bf C}_{V}(R_{j})=1$ for $i\neq j$, because $\langle
R_{i},R_{j}\rangle=H$ and $V$ is a faithful $H$-module. Thus, writing $|{\bf
C}_{V}(R_{i})|=p^{a}$, we have $a\leq[n/2]$. Define
$f(p,n)=p^{n}-4p^{[n/2]}-21$ and observe that, writing
$V^{\ast}=V\smallsetminus\\{1_{K}\\}$,

(2.1) $|V^{\ast}|-|\bigcup_{i=1}^{4}{\bf
C}_{V^{\ast}}(R_{i})|=(p^{n}-1)-4(p^{a}-1)$

is larger than $|H|=24$ if $f(n,p)>0$. So, $H$ has at least two regular orbits
on $V$ whenever $f(p,n)>0$. Note that $n$ is even, as $V$ is direct sum of
faithful irreducible $Q$-modules, and in particular that $n\geq 2$.

One can check that $f(p,n)>0$ for all pairs $(p,n)$ with $p>7$ and $n\geq 2$,
or such that $n\geq 4$ (and $p\neq 2$), thus proving the claim.

Now, if $(p,n)=(3,2)$ then $G\cong G_{1}=\mathrm{SmallGroup}(216,153)$ (in the
GAP library of small groups [BEOH23]) and satisfies Hypothesis C. If
$(p,n)=(5,2)$, then $G\cong G_{2}=\mathrm{SmallGroup}(600,150)$ (in [BEOH23])
and satisfies Hypothesis C. In fact, $G_{2}$ acts regularly on
$\operatorname{Irr}(K)\smallsetminus\\{1_{K}\\}$. If $(p,n)=(7,2)$, then $G$
is either $\mathrm{SmallGroup}(1176,214)$ or $\mathrm{SmallGroup}(1176,215)$
(in [BEOH23]), and in both cases they do not satisfy Hypothesis C.

We have shown that if $K$ is a $p$-group, then either $p=3$ and
$G/\Phi(K)\cong G_{1}\cong(C_{3}\times C_{3})\rtimes{\rm SL}_{2}(3)$, or $p=5$
and $G/\Phi(K)\cong G_{2}\cong(C_{5}\times C_{5})\rtimes{\rm SL}_{2}(3)$.

(III.b) Assuming that $K$ is nilpotent, we now show that $G\cong G_{1}$ or
$G\cong G_{2}$.

First we show that $K$ is a $p$-group, and so $(p,G/\Phi(K))=(3,G_{1})$ or
$(5,G_{2})$ by the conclusion of (III.a). Indeed, suppose that $K$ is not a
$p$-group. By the conclusion of (III.a) applied to $G/{\bf O}_{p}(K)$ for each
prime divisor $p$ of $|K|$, we see that $G/\Phi(K)\cong(A\times B)\cdot{\rm
SL}_{2}(3)$, with $A\cong C_{3}\times C_{3}$ and $B\cong C_{5}\times C_{5}$;
moreover, $G/K$ acts regularly on
$\operatorname{Irr}(B)\smallsetminus\\{1_{B}\\}$. It follows that, for any
$\alpha\in\operatorname{Irr}(A)$ and any
$1_{B}\neq\beta\in\operatorname{Irr}(B)$, $\alpha\times\beta$ has stabilizer
$A\times B=K/\Phi(K)$ in $G/\Phi(K)$. Thus $G/K$ has $9$ regular orbits on
$\operatorname{Irr}(A\times B)$, and so, by the third paragraph in (III.a),
$G/\Phi(K)$ does not satisfy Hypothesis C.

Now, if $\Phi(K)=1$, then we are done. We may therefore assume that
$X=\Phi(K)\neq 1$ is minimal normal in $G$. In particular, $X\leq{\bf Z}(K)$.
If $K$ is abelian, then it is an indecomposable $G/K$-module (as $K/X$ is an
irreducible $G/K$-module), so it is homocyclic and by taking a suitable
quotient of $G$ we can assume $K\cong C_{p^{2}}\times C_{p^{2}}$, where
$p\in\\{3,5\\}$. If $p=3$, then one easily checks in [GAP21] that $G$ is
either $\mathrm{SmallGroup}(1944,2289)$ or $\mathrm{SmallGroup}(1944,2290)$;
but in both case there exist three characters in ${\rm Irr_{+}}(G)$ of degree
$24$, against Hypothesis C. If $p=5$, then (as $K/X$ and $X$ are isomorphic
$G/K$-modules), $|{\bf C}_{K}(R)|=5^{2}$ for $R\in{\rm Syl}_{3}(G)$ and,
recalling that $Q$ acts fixed point freely on the abelian group $K$, a
counting argument similar to 2.1 proves the existence of more than four
regular orbits of $G/K$ on $K$, so more than four irreducible characters of
$G$ of degree $24$, a contradiction. Therefore, $K^{\prime}\neq 1$ and hence
$X=K^{\prime}$. Since $K/X$ is $2$-generated, the same is true for $K$ and
hence $X=K^{\prime}$ is cyclic, so $|X|=p$, $K$ is an extraspecial group of
order $p^{3}$, $p\in\\{3,5\\}$. Moreover, $G/{\bf C}_{G}(X)$ is an abelian
group of order dividing $|G/G^{\prime}|=3$, so $X\leq{\bf Z}(G)$. Thus, the
$p-1$ irreducible characters of degree $p$ of $K$ extend to $G$ by [Isa76,
Corollary 11.31], as $G/K$ has either cyclic or quaternion Sylow subgroups),
and hence by Gallagher’s theorem $G$ has $3(p-1)\geq 6$ irreducible characters
of degree $p$, a contradiction.

(III.c) We now assume that $K$ is not nilpotent and derive a contradiction. By
considering a suitable factor group of $G$, we can assume that there is a
subgroup $F\leq N$, $F$ is minimal normal in $G$, such that $G/F\cong G_{i}$
for $i\in\\{1,2\\}$. Write $|F|=q^{m}$ and observe that $q\neq p$, where $p=3$
if $i=1$, and $p=5$ if $i=2$. For $M\in{\rm Syl}_{p}(K)$, ${\bf C}_{F}(M)={\bf
C}_{F}(K)=1$, by the minimality of $F$, and as before, $H={\bf N}_{G}(M)$ is a
complement of $F$ in $G$. Let $L=N\cap H$; then $L\cong(C_{p}\times
C_{p})\rtimes Q_{8}$ has index $3$ in $H$. In order to finish the proof, we
will show that then $L$ has at least $14$ regular orbits on $V=\widehat{F}$.
In fact, the regular $L$-orbits are permuted by the action of
$\overline{H}=H/L\cong C_{3}$, and the ones that are not fixed by
$\overline{H}$ give, in groups of three, regular orbits of $H$ on $V$. If,
instead, $\Delta$ is a regular $L$-orbit in $V$ such that $\Delta$ is
$\overline{H}$-invariant, then $\Delta$ is an $H$-orbit of size $|L|$. Thus,
for $\lambda\in\Delta$, $|H_{\lambda}|=3$ and by Clifford Correspondence ${\rm
Irr}(G|\lambda)$ contains three irreducible characters of degree $|L|$. Hence,
if $L$ has more than $14=4.3+2$ orbits on $V$, then $G$ has more than four
irreducible characters of degree either $|H|$ or of degree $|L|$, against
Hypothesis C.

Assume first $p=3$: so $L$ has four subgroups of order $3$ and nine subgroups
of order $2$. These are the only subgroups of prime order of $L$ and hence
every element in $F$ that is not centralized by any of them is in a regular
$L$-orbit. If $X\leq L$, $|X|=3$, then $|{\bf C}_{V}(X)|\leq q^{m/2}$ because
two distinct subgroups of order $3$ generate $K\cap H\cong C_{3}\times C_{3}$,
which acts faithfully on $V$. If $X\leq L$ and $|X|=2$, then $|{\bf
C}_{V}(X)|=q^{m/2}$ by [Isa76, Theorem 15.16]. Moreover, in characteristic
$q\neq 3$, all faithful absolutely irreducible representations of $H\cong
G_{1}$ have degree $8$, as one can check by looking at its (ordinary and)
Brauer character tables, so $m$ is a multiple of $8$. Now, the function

$g(q,m)=(q^{m}-1)-(9+4)\cdot(q^{m/2}-1)-14|L|$

is $>0$ for all primes $q$ and $m\geq 8$, except for the pair $(q,m)=(2,8)$.
Using [GAP21], one checks that $H$ has a unique faithful (absolutely)
irreducible module $F_{0}$ of dimension $8$ over $\mathbb{F}_{2}$ and that the
semidirect product $G_{0}=F_{0}H$ has $6$ irreducible characters of degree
$72$, so does not satisfy Hypothesis C. Hence, $p\neq 3$.

Finally, if $p=5$ then $L$ has $6$ subgroups of order $5$ and $25$ subgroups
of order $2$. As in the previous paragraph, one checks that for $X\leq L$,
$|X|\in\\{2,5\\}$, $|{\bf C}_{F}(X)|\leq|F|^{\frac{1}{2}}$. Moreover, from the
Brauer character tables of $H\cong G_{2}$ it turns out that, for every
characteristic $q\neq 5$, $H$ has just one faithful absolutely irreducible
representation, which has dimension $24$. So $m$ is a multiple of $24$, and
one easily checks that

$h(q,m)=(q^{m}-1)-(6+25)\cdot(q^{m/2}-1)-14|L|$

is positive for all primes $q$ and integers $m\geq 24$, concluding the proof.
∎

##  3\. Almost Simple groups

######  Theorem 3.1.

If $G$ is an almost simple group and $G$ satisfies Hypothesis C, then $G$ is
one of the groups in the list $\mathcal{L}$ defined in (1.1).

###### Proof.

(a) Let $S$ denote the socle of $G$, so that $S\lhd
G\leq\operatorname{Aut}(S)$. First we consider the case $S={\sf A}_{n}$ with
$n\geq 5$. Since the cases $5\leq n\leq 12$ can be checked directly using
[CCN+85], we will assume that $n\geq 13$, and hence $G=S$ since ${\sf S}_{n}$
does not satisfy Hypothesis C. To rule out such groups $G$, it suffices to
find a partition $\lambda\vdash n$ which is self-associate, such that the
Young diagram of $\lambda$ has $k$ nodes on the main diagonal and $k\equiv
n\pmod{4}$. Indeed, in this case, the character
$\chi\in\operatorname{Irr}({\sf S}_{n})$ labeled by $\lambda$ splits upon
restriction to ${\sf A}_{n}$ into the sum $\chi^{+}+\chi^{-}$ of two real-
valued irreducible characters of ${\sf A}_{n}$, which are ${\sf
S}_{n}$-conjugate and hence of the same degree and same Frobenius–Schur
indicator, see [JK81, Theorem 2.5.13]. Now, if $n\equiv 1\pmod{4}$, we take
$\lambda=((n+1)/2,1^{(n-1)/2})$, the symmetric hook partition of $n$. If
$n\equiv 2\pmod{4}$, take $\lambda=(n/2,2,1^{n/2-2})$, the symmetric hook
partition of $n-1$ augmented by a node in the second row. If $n\equiv
3\pmod{4}$, take $\lambda=((n-3)/2,3,3,1^{(n-9)/2})$, the symmetric hook
partition of $n-4$ augmented by a $2\times 2$-square in the second and third
rows. If $n\equiv 0\pmod{4}$, take $\lambda=((n-8)/2,3,3,3,1^{(n-16)/2})$, the
symmetric hook partition of $n-9$ augmented by a $3\times 3$-square in the
second, third, and fourth rows.

(b) Checking the almost simple groups with sporadic socle $S$ using [CCN+85],
we may now assume that $S$ is a simple group of Lie type. Next we observe that
$[{\mathbb{Q}}(\chi):{\mathbb{Q}}]\leq 2$ for any
$\chi\in\operatorname{Irr}(G)$. Indeed, the $\Gamma$-orbit of $\chi$ has
length $|\Gamma|$, where
$\Gamma:={\mathrm{Gal}}({\mathbb{Q}}(\chi)/{\mathbb{Q}})$. Clearly, any Galois
conjugate of $\chi$ has the same degree and Frobenius–Schur indicator as of
$\chi$. Hence Hypothesis C implies that
$2\geq|\Gamma|=[{\mathbb{Q}}(\chi):{\mathbb{Q}}]$.

Thus $G$ is a quadratic-rational group, in the sense of [Tre17]. Therefore,
all the possibilities for $S$ are listed in [Tre17, Theorem 1.2]; furthermore,
$G/S$ is a solvable group with no normal subgroup of index $2$ or $5$ (by
Hypothesis C applied to $G/S$). All such groups $G$ can be checked using GAP.
∎

##  4\. Modules and the Almost Simple Groups

######  Theorem 4.1.

Suppose that $\tilde{G}$ is a finite group having a unique minimal normal
subgroup $N$ such that $N$ is an irreducible faithful module for
$G=\tilde{G}/N$ and $G$ is one of the groups in the list $\mathcal{L}$ defined
in (1.1). Then $\tilde{G}$ does not satisfy Hypothesis C.

###### Proof.

By assumption, $N$ is an irreducible faithful $G$-module of dimension $n$ over
a prime field $\mathbb{F}=\mathbb{F}_{p}$ with $p$ elements. Viewing
$\operatorname{Irr}(N)$ as a vector space over $\mathbb{F}$, we can identify
it with the dual module $M=N^{*}$. By Clifford theory, each regular orbit of
$G$ on $M$ yields an irreducible character of degree $|G|$ of $\tilde{G}$, see
[Isa76, Theorem 6.11]. Thus we are done if we show that $G$ has at least five
regular orbits on $M$. We use the techniques from [FOS16] and [FMOW19] to
derive lower bounds for the number $R(G,M)$ of regular orbits, as follows.

By construction, $M$ is an irreducible faithful $G$-module of dimension $n$
over $\mathbb{F}$. Let $P$ be a set that contains one generator for each
cyclic subgroup of prime order in $G$. Then

$M=\bigl{\\{}v\in
M;|\operatorname{Stab}_{G}(v)|=1\bigr{\\}}\cup\left(\bigcup_{g\in
P}\operatorname{Fix}_{M}(g)\right)$

implies

$|M|\leq|G|\cdot R(G,M)+\sum_{g\in P}|\operatorname{Fix}_{M}(g)|.$

Let $T$ denote the socle of $G$. For a nonidentity element $g\in G$, define
$r(g)$ to be the minimal number of $T$-conjugates of $g$ that generate
$\langle T,g\rangle$, and set

$r(G)=\max\\{r(g);g\in G^{\times}\\}.$

Then [FOS16, Lemma 3.4] yields

$\dim_{\mathbb{F}}(\operatorname{Fix}_{M}(g))\leq\dim_{\mathbb{F}}(M)(1-1/r(g)).$

The values $r(g)$ for the groups listed in the theorem are given by [FMOW19,
Theorem 1.3] for sporadic simple groups, and have been computed directly for
the other groups in question; they can be read off from Table 1, in the
following sense: We always have $r(g)\geq 3$ if $g$ has order $2$, and
$r(g)\geq 2$ otherwise; Table 1 lists all cases where the inequality is
strict.

$\begin{array}[]{|l|l|l|l|}\hline\cr G&r(g)=3&r(g)=4&r(g)=5\\\ \hline\cr{\sf
A}_{8}&&2A,2B,3A&\\\ {\rm M}_{22}&&2B&\\\ {\rm SU}_{3}(3)&3A&2A&\\\ {\rm
McL}&3A&&\\\ {\rm O}_{8}^{+}(2).3&3D,3E,3F&3ABC&2A,2BCD\\\
\hline\cr\end{array}$ Table 1. Exceptional values of $r(g)$.

It follows that

$|G|\cdot R(G,M)\geq p^{n}-\sum_{g\in P}p^{n(1-1/r(g))}\geq
p^{n(1-1/r(G))}\left(p^{n/r(G)}-|P|\right).$

Thus $R(G,M)\geq 5$ holds except for only finitely many pairs $(p,n)$.

We define $D(G)$ to be the smallest integer $n$ such that

$5|G|\geq 2^{n(1-1/r(G))}\left(2^{n/r(G)}-|P|\right)$

holds. Then $R(G,M)\geq 5$ holds for any $G$-module $M$ of dimension larger
than $D(G)$ over a finite prime field.

The Brauer characters of those $G$-modules $M$ for which this crude estimate
does not suffice are listed in [JLPW95] and (if the characteristic $p$ does
not divide $|G|$) in [CCN+85]. For each $p^{\prime}$-element $g$, we can
compute the exact value of $\dim_{\mathbb{F}}(\operatorname{Fix}_{M}(g))$ from
the Brauer character of $M$, which yields better estimates for $R(G,M)$ and
rules out many cases. The modules for which this does not establish at least
five regular orbits are listed in Table 2. Direct computations with the table
of marks of $G$ and with matrices that describe the $G$-action on $M$ yield
the orbit lengths. Finally, if this is not sufficient, we compute the possible
extensions $\tilde{G}$ of $G$ by $N\cong M^{*}$, and verify directly that
these groups $\tilde{G}$ do not satisfy Hypothesis C.

For example, let $G$ be the Thompson group Th. We have

$r(G)=3\mbox{ and }|P|=675\,176\,077\,846\,831,$

thus $p^{n/r(G)}-|P|>0$ if $n\geq 148$ holds, and $2^{2n/3}(2^{n/3}-|P|)>5|G|$
in this case; hence $D(G)=148$. Since the minimal degree of a nontrivial
irreducible representation of $G$ in any characteristic is $248$ (see
[Jan05]), we have shown that the claim holds for the group Th.

For the group $G=\mathrm{O}_{8}^{+}(2).3$, we compute $D(G)=113$, thus the
five modules listed in Table 2 must be considered.

  * •

The $24$-dimensional representation over $\mathbb{F}_{2}$ is induced from
$\mathrm{O}_{8}^{+}(2)$, thus we consider the restriction of the possible
group extensions (there are two, a split and a non-split one) to the index $3$
subgroup $2^{24}.\mathrm{O}_{8}^{+}(2)$. The $24$-dimensional module splits as
a direct sum of three $8$-dimensional ones, thus there are three factor groups
$2^{8}.\mathrm{O}_{8}^{+}(2)$, which are permuted cyclicly by the automorphism
of order three. The groups $2^{8}.\mathrm{O}_{8}^{+}(2)$ (there are two, a
split and a non-split one) have the character degree $2835$ with multiplicity
$6$, which yields the degree $3\cdot 2835$ with multiplicity at least $6$ for
$2^{24}.\mathrm{O}_{8}^{+}(2).3$.

  * •

The $26$-dimensional representation over $\mathbb{F}_{2}$ extends from
$\mathrm{O}_{8}^{+}(2)$, and $\mathrm{O}_{8}^{+}(2)$ has a unique orbit of
length $1575$ on the module. Hence also $\mathrm{O}_{8}^{+}(2).3$ has this
orbit. The corresponding $H^{2}$ is trivial, thus it is enough to establish a
big enough degree multiplicity among the ordinary irreducible characters of
the point stabilizer, which is an involution centralizer of order $331776$.
This subgroup has the irreducible degree $16$ with multiplicity $6$, which
yields multiplicity at least $6$ for the degree $1575\cdot 6$ in
$2^{26}.\mathrm{O}_{8}^{+}(2).3$.

  * •

The $52$-dimensional representation over $\mathbb{F}_{2}$ extends from
$\mathrm{O}_{8}^{+}(2)$, with restriction to this subgroup being twice the
$26$-dimensional representation which we dealt with above. This restriction
has $25493462$ regular orbits, which is enough for $\mathrm{O}_{8}^{+}(2).3$.

  * •

The $G$-representations of the degrees $28$ and $48$ over $\mathbb{F}_{3}$ are
extensions from $\mathrm{O}_{8}^{+}(2)$. The restrictions to this normal
subgroup have $127877$ and $457929531422105$ regular orbits, which yields huge
degree multiplicities also for $\mathrm{O}_{8}^{+}(2).3$.

In this case, we chose to work with restrictions to $\mathrm{O}_{8}^{+}(2)$
since the table of marks of this simple subgroup is available, but the table
of marks for $\mathrm{O}_{8}^{+}(2).3$ is not.

For the other groups $G$ in question, Table 2 describes the arguments that
exclude the modules, as follows.

$\begin{array}[]{|l|r|r|r|r|r|r|}\hline\cr
G&D(G)&p&M&R_{1}(R_{2})&H^{2}&\textrm{deg.}\\\ \hline\cr{\sf
A}_{8}&45&2&\texttt{4a}&&1&+105^{3}\\\ &&&\texttt{4b}&&1&+105^{3}\\\
&&&\texttt{6a}&&0&35^{5}\\\ &&&\texttt{14a}&&2&210^{5}\\\
&&&\texttt{20a}&37&&\\\ &&&\texttt{20b}&37&&\\\ &&3&\texttt{7a}&&1&28^{5}\\\
&&&\texttt{13a}&23&&\\\ &&5&\texttt{7a}&&0&28^{9}\\\ &&7&\texttt{7a}&(15)&&\\\
&&11&\texttt{7a}&240&&\\\ &&13&\texttt{7a}&1122&&\\\ \hline\cr{\rm
SL}_{3}(2)&18&2&\texttt{3a}&&1&+21^{2}\\\ &&&\texttt{3b}&&1&+21^{2}\\\
&&&\texttt{8a}&&0&24^{7}\\\ &&3&\texttt{6a}&&0&21^{24}\\\
&&&\texttt{3ab}&&0&42^{16}\\\ &&7&\texttt{3a}&&1&24^{14}\\\ \hline\cr{\rm
M}_{11}&30&2&\texttt{10a}&&1&55^{5}\\\ &&3&\texttt{5a}&&0&+110^{2}\\\
&&&\texttt{5b}&&0&220^{5}\\\ &&&\texttt{10a}&&0&55^{9}\\\
&&&\texttt{10b}&5&&\\\ &&&\texttt{10c}&5&&\\\ \hline\cr{\rm
M}_{22}&64&2&\texttt{10a}&&&+1155^{2}\\\ &&&\texttt{10b}&&&+1155^{2}\\\
&&3&\texttt{21a}&20241&&\\\ \hline\cr{\rm
M}_{23}&59&2&\texttt{11a}&&1&+253^{2}\\\ &&&\texttt{11b}&&0&+253^{3}\\\
\hline\cr\end{array}\ \begin{array}[]{|l|r|r|r|r|r|r|}\hline\cr
G&D(G)&p&M&R_{1}(R_{2})&H^{2}&\textrm{deg.}\\\ \hline\cr{\rm
M}_{24}&68&2&\texttt{11a}&&0&+239085^{4}\\\ &&&\texttt{11b}&&1&+239085^{2}\\\
&&&\texttt{44a}&69255&&\\\ &&&\texttt{44b}&69173&&\\\ &&3&\texttt{22a}&34&&\\\
\hline\cr{\rm SU}_{3}(3)&38&2&\texttt{6a}&&1&189^{6},+378^{2}\\\
&&&\texttt{14a}&&0&252^{13}\\\ &&3&\texttt{3ab}&&0&224^{9}\\\
&&&\texttt{6ab}&40&&\\\ &&&\texttt{7a}&&2&672^{9},756^{8}\\\
&&5&\texttt{6a}&&0&756^{8}\\\ &&&\texttt{7b}&&0&108^{8}\\\
&&&\texttt{7c}&&0&108^{8}\\\ &&7&\texttt{6a}&16&&\\\ \hline\cr{\rm
McL}&77&2&\texttt{22a}&&1&311850^{6}\\\ &&3&\texttt{21a}&(10)&&\\\
\hline\cr{\rm Th}&148&&\textrm{(none)}&&&\\\ \hline\cr{\rm
SL}_{2}(8).3&24&2&\texttt{6a}&&1&63^{6}\\\ &&&\texttt{12a}&&0&63^{6}\\\
&&&\texttt{8a}&&0&36^{6}\\\ &&&\texttt{8bc}&34&&\\\
&&3&\texttt{7a}&&2&56^{9}\\\ \hline\cr{\rm
O}_{8}^{+}(2).3&113&2&\texttt{24a}&&&\\\ &&&\texttt{26a}&&&\\\
&&&\texttt{26bc}&&&\\\ &&3&\texttt{28a}&&&\\\ &&3&\texttt{48a}&&&\\\
\hline\cr\end{array}$ Table 2. Modules to be considered.

  * •

The column with header $p$ lists the characteristics for which at least one
irreducible $\mathbb{F}G$-module $M$ exists such that the character-theoretic
lower bound for the number of regular orbits is less than $5$.

  * •

The column with header $M$ lists the irreducible modules to be considered.
Names consisting of an integer $n$ followed by one letter a, b, etc. denote
the first, second, etc. absolutely irreducible module of dimension $n$, names
involving several letters denote modules that decompose into a sum of the
corresponding absolutely irreducible modules of dimension $n$ over an
extension field; for example 3ab means a module that is irreducible over the
prime field with $p$ elements, of dimension $6$, and decomposes into two
absolutely irreducible modules over the field with $p^{2}$ elements.

  * •

The column with header $R_{1}(R_{2})$ contains the number of regular orbits,
provided this number is at least $5$, or in brackets the number of orbits of
length $|G|/2$, provided this number is at least $9$. Note that each orbit of
length $|G|/2$ corresponds to an inertia subgroup which has only the character
degrees $1$ and $2$, hence $9$ orbits yield one character degree with
multiplicity at least $5$.

The values in this column have been computed from the known table of marks of
$G$, together with matrix generators describing $M$.

  * •

If the column $R_{1}(R_{2})$ does not exclude the module in question then the
column with header $H^{2}$ contains the dimension of the corresponding $H^{2}$
group, and the column with header deg. contains a description of character
degrees that exclude the module. In this column deg., the entry $d^{n}$ means
that the character degree $d$ occurs with multiplicity $n$, and the entry
$+d^{n}$ means that the degree $d$ occurs with multiplicity $n$ for characters
with indicator $+$.

In the computations, we used the character tables and tables of marks from the
GAP packages [Bre24] and [MNP24], respectively. The matrix representations
were taken from the Atlas of Group Representations [WWT+] or derived from
representations available there. Cohomology groups and group extensions were
computed using implementations in [Hol23, GAP21, OSC24]. Permutation
representations of small degrees of the group extensions were obtained using
ideas from [Hul01], and [BCP97] was used to compute the character tables. ∎

##  5\. Non-Solvable groups

######  Lemma 5.1.

Let $S$ be a non-abelian simple group and $A={\rm Aut}(S)$. Then, there exists
a non-principal character $\phi\in{\rm Irr_{+}}(S)$ having an extension
$\hat{\phi}\in{\rm Irr_{+}}(A)$.

###### Proof.

If $S$ is a simple group of Lie type, not isomorphic to
${}^{2}F_{4}(2)^{\prime}$, we let $\phi$ be the Steinberg character of $S$. It
is known ([Fei93, Theorem B]) that $\phi$ has an extension $\hat{\phi}\in{\rm
Irr_{+}}({\rm Aut}(S))$.

For $S={\sf A}_{n}$, $n\geq 5$, $n\neq 6$ (${\sf A}_{6}\cong{\rm PSL}_{2}(9)$)
we consider the (unique) character $\phi\in{\rm Irr}({\sf A}_{n})$ such that
$\phi(1)=n-1$; then it is well known that $\phi$ has an extension
$\hat{\phi}\in{\rm Irr_{+}}({\rm Aut}(S))$.

If $S$ is a sporadic simple group or the Tits group, we choose the unique
$\phi\in{\rm Irr}(S)$ of degree $d$, where the pairs $(S,d)$ are as follows
(see [CCN+85]):

$({\rm M}_{11},11)$, $({\rm M}_{12},45)$, $({\rm M}_{22},55)$, $({\rm
M}_{23},22)$, $({\rm M}_{24},23)$, $({\rm J}_{1},209)$, $({\rm J}_{2},36)$,
$({\rm J}_{3},324)$, $({\rm J}_{4},889111)$, $({\rm Co}_{1},276)$, $({\rm
Co}_{2},23)$, $({\rm Co}_{3},23)$, $({\rm Fi}_{22},78)$, $({\rm
Fi}_{23},782)$, $({\rm Fi}^{\prime}_{24},8671)$, $({\rm HS},22)$, $({\rm
McL},22)$, $({\rm He},680)$, $({\rm Ru},406)$, $({\rm Suz},143)$, $({\rm
O^{\prime}N},10944)$, $({\rm HN},760)$, $({\rm Ly},45694)$, $({\rm Th},248)$,
$({\rm B},4371)$, $({\rm M},196883)$, $({}^{2}{\rm F}_{4}(2)^{\prime},78)$. ∎

For a finite set $X$ and an integer $1\leq k\leq|X|$, we denote by

$X^{[k]}=\\{Y\subseteq X\mid|Y|=k\\}$

the set consisting of the subsets of cardinality $k$ of $X$. Given an action
of a group $G$ on a set $X$ and a subset $Y\in X^{[k]}$, we denote by
$G_{Y}=\\{g\in G\mid Y^{g}=Y\\}$ the stabilizer of $Y$ in the natural action
of $G$ on $X^{[k]}$.

The following result will be useful in dealing with some of the cases arising
in the proof of Theorem A.

######  Lemma 5.2.

Let $G$ be a group and let $M$ be a non-solvable minimal normal subgroup of
$G$ such that ${\bf C}_{G}(M)=1$. Writing $M=S_{1}\times
S_{2}\times\cdots\times S_{n}$, with $S_{i}$ isomorphic non-abelian simple
groups, we consider the action of $G$ on $X=\\{1,2,\ldots,n\\}$ defined, for
$g\in G$ and $i\in X$, by

$S_{i^{g}}=(S_{i})^{g}\;.$

If $G$ satisfies Hypothesis C, then

  1. (a)

For every non-empty $Y\subseteq X$, either $G_{Y}$ is a perfect group or
$|G_{Y}:G_{Y}^{\prime}|=3$.

  2. (b)

For every $1\leq k\leq n$, the orbits of $G$ on $X^{[k]}$ have distinct sizes.

###### Proof.

Let $X=\\{1,2,\ldots,n\\}$, $S=S_{1}$ and $A={\rm Aut}(S)$. Since ${\bf
C}_{G}(M)=1$, $G$ is isomorphic to a subgroup of ${\rm Aut}(M)$. Hence, we can
identify $G$ with a subgroup of the wreath product $W=A\wr{\mathrm{Sym}}(X)$
([Rob93, 3.3.20]) and the action of $W$ on $X$, defined by
$S_{i^{w}}=(S_{i})^{w}$ for $w\in W$ and $i\in X$, extends the action of $G$
on $X$. As $M$ is a minimal normal subgroup of $G$, $G$ acts transitively on
$X$ and hence there exist elements $x_{1}=1,x_{2},\ldots,x_{n}\in G$ such that
$S_{i}=S^{x_{i}}$. So, the base group of $W$ is $B=A_{1}\times
A_{2}\times\cdots\times A_{n}$, where $A_{i}=A^{x_{i}}$.

By Lemma 5.1 there exists a character $\phi\in{\rm Irr_{+}}(S)$ having an
extension $\hat{\phi}\in{\rm Irr_{+}}(A)$. So,
$\hat{\phi}_{i}=\hat{\phi}^{x_{i}}\in{\rm Irr_{+}}(A_{i})$ extends
$\phi_{i}=\phi^{x_{i}}\in{\rm Irr}(S_{i})$, for every $i\in X$.

Let $Y$ be a non-empty subset of $X$. We consider the character

$\theta=\prod_{i\in X}\alpha_{i}\in{\rm Irr}(M)\text{, \ where
}\alpha_{i}=\begin{cases}\phi_{i}&\text{ if }i\in Y\\\ 1_{S_{i}}&\text{ if
}i\in X\smallsetminus Y\,.\end{cases}$

Hence, the character

$\psi=\prod_{i\in X}\beta_{i}\in{\rm Irr}(B)\text{, \ where
}\beta_{i}=\begin{cases}\hat{\phi}_{i}&\text{ if }i\in Y\\\ 1_{A_{i}}&\text{
if }i\in X\smallsetminus Y\end{cases}$

is an extension of $\theta$ to $B$ and $\psi\in{\rm Irr_{+}}(B)$.

By [Hup98, Lemma 25.5(a)], the set stabilizer $W_{Y}$ coincides with the
inertia subgroup $I_{W}(\psi)$. Hence, as each character $\beta_{i}$ is
afforded by a real representation of $A_{i}$, the argument used in the proof
of [Hup98, Lemma 25.5(b)] shows that $\psi$ has an extension
$\hat{\psi}\in{\rm Irr_{+}}(W_{Y})$.

Since $\theta=\psi_{M}$, we have $W_{Y}=I_{W}(\psi)\leq I_{W}(\theta)$. On the
other hand, if $w\in I_{W}(\theta)$, then $w$ normalizes ${\rm
ker}(\theta)=\prod_{i\in X\smallsetminus Y}S_{i}$ and hence $w\in
W_{X\smallsetminus Y}=W_{Y}$. Therefore, $W_{Y}=I_{W}(\theta)$ and

$G_{Y}=G\cap W_{Y}=G\cap I_{W}(\theta)=I_{G}(\theta).$

Let $\hat{\theta}=\hat{\psi}_{G_{Y}}$ be the restriction of $\hat{\psi}$ to
$G_{Y}$. As $\hat{\theta}_{M}=\hat{\psi}_{M}=\theta\in{\rm Irr}(M)$,
$\hat{\theta}$ is an irreducible character of $G_{Y}$ and, since
$\hat{\psi}\in{\rm Irr_{+}}(W_{Y})$, we deduce that $\hat{\theta}\in{\rm
Irr_{+}}(G_{Y})$.

Now, if $|G_{Y}/G_{Y}^{\prime}|$ is even, then there exists a linear character
$\lambda$ of $G_{Y}$ such that $o(\lambda)=2$. Hence, by Gallagher’s theorem
and [Isa76, Lemma 4.8] $\hat{\theta}$ and $\lambda\hat{\theta}$ are distinct
characters in ${\rm Irr_{+}}(G_{Y})$, and by Clifford correspondence
$\hat{\theta}^{G}$ and $(\lambda\hat{\theta})^{G}$ are distinct characters of
the same degree in ${\rm Irr_{+}}(G)$, against Hypothesis C.

On the other hand, if $|G_{Y}/G_{Y}^{\prime}|\geq 5$, then Clifford
correspondence yields at least five characters of the same degree in ${\rm
Irr}(G|\hat{\theta})$, a contradiction again. Hence,
$|G_{Y}:G_{Y}^{\prime}|\in\\{1,3\\}$ and part (a) is proved.

So far, we have shown that for every non-empty subset $Y$ of $X$ there exists
a character $\chi_{Y}\in{\rm Irr_{+}}(G)$ such that

(5.1) $\chi_{Y}(1)=\phi(1)^{|Y|}|G:G_{Y}|.$

Moreover, the irreducible constituents of the restriction of $\chi_{Y}$ to $M$
have kernels of the form

$\prod_{i\in X\smallsetminus Y^{g}}S_{i}$

for some $g\in G$. Hence, if $Y_{1},Y_{2}\in X^{[k]}$ ($1\leq k\leq n$) are
sets belonging to different $G$-orbits, then $\chi_{Y_{1}}\neq\chi_{Y_{2}}$.

As $|G:G_{Y}|$ is the size of the orbit of $Y\in X^{[k]}$ in the action of $G$
on $X^{[k]}$, part (b) follows immediately from (5.1) and Hypothesis C. ∎

######  Theorem 5.3.

If $G$ is a non-solvable group and $G$ satisfies Hypothesis C, then $G$ is one
of the groups in the list $\mathcal{L}$ defined in (1.1).

###### Proof.

We work by induction on $|G|$.

(I) We first assume that there exists a non-solvable minimal normal subgroup
$M$ of $G$, and we start by showing that ${\bf C}_{G}(M)=1$.

Suppose, working by contradiction, that $C={\bf C}_{G}(M)\neq 1$. Let $C_{0}$
be a minimal normal subgroup of $G$, with $C_{0}\leq C$. As $G/C_{0}$ is non-
solvable and satisfies Hypothesis C, by induction $G/C_{0}$ is a group in
$\mathcal{L}$. In particular, $G/C_{0}$ is almost-simple, so $C=C_{0}$ is
minimal normal in $G$, $G/C$ has socle $MC/C\cong M$ and $|G:MC|\in\\{1,3\\}$.

Assuming $G=M\times C$, then $C\cong G/M$ satisfies Hypothesis C. If $C$ is
solvable, then $C\cong C_{3}$ and hence, as $M$ (by direct inspection of the
socles of the groups in $\mathcal{L}$ or by [BCH92, Lemma 1]) has at least two
irreducible characters with the same degree, $G$ has at least six irreducible
characters with the same degree, a contradiction. Hence, both $M$ and $C$ are
simple groups belonging to the list $\mathcal{L}$. We observe that $M$ and $C$
are non-isomorphic, as otherwise $G$ has irreducible characters with the same
degree and Frobenius–Schur indicator, but with different kernels, a
contradiction. Every simple group in $\mathcal{L}$, except for ${\rm
SL}_{3}(2)$, ${\rm M}_{22}$, ${\rm McL}$ and ${\rm Th}$, has three irreducible
characters with the same degree and, as $G$ cannot have six characters of the
same degree, we deduce that $M$ and $C$ are (non-isomorphic) groups belonging
to the sublist $\\{{\rm SL}_{3}(2),{\rm M}_{22},{\rm McL},{\rm Th}\\}$. Now,
one checks that for each of the six possibilities for $G=M\times C$, ${\rm
Irr}(G)$ contains four non-real characters with the same degree, against
Hypothesis C.

Assuming instead that $|G:MC|=3$, then either $M\cong{\rm PSL}_{2}(8)$ or
$M\cong{\rm O}_{8}^{+}(2)$. Hence, there are three characters
$\phi_{1},\phi_{2},\phi_{3}\in{\rm Irr}(M)$ with the same degree, that are
transitively permuted by the outer automorphism group of order $3$ of $M$. So,
for every $\theta\in{\rm Irr}(C)$, $I_{G}(\phi_{i}\times\theta)=MC$ and hence
$(\phi_{i}\times\theta)^{G}\in{\rm Irr}(G)$. If $M\cong{\rm SL}_{2}(8)$,
$\phi_{i}(1)=7$ and $\phi_{i}$ is non-real, while if $M\cong{\rm
O}_{8}^{+}(2)$, $\phi_{i}(1)=35$ and $\phi_{i}$ is real, for $i=1,2,3$. We
observe that $G/M$ satisfies Hypothesis C and that $CM/M$ is a minimal normal
subgroup of index three of $G/M$. If $C$ is solvable, then by Theorem 2.5
either $G/M\cong C_{7}\rtimes C_{3}$ and $C\cong C_{7}$, or $G/M\cong{\sf
A}_{4}$ and $C\cong C_{2}\times C_{2}$. If $C$ is non-solvable, then by
induction either $C\cong{\rm PSL}_{2}(8)$ or $C\cong{\rm O}_{8}^{+}(2)$. In
any case, there are at least three characters
$\theta_{1},\theta_{2},\theta_{3}\in{\rm Irr}(C)$ having the same degree and
either all non-real or all real. It follows that the characters
$\phi_{i}\times\theta_{j}$, for $1\leq i,j\leq 3$, are either nine non-real,
or nine real, irreducible characters with the same degree of $MC$. As
$I_{G}(\phi_{i}\times\theta_{j})=MC$, recalling [IN12, Lemma 2.1] we conclude
that $G$ has either at least three non-real, or at least three real,
irreducible characters with the same degree, against Hypothesis C.

So far, we have proved that ${\bf C}_{G}(M)=1$ and hence $M$ is the only
minimal normal subgroup of $G$. Writing

$M=S_{1}\times\cdots\times S_{n},$

with $S_{i}\simeq S$ isomorphic non-abelian simple groups, we define

$N=\bigcap_{i=1}^{n}{\bf N}_{G}(S_{i}),\mbox{ and }H=G/N.$

So, $H$ is a transitive permutation group on $X=\\{1,\ldots,n\\}$. If $n=1$,
then $G$ is an almost simple group and $G\in\mathcal{L}$ by Theorem 3.1. We
now assume, aiming at a contradiction, that $n\neq 1$. We observe that then,
in particular, $H\neq 1$. We have two cases.

(I.a) We first assume that $G/M$ is solvable. Hence, for every non-empty
subset $Y$ of $X$, by part (a) of Lemma 5.2 either $G_{Y}=M$ or
$[G_{Y}:G_{Y}^{\prime}]=3$. By Theorem 2.5, both $G/M$ and $H=G/N$ belong to
the list

$\begin{array}[]{rl}\mathcal{S}=&\bigl{\\{}C_{3},C_{7}\rtimes
C_{3},(C_{2}\times C_{2}\times C_{2})\rtimes(C_{7}\rtimes C_{3}),{\sf
A}_{4},\\\ &{\rm SL}_{2}(3),(C_{3}\times C_{3})\rtimes{\rm
SL}_{2}(3),(C_{5}\times C_{5})\rtimes{\rm SL}_{2}(3)\bigr{\\}}.\end{array}$

We remark that, as $H_{Y}$ is a quotient of $G_{Y}$,

(5.2) $\forall\ Y\subseteq X,Y\neq\varnothing,\text{ \ either \ }H_{Y}=1\text{
\ or \ }|H_{Y}:H_{Y}^{\prime}|=3\,.$

Let $L=H_{x}$, for some $x\in X$, be a point stabilizer in $H$; so
$n=|X|=|H:L|$. We first show that $H$ does not act regularly on $X$, i.e. that
$L\neq 1$. To prove this, we assume that $H$ acts regularly on $X$ and we
consider a subset $Y=\\{x_{1},x_{2}\\}\in X^{[2]}$. For $h\in H_{Y}$, then
$h^{2}$ fixes $x_{i}$, for $i=1,2$, so $h^{2}=1$. It follows that $H_{Y}$ is
an elementary abelian group, and hence $H_{Y}=1$ by (5.2). Thus, $H$ has
$(n-1)/2$ orbits of size $n$ on $X^{[2]}$ and part (b) of Lemma 5.2 yields
$n=3$ and $H\cong C_{3}$. Moreover, as $G/M\in\mathcal{S}$ and $G_{Y}/M=N/M$,
part (a) of Lemma 5.2 implies that $N=M$. In particular, for every
$\alpha\in{\rm Irr}(S)$, $\alpha\times\alpha\times\alpha\in{\rm Irr}(M)$ has
three extensions to ${\rm Irr}(G)$. Since $S$ has at least two irreducible
characters with the same degree by [BCH92, Lemma 1], then $G$ has at least six
irreducible characters with the same degree, a contradiction.

Hence, $L\neq 1$ and, as $H$ acts faithfully on $X$, the normal core $L_{H}$
of $L$ in $H$ is trivial. Recalling that $L<H$ since $n\neq 1$ and that
$|L/L^{\prime}|=3$ by (5.2), we conclude that

$|X|\in N(H)=\bigl{\\{}|H:L|\mid 1\neq L<H,|L/L^{\prime}|=3\text{ and
}L_{H}=1\bigr{\\}}.$

In order to reduce the number of cases for the relevant transitive actions of
the groups in $\mathcal{S}$, we introduce the set

$W(H)=\bigl{\\{}|H:T|\mid T<H\text{ and
}|T/T^{\prime}|=3\\}\cup\\{|H|\bigr{\\}}$

and we claim that, for every $0<k<n$, the size $|H:H_{Y}|$ of the $H$-orbit of
$Y\in X^{[k]}$ belongs to $W(H)$. In fact, $H_{Y}<H$ as $H$ is transitive on
$X$ and $Y\neq X$, so the claim follows by (5.2). Moreover, for $0<k<n$ and
$Y\in X^{[k]}$, $H_{Y}<H$ as $H$ is transitive on $X$.

Thus, recalling that $n=|X|\in N(H)$, part (b) of Lemma 5.2 yields that, for
every $k$ such that $0<k<n$,

(5.3) $|X^{[k]}|={n\choose k}\text{ is a sum of distinct elements of }W(H).$

Since $H\not\cong C_{3}$ as $H$ acts non-regularly on $X$, the triples
$(H,N(H),W(H))$ are listed in Table 3.

$\begin{array}[]{|l|r|r|r|r|r|r|}\hline\cr H&C_{7}\rtimes
C_{3}&C_{2}^{3}\rtimes(C_{7}\rtimes C_{3})&{\sf A}_{4}&{\rm
SL}_{2}(3)&C_{3}^{2}\rtimes{\rm SL}_{2}(3)&C_{5}^{2}\rtimes{\rm SL}_{2}(3)\\\
\hline\cr
N(H)&\\{7\\}&\\{8,14,56\\}&\\{4\\}&\\{8\\}&\\{9,72\\}&\\{25,200\\}\\\
\hline\cr
W(H)&\\{7,21\\}&\\{8,14,56,168\\}&\\{4,12\\}&\\{8,24\\}&\\{9,72,216\\}&\\{25,200,600\\}\\\
\hline\cr\end{array}$ Table 3.

One easily checks that (5.3) is not satisfied for $k=3$ if $H\cong
C_{7}\rtimes C_{3}$, while it is not satisfied for $k=2$ in all other cases, a
contradiction.

(I.b) We assume now that $G/M$ is non-solvable. Since $G/M$ satisfies
Hypothesis C, by induction $G/M\in\mathcal{L}$. We observe that $N/M$, being
isomorphic to a subgroup of a direct product of outer automorphism groups of a
simple group, is a solvable normal subgroup of $G/M$. Hence, $N=M$ and
$H=G/M$.

In order to reduce the number of cases for the action of $H$ on $X$, we
consider the function

$b(|H|)=\max\Bigl{\\{}a\in d(|H|)\mid\sum_{1\neq d\in
d(|H|)}d\geq{a\choose\lfloor\frac{a}{2}\rfloor}\Bigr{\\}}$

where $d(|H|)$ is the set of the positive divisors of $|H|$. By the orbit
formula and part (b) of Lemma 5.2, setting $m=\lfloor\frac{|X|}{2}\rfloor$,
$|X^{[m]}|$ is a sum of distinct elements (orbit sizes) $d_{i}\in d(|H|)$.
Moreover, every $d_{i}\neq 1$, because $H$ has no fixed points in $X^{[m]}$
since $H$ is transitive on $X$ and $0<m<|X|$.

We deduce that $|X|\leq b(|H|)$ and hence $|X|$ belongs to the set

$N_{1}(H)=\bigl{\\{}c\cdot|H:L|\mid c\in\mathbb{Z}_{+},L\text{ is a maximal
subgroup of }H\text{ and }c\cdot|H:L|\leq b(|H|)\bigr{\\}}.$

We remark that, for $H\in\mathcal{L}$, both $b(|H|)$ and $N_{1}(H)$ can be
computed using the information in [CCN+85] and that the set stabilizers
$H_{Y}$ for the relevant actions of $H$ on $X^{[k]}$, for $|X|\in N_{1}(H)$
and $0<k<|X|$, can be computed in [GAP21]. In particular, we get that
$N_{1}(H)=\varnothing$ for

$H\in\bigl{\\{}{\rm SU}_{3}(3),{\rm McL},{\rm Th},{\rm
O}_{8}^{+}(2).3\bigr{\\}}.$

For the remaining cases, we refer to Table 4.

$\begin{array}[]{|l|r|r|r|r|r|r|r|}\hline\cr H&{\sf A}_{8}&{\rm
PSL}_{3}(2)&{\rm M}_{11}&{\rm M}_{22}&{\rm M}_{23}&{\rm M}_{24}&{\rm
PSL}_{2}(8).3\\\ \hline\cr b(|H|)&18&8&16&22&28&32&14\\\ \hline\cr
N_{1}(H)&\\{8,15\\}&\\{7,8\\}&\\{11,12\\}&\\{22\\}&\\{23\\}&\\{24\\}&\\{9\\}\\\
\hline\cr\end{array}$ Table 4.

For $(H,|X|)\in\\{({\rm PSL}_{3}(2),7),({\rm M}_{11},11),({\rm M}_{11},12)\\}$
and $Y=\\{x\\}\subseteq X$, $H_{Y}$ has a quotient of order $2$, a
contradiction by part (a) of Lemma 5.2. In all other cases, for $Y\in
X^{[2]}$, $H_{Y}$ has a quotient of order $2$, again a contradiction.

(II.) We can now assume that $G$ has no non-solvable minimal normal subgroups.
So, $M$ is solvable and $G/M$ is a non-solvable group that satisfies
Hypothesis C. Hence, by induction $G/M$ is an almost-simple group belonging to
the list $\mathcal{L}$. Let $K/M$ be the socle of $G/M$.

We observe that $M\leq K^{\prime}$. In fact,
$K^{\prime}M/M=(K/M)^{\prime}=K/M$, hence $K=K^{\prime}M$, and
$K^{\prime}\trianglelefteq\,G$, so if $M\not\leq K^{\prime}$, then $M\cap
K^{\prime}=1$ and $K^{\prime}\cong K/M$ is a non-solvable minimal normal
subgroup of $G$, a contradiction.

If $M<{\bf C}_{G}(M)$, then $M\leq{\bf Z}(K)$ and hence $K$ is a quotient of
the Schur cover of $K/M$. If $G=K$, i.e. $G/M$ is a simple group in
$\mathcal{L}$, by direct inspection using [CCN+85], we see that all the groups
arising in this way do not satisfy Hypothesis C. Since the Schur multiplier of
${\rm SL}_{2}(8)$ is trivial, the only case left is $G/M\cong{\rm
O}_{8}^{+}(2).3$. Hence, $|M|=2^{2}$ (as the non-trivial elements of the Schur
multiplier of ${\rm O}_{8}^{+}(2)$ are transitively permuted by the subgroup
of order three of its outer automorphism group) and $G$ has three real
irreducible characters of degree $3024$, a contradiction.

Therefore, $M$ is the unique minimal normal subgroup of $G$ and it is a
faithful non-trivial irreducible $G/M$-module. Hence, an application of
Theorem 4.1 gives the final contradiction, concluding the proof. ∎

Now Theorem A follows immediately from Theorem 2.5 and Theorem 5.3.

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$\displaystyle\chi^{\prime}\Big{(}f(a),f(b)\Big{)}=\frac{\omega(1)\cdot\omega(1)^{ab}}{\omega(1)^{a}\cdot\omega(1)^{b}}\cdot\chi(a,b)^{h}\;,$

They can also be combined with Equations 74 and 77 to imply that

(96) $\displaystyle\omega(1)^{2}\;=\;\big{(}\omega(1)^{2}\big{)}^{g}\,.$

Finally Equations 92, 93, 94, and 96 can be used to reduce Equation 80 to
derive

(97)
$\displaystyle\frac{\tau^{\prime}}{\tau}\;=\;\frac{\omega(1)}{\omega(1)^{g}}\;.$

By setting $\omega(1)=\lambda$, the reader will recognize Equations 95 and 97
as Conditions 66 and 67 respectively from the statement of the theorem. Thus
we have shown that a generic equivalence forces $g=g^{\prime}$, gives rise to
the data $(f,h,\lambda)$ stated in the theorem, and makes Conditions 66 and 67
necessary.

We now turn to the question of sufficiency. Suppose that $g=g^{\prime}$ and
that the data $(f,h,\lambda)$ are given. Then the pair $(f,h)$ uniquely
determines the underlying functor
$F:\mathcal{C}_{\mathbb{C}}(g,\tau,\chi)\to\mathcal{C}_{\mathbb{C}}(g,\tau^{\prime},\chi^{\prime})$.
Define the functions

$\displaystyle J_{a,b}=\textsf{id}_{a\otimes b}\;,\hskip
11.38109ptJ_{a,m}=\textsf{id}_{a\otimes m}\;,\hskip
11.38109ptJ_{m,a}=\left(\frac{\lambda}{\lambda^{a}}\right)\otimes\textsf{id}_{a}\;,\hskip
11.38109ptJ_{m,m}=\textsf{id}_{m}\otimes\lambda\;.$

These are the components of a monoidal structure map, and are clearly
isomorphisms. This monoidal structure map $J$ can be described in terms of
complex valued coordinate functions $(\theta,\varphi,\psi,\omega)$ as follows

$\displaystyle\theta(a,b)=1\;,\hskip 17.07164pt\varphi(a)=1\;,\hskip
17.07164pt\psi(a)=\frac{\lambda}{\lambda^{a}}\;,\hskip
17.07164pt\omega(a)=\lambda\;.$

The fact that these coefficient functions satisfy Equations 72 through 79 is
easy to check. We comment that Condition 66 is used to prove Equation 74,
Condition 67 is used to prove Equation 76 and 78, and both Conditions 66 and
67 are necessary to prove Equations 77 and 79. Thus the pair $(F,J)$ is a
monoidal equivalence, and the theorem is proven.

∎

######  Example 6.12.

The simplest dihedral group is the group $G=D_{2\cdot
1}\cong\mathbb{Z}/2\mathbb{Z}$. This corresponds to the case where $A$ is the
trivial group. Let us denote the nontrivial element of $G$ by $w$. By using
Theorem 6.11, we find that there are are exactly four categories with complex
$m$, and group $G$. Upon base extension to $\mathbb{C}$, these categories
become pointed, with fusion rules corresponding to $\mathbb{Z}/4\mathbb{Z}$ or
$(\mathbb{Z}/2\mathbb{Z})^{2}$.

When $g$ conjugates, the symmetry of $\chi$ implies that $\chi(w,w)=\pm 1$.
Condition 67 implies that it is possible for $\tau^{\prime}=-\tau$ by setting
$\lambda=\pm i$. This shows that there is an equivalence
$\mathcal{C}(g,\chi,\tau)\simeq\mathcal{C}(g,\chi,-\tau)$. Thus the only
relevant invariant of these categories is the number $\chi(w,w)$. The case
where $\chi(w,w)=1$ becomes
$\mathbb{C}\text{-}\mathrm{Vec}_{(\mathbb{Z}/2\mathbb{Z})^{2}}$ upon extension
to $\mathbb{C}$, while the case where $\chi(w,w)=-1$ becomes
$\mathbb{C}\text{-}\mathrm{Vec}_{(\mathbb{Z}/2\mathbb{Z})^{2}}^{\zeta\xi}$
upon extension to $\mathbb{C}$, where

$\zeta\xi(a^{i}b^{j},a^{k}b^{\ell},a^{m}b^{n})=(-1)^{ikm+j\ell n}\,$

is a representative cocycle for the associator. Using the descent theory of
[EG12], it can be verified that these are the only two cohomology classes in
$H^{3}\big{(}(\mathbb{Z}/2\mathbb{Z})^{2};\mathbb{C}^{\times}\big{)}$ for
which the corresponding pointed category admits a real form.

When $g$ doesn’t conjugate, Condition 67 implies that $\tau^{\prime}=\tau$.
Theorem 6.11 implies that the scalar $\chi(w,w)$ can always be normalized to
be $1$ by choosing any $\lambda$ such that $\lambda^{4}=\chi(w,w)^{-1}$. Thus
the only relevant invariant of these categories is $\tau=\pm\tfrac{1}{2}$. The
case where $\tau=\tfrac{1}{2}$ becomes
$\mathbb{C}\text{-}\mathrm{Vec}_{\mathbb{Z}/4\mathbb{Z}}$ upon extension to
$\mathbb{C}$, while the case where $\tau=-\tfrac{1}{2}$ becomes
$\mathbb{C}\text{-}\mathrm{Vec}_{\mathbb{Z}/4\mathbb{Z}}^{\upsilon^{2}}$,
where

$\upsilon^{2}(a^{i},a^{j},a^{k})=(-1)^{i\cdot\lfloor\frac{j+k}{4}\rfloor}$

represents the unique cohomology class of order two in
$H^{3}(\mathbb{Z}/4\mathbb{Z};\mathbb{C}^{\times}\big{)}$. Descent theory
again verifies that these are the only two associators that allow the category
to have a real form.

######  Example 6.13.

When $A=\mathbb{Z}/n\mathbb{Z}$, the group is $G=D_{2\cdot n}$, the dihedral
group of order $2n$. If $a\in A$ is a generator, then $\chi(a,a)$ must be a
primitive $n^{\text{th}}$ root of unity. The symmetry condition then implies
that $g$ is allowed to conjugate only in the case where
$A=\mathbb{Z}/2\mathbb{Z}$, and for all other cases $g$ must not conjugate.

Note that this restriction is only for classical dihedral groups. For
generalized dihedral groups where $A$ is non-cyclic, there are typically more
options.

##  7\. Analysis of Complex Galois case

The final case under consideration is where all simple objects are complex,
and $m$ is the unique Galois nontrivial simple object. The analysis follows
the same pattern as before, with care taken now to keep track of each time the
Galois action of $m$ is applied. The pentagon equations are as follows

(98) $\displaystyle\delta\alpha$ $\displaystyle=1$ (99)
$\displaystyle\delta\alpha_{3}$ $\displaystyle=\alpha^{-1}$ (100)
$\displaystyle\delta\alpha_{1}$ $\displaystyle=\overline{\alpha}$ (101)
$\displaystyle\alpha_{2}(a,bc)$ $\displaystyle=\alpha_{2}(a,c)\alpha_{2}(a,b)$
(102) $\displaystyle\alpha_{2}(ab,c)$
$\displaystyle=\alpha_{2}(b,c)\alpha_{2}(a,c)$ (103)
$\displaystyle\overline{\alpha(a,b,b^{-1}a^{-1}c)}\beta_{1}(ab,c)$
$\displaystyle=\beta_{1}(b,a^{-1}c)\beta_{1}(a,c)\overline{\alpha_{3}(a,b)}$
(104) $\displaystyle\beta_{3}(ab,c)\overline{\alpha(cb^{-1}a^{-1},a,b)}$
$\displaystyle=\alpha_{1}(a,b)\beta_{3}(b,c)\beta_{3}(a,cb^{-1})$ (105)
$\displaystyle\beta_{2}(b,c)$
$\displaystyle=\beta_{2}(b,a^{-1}c)\overline{\alpha_{2}(a,b)}$ (106)
$\displaystyle\beta_{2}(a,c)$
$\displaystyle=\alpha_{2}(a,b)\beta_{2}(a,cb^{-1})$ (107)
$\displaystyle\beta_{1}(a,c)\beta_{3}(b,c)$
$\displaystyle=\beta_{3}(b,a^{-1}c)\overline{\alpha(a,a^{-1}cb^{-1},b)}\beta_{1}(a,cb^{-1})$
(108) $\displaystyle\beta_{2}(a,c)\beta_{2}(b,c)$
$\displaystyle=\alpha_{3}(a,b)\beta_{2}(ab,c)\overline{\alpha_{1}(a,b)}$ (109)
$\displaystyle\alpha_{2}(a,c)\gamma(c,b)$
$\displaystyle=\overline{\beta_{1}(a,b)}\alpha_{3}(a,a^{-1}b)\gamma(c,a^{-1}b)$
(110) $\displaystyle\alpha_{2}(b,a)\gamma(c,b)$
$\displaystyle=\beta_{3}(a,c)\alpha_{1}(ca^{-1},a)\gamma(ca^{-1},b)$ (111)
$\displaystyle\alpha_{1}(a,c)\gamma(c,b)$
$\displaystyle=\overline{\beta_{2}(a,b)}\beta_{1}(a,ac)\gamma(ac,b)$ (112)
$\displaystyle\alpha_{3}(b,a)\gamma(c,b)$
$\displaystyle=\beta_{2}(a,c)\overline{\beta_{3}(a,ba)}\gamma(c,ba)$ (113)
$\displaystyle\delta_{d,ba^{-1}}\beta_{3}(a,b)\beta_{1}(ba^{-1},b)$
$\displaystyle=\sum_{c}\beta_{2}(c,b)\overline{\gamma(c,d)}\gamma(a,c)$

The change of basis formulas are

(114) $\displaystyle\alpha^{\prime}$ $\displaystyle=\delta\theta\cdot\alpha$
(115) $\displaystyle\alpha_{1}^{\prime}(a,b)$
$\displaystyle=\frac{\overline{\psi}(ab)\theta(a,b)}{\overline{\psi}(a)\overline{\psi}(b)}\cdot\alpha_{1}(a,b)$
(116) $\displaystyle\alpha_{2}^{\prime}(a,b)$ $\displaystyle=\alpha_{2}(a,b)$
(117) $\displaystyle\alpha_{3}^{\prime}(a,b)$
$\displaystyle=\frac{\varphi(b)\varphi(a)}{\varphi(ab)\theta(a,b)}\cdot\alpha_{3}(a,b)$
(118) $\displaystyle\beta_{1}^{\prime}(a,b)$
$\displaystyle=\frac{\omega(a^{-1}b)\theta(a,a^{-1}b)}{\overline{\varphi}(a)\omega(b)}\cdot\beta_{1}(a,b)$
(119) $\displaystyle\beta_{2}^{\prime}(a,b)$
$\displaystyle=\frac{\varphi(a)}{\overline{\psi}(a)}\cdot\beta_{2}(a,b)$ (120)
$\displaystyle\beta_{3}^{\prime}(a,b)$
$\displaystyle=\frac{\overline{\psi}(a)\overline{\omega}(b)}{\theta(ba^{-1},a)\overline{\omega}(ba^{-1})}\cdot\beta_{3}(a,b)$

The sequence of deductions that follow is only a superficial modification of
the original argument of [TY98], but we include it here for completeness.

By setting $\theta=\alpha_{3}$, and $\varphi\equiv 1$ in Equation 117, we may
assume that $\alpha\equiv 1$ and $\alpha_{3}\equiv 1$. Setting
$\overline{\psi}(a)=\beta_{2}(a,1)$, in Equation 119 allows us to assume that
$\beta_{2}(a,1)\equiv 1$.

Next, we can substitute $b=1$ into Equation 118 to find that

$\beta_{1}^{\prime}(a,1)=\frac{\omega(a^{-1})\theta(a,a^{-1})}{\omega(1)}\cdot\beta_{1}(a,1)\,.$

By rearranging this formula, we find that if we set

$\omega(a^{-1}):=\frac{\omega(1)}{\theta(a,a^{-1})\beta_{1}(a,1)}\,,$

this will allow us to assume that $\beta_{1}(a,1)\equiv 1$. Upon substituting
$c=1$ into Equation 103, this new normalization shows that $\beta_{1}\equiv
1$.

Setting $a=c$ in Equation 105, and $b=c$ in Equation 106 imply that

$\alpha_{2}(b,a)\;=\;\beta_{2}(b,a)\;=\;\overline{\alpha_{2}(a,b)}\,.$

By Equations 101 and 102, $\beta_{2}$ is a bicharacter, and thus Equation 108
implies that $\alpha_{1}\equiv 1$.

Equation 107 implies that $\beta_{3}(a,b)=\beta_{3}(a,1)$, and by setting
$a=b$ in Equation 109, we find that

$\gamma(a,b)=\alpha_{2}(a,b)\gamma(a,1)\,.$

We can combine these with Equations 110 and 111 to obtain

$\gamma(c,1)\;=\;\beta_{3}(a,1)\gamma(ca^{-1},1)\qquad\gamma(c,1)\;=\;\gamma(ac,1)\,,$

which shows that $\beta_{3}\equiv 1$ and

$\gamma(a,b)\;=\;\alpha_{2}(a,b)\gamma(1,1)\,.$

With all these observations in place, Equation 113 reduces to

$\delta_{d,ba^{-1}}\;=\;\gamma(1,1)^{2}\sum_{c\in
G}\alpha_{2}\big{(}c,ba^{-1}d^{-1}\big{)}\,,$

which is equivalent to $\alpha_{2}$ being nondegenerate, and
$\gamma(1,1)^{2}|G|=1$.

######  Theorem 7.1.

Let $\tau=\nicefrac{{\pm 1}}{{\sqrt{|G|}}}$, and let $\chi:G\times
G\to\mathbb{C}^{\times}$ be a nondegenerate skew-symmetric bicharacter on $G$.

A triple of such data gives rise to a non-split Tambara-Yamagami category
$\mathcal{C}_{\overline{\mathbb{C}}}(G,\tau,\chi)$, with
$\mathrm{End}(X)\cong\mathbb{C}$ for every simple object $X$. Furthermore, all
equivalence classes of such categories arise in this way. Two categories
$\mathcal{C}_{\overline{\mathbb{C}}}(G,\tau,\chi)$ and
$\mathcal{C}_{\overline{\mathbb{C}}}(G^{\prime},\tau^{\prime},\chi^{\prime})$
are equivalent if and only if $\tau=\tau^{\prime}$, and there exist
isomorphisms:

  1. i

$f:G\to G^{\prime}$, and

  2. ii

$(-)^{h}:\mathbb{C}\to\mathbb{C}$, either the identity or complex conjugation,

subject to the condition that for all $a,b\in G$:

(121) $\displaystyle\chi^{\prime}\Big{(}f(a),f(b)\Big{)}=\chi(a,b)^{h}\;.$

###### Proof.

The first statement follows from a construction of the category
$\mathcal{C}_{\overline{\mathbb{C}}}(\tau,\chi)$, and this is completely
analogous to the construction in [TY98]. The ‘furthermore’ statement follows
from the reduction process outlined above.

The main subtlety in the equivalence classification comes from the fact that
these categories are only fusion over $\mathbb{R}$ and not over $\mathbb{C}$.
This means that functors are allowed to act by nontrivial homomorphisms on the
algebras $\mathrm{End}(X)$. The equivalences are monoidal, and the naturality
property of the monoidal structure map dictates that all of the algebra
homomorphisms $F:\mathrm{End}(X)\to\mathrm{End}\big{(}F(X)\big{)}$ are
completely determined by what happens on $\mathrm{End}(\mathds{1})$. Since
this map is allowed to be either the identity or complex conjugation, the
result follows. ∎

######  Remark 7.2.

The reader may recognize this as a skew-symmetric analogue of the classical
Tambara-Yamagami classification. Nondegenerate skew-symmetric bicharacters on
finite abelian groups were classified in [Wal63]. The classification involves
the familiar ‘hyperbolic’ bicharacters on the $p$-primary summands, with extra
possibilities for the case when $p=2$.

######  Example 7.3.

Let $G=\mathbb{Z}/2\mathbb{Z}=\langle t\rangle$. The bicharacter
$\chi(t,t)=-1$ is nondegenerate, and simultaneously symmetric and skew-
symmetric. Thus in addition to giving rise to classical split Tambara-Yamagami
categories, it can also be used to produce
$\mathcal{C}_{\overline{\mathbb{C}}}(G,\tau,\chi)$. The only difference
between the two constructions is the Galois nontriviality of $m$.

######  Example 7.4.

Let $G=(\mathbb{Z}/4\mathbb{Z})^{2}=\langle x,y\rangle$. Define a skew-
symmetric bicharacter by the formulae

$\chi(x,x)=1\,,\qquad\chi(x,y)=i\,,\qquad\chi(y,y)=-1\,.$

This is nondegenerate, so $\chi$ can be used to produce
$\mathcal{C}_{\overline{\mathbb{C}}}(G,\tau,\chi)$. This type of bicharacter
has no nondegenerate analogues for odd $p$-primary groups.

## References

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  * [BL04] John C. Baez and Aaron D. Lauda  “Higher-dimensional algebra. V. 2-groups”  In _Theory Appl. Categ._ 12, 2004, pp. 423–491
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  * [San22] Sean C. Sanford  “Fusion Categories over Non-algebraically Closed Fields” Copyright - Database copyright ProQuest LLC; ProQuest does not claim copyright in the individual underlying works; Last updated - 2022-11-23  In _ProQuest Dissertations and Theses_ , 2022, pp. 172  URL: https://proxyiub.uits.iu.edu/login?qurl=https%5C%3A%5C%2F%5C%2Fwww.proquest.com%5C%2Fdissertations-theses%5C%2Ffusion-categories-over-non-algebraically-closed%5C%2Fdocview%5C%2F2715420938%5C%2Fse-2
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As the susceptibility sensitively depends on $\Omega$ at the resonance
condition $\Delta\omega_{C}=0$, the superhet detector operators under these
circumstances, in which

$\displaystyle
c_{2}=i\frac{\lambda_{p}}{2\pi}\left(\gamma_{2}+\frac{\Omega_{C}^{2}\gamma}{\Omega^{2}+\gamma^{2}}\right).$
(S99)

Based on Eqs. (S79), (S81) and (S98), we can now evaluate the SNR for the
operator $\hat{\mathcal{X}}$. Recalling that the total Rabi coupling between
the Rydberg states consists of the local oscillator and the axion-induced
dipole transitions $\Omega=\Omega_{LO}+\Omega_{a}$, we find that the signal
for small $\Omega_{a}$ is given by

$\displaystyle\delta\left<\hat{\mathcal{X}}\right>$ $\displaystyle=$
$\displaystyle\Omega_{a}\frac{d}{d\Omega_{a}}\left.\left<\hat{\mathcal{X}}\right>\right|_{\Omega_{a}=0}$
(S100) $\displaystyle=$ $\displaystyle
2\cdot\frac{1}{\left(\gamma_{2}+\frac{\omega_{p}\sigma}{c}\Gamma+\frac{\Omega_{C}^{2}\gamma}{\Omega_{LO}^{2}+\gamma^{2}}\right)^{2}}\left(1+\Gamma^{\prime}\right)\frac{\Omega_{C}^{2}\gamma}{\left(\Omega_{LO}^{2}+\gamma^{2}\right)^{2}}2\Omega_{LO}\cdot
N\Omega_{p}\Omega_{a}t_{\text{m}},$

where we have defined $\Gamma=\Gamma\left(\zeta/(\sqrt{2}\sigma)\right)$ with
$\zeta=\gamma_{2}+\gamma\Omega_{c}^{2}/\left(\Omega_{LO}^{2}+\gamma^{2}\right)$
and $\sigma^{2}=k_{b}T/m_{R}$. Moreover, we have introduced
$\Gamma^{\prime}=\partial_{z}\Gamma(z)\mid_{z\rightarrow\zeta/(\sqrt{2}\sigma)}$.
Likewise, we can evaluate the variance for an ensemble of atoms

$\displaystyle\left<\text{Var}\hat{\mathcal{X}}\right>$ $\displaystyle=$
$\displaystyle\left<\hat{\mathcal{X}}^{2}\right>=\frac{4t_{\text{m}}N}{{\gamma_{b}}+\frac{\omega_{p}\sigma}{c}\Gamma+\frac{\Omega_{C}^{2}\gamma}{\Omega_{LO}^{2}+\gamma^{2}}},$
(S101)

where we have approximated the Rabi coupling between the Rydberg states by the
local oscillator $\Omega\rightarrow\Omega_{L0}$. Putting everything together,
the SNR reads as

$\displaystyle SNR$ $\displaystyle\equiv$
$\displaystyle\frac{\delta\left<\hat{\mathcal{X}}\right>}{\left<\text{Var}\hat{\mathcal{X}}\right>^{\frac{1}{2}}}$
(S102) $\displaystyle=$
$\displaystyle\sqrt{N}\Omega_{a}\sqrt{T_{c}t_{\text{m}}}=\sqrt{N}\Omega_{a}\tau,$
(S103)

where we have defined the effective measurement time $\tau=\sqrt{T_{c}\cdot
t_{\text{m}}}$ in terms of the effective coherence time

$\displaystyle T_{c}$ $\displaystyle=$
$\displaystyle\frac{4\Omega_{p}^{2}\Omega_{C}^{4}}{\left[\left(\gamma_{2}+\frac{\omega_{p}\sigma}{c}\Gamma\right)\left(\Omega_{LO}^{2}+\gamma^{2}\right)+\Omega_{C}^{2}\gamma\right]^{3}}\frac{\gamma^{2}\Omega_{LO}^{2}}{\left(\Omega_{LO}^{2}+\gamma^{2}\right)}\left(1+\Gamma^{\prime}\right)^{2}.$
(S104)

Finally, we recall that we have considered $\Omega$ as quasistatic throughout
the derivations. Taking the time dependence
$\Omega(t)=\Omega_{LO}+\Omega_{a}e^{i(\omega_{a}-\omega_{LO})t}$ into account,
we find that the measurement signal will slowly oscillate with frequency
$\omega_{a}-\omega_{LO}$ and amplitude in Eq. (S100). Thus, the SNR will
remain unchanged in the case of heterodyning.