Fake Galois Actions

TL;DR

This paper establishes the existence of fake Galois actions on irreducible Brauer characters for universal covers of non-abelian simple groups, aiding the proof of an $ ext{l}$-modular Glauberman-Isaacs correspondence.

Contribution

It proves the existence of fake Galois actions for all non-abelian simple groups' universal covers, a key step in modular representation theory.

Findings

Existence of fake Galois actions for all non-abelian simple groups' universal covers.

Provides an inductive condition for $ ext{l}$-modular Glauberman-Isaacs correspondence.

Advances understanding of modular representation theory of finite groups.

Abstract

We prove that for all non-abelian finite simple groups $S$, there exists a fake mth Galois action on IBr$(X)$ with respect to $X \lhd X \rtimes $ Aut$(X)$, where $X$ is the universal covering group of $S$ and $m$ is any non-negative integer coprime to the order of $X$. This is one of the two inductive conditions needed to prove an $\ell$-modular analogue of the Glauberman-Isaacs correspondence.

Equations

$$(G,N,\theta {)}^{(m)}\approx (G,N,{\theta }^{\prime })$$

$$\alpha (g,g^{\prime }{)}^m={\alpha }^{\prime }(g,g^{\prime })$$

$$f_m:\text{ IBr}(N)\to \text{ IBr}(N)$$

$$(G_{\theta },N,\theta {)}^{(m)}\approx (G_{\theta },N,f_m(\theta )).$$

$$U=\{(g,h)\in G_1\times H_1\mid \phi (g{G}_2)=h{H}_2\},$$

$$1\to G_2\times 1\to U\to H_1\to 1.$$

$$\mathcal{P}(g)={\mathcal{D}}_1(h_1){\mathcal{P}}_2(h_2)$$

$$\alpha (g,g^{\prime })={\lambda }_{h_2}(h_1^{\prime }){\alpha }_2(h_2,h_2^{\prime }{)}^{-1}$$

$$\mathcal{P}(g)\mathcal{P}(g^{\prime })$$

$$={\mathcal{D}}_1(h_1){\mathcal{D}}_1(h_1^{\prime }{)}^{{\mathcal{P}}_2(h_2{)}^{-1}}{\mathcal{P}}_2(h_2){\mathcal{P}}_2(h_2^{\prime }).$$

$$\begin{array}[]{rrclccrrcl}\nu_{1}:&H_{1}&\to&GL_{\theta(1)}({\mathbb{F}})&&&\nu_{2}:&H_{1}&\to&GL_{\theta(1)}({\mathbb{F}})\\ &h&\mapsto&{\mathcal{D}}_{1}(h)^{{\mathcal{P}}_{2}(h_{2})^{-1}}&&&&h&\mapsto&{\mathcal{D}}_{1}(h_{2}hh_{2}^{-1})\end{array}$$

$$\mathcal{P}(g)\mathcal{P}(g^{\prime })$$

$$={\mathcal{D}}_1(h_1)({\nu }_2{\lambda }_{h_2})(h_1^{\prime }){\mathcal{P}}_2(h_2){\mathcal{P}}_2(h_2^{\prime })$$

$$={\lambda }_{h_2}(h_1^{\prime }){\alpha }_2(h_2,h_2^{\prime }{)}^{-1}\mathcal{P}(g{g}^{\prime }).$$

$${\tilde{f}}_m:\text{ IBr}({\tilde{\text{G}}}^F)\to \text{ IBr}({\tilde{\text{G}}}^F)$$

$${\tilde{f}}_m:\mathrm{IBr}({\tilde{\text{G}}}^F)$$

$$\chi$$

$${\tilde{f}}_m(\lambda \chi )={\tilde{\Theta }}^{-1}(\tilde{\Theta }(\chi \lambda {)}^{\sigma })={\tilde{\Theta }}^{-1}((\hat{z}\tilde{\Theta }(\chi ){)}^{\sigma })=d^1(\hat{z}{)}^m{\tilde{\Theta }}^{-1}(\tilde{\Theta }(\chi {)}^{\sigma })={\lambda }^m{\tilde{f}}_m(\chi ).$$

$$D_{\theta }=\{d\in D\mid {\chi }^d=\chi \lambda \text{ for some }\lambda \in \text{ IBr}({\tilde{\text{G}}}^F/{\text{G}}^F)\}.$$

$$f_m(\theta ):={\chi }_0^{gd}.$$

$$D_{{\theta }_0}=\left\{d\in D\mid {\tilde{\theta }}_0^d={\tilde{\theta }}_0\lambda \text{ for some }\lambda \in \text{ IBr}({\tilde{\text{G}}}^F/{\text{G}}^F)\right\}.$$

$$D_{{\theta }_0}=$$

$$=$$

$$(({\tilde{\text{G}}}^F⋊D{)}_{\theta },{\text{G}}^F,\theta {)}^{(m)}\approx (({\tilde{\text{G}}}^F⋊D{)}_{\theta },{\text{G}}^F,f_m(\theta ))$$

$$\mathcal{P}(x)={\mathcal{D}}_1(h_1){\mathcal{D}}_2(h_2),\text{mbox}and$$

$$\alpha (x,x^{\prime })={\lambda }_{h_2}(h_1^{\prime }),$$

$${\mathcal{P}}^{\prime }(x)={\mathcal{D}}_1^{\prime }(h_1){\mathcal{D}}_2^{\prime }(h_2),\text{mbox}and$$

$${\alpha }^{\prime }(x,x^{\prime })={\lambda }_{h_2}^{\prime }(h_1^{\prime }),$$

$${\tilde{f}}_m({\tilde{\theta }}_1){\tilde{\lambda }}_{h_2}^{\prime -1}={\tilde{f}}_m({\tilde{\theta }}_1{)}^{h_2}={\tilde{f}}_m({\tilde{\theta }}_1{\tilde{\lambda }}_{h_2}^{-1})={\tilde{f}}_m({\tilde{\theta }}_1){\tilde{\lambda }}_{h_2}^{-m}.$$

$$\text{Diag}({\text{G}}^F)\cong {\tilde{\text{G}}}^F/(\text{Z}(\tilde{\text{G}}{)}^F{\text{G}}^F)\cong {\mathcal{L}}^{-1}(\text{Z}(\text{G}))/(\text{Z}(\text{G}){\text{G}}^F).$$

$$A/{\text{G}}^F={\mathcal{L}}^{-1}(\text{Z}(\text{G}))/{\text{G}}^F⋊⟨F_0,\Gamma ⟩\cong \text{Z}(\text{G})⋊⟨F_0,\Gamma ⟩,$$

$$\delta :A/{\text{G}}^F\twoheadrightarrow \mathrm{Out}({\text{G}}^F).$$

Full text

The first author gratefully acknowledges financial support received from a London Mathematical Society 150th Anniversary Postdoctoral Mobility Grant, and from the DFG project SFB-TRR 195.

This material is based upon work supported by the National Science Foundation under Grant No. DMS-1440140 while the second named author was in residence at the Mathematical Sciences Research Institute in Berkeley, California, during the Spring 2018 semester. The second author’s research was conducted in the framework of the research training group GRK 2240: Algebro-geometric Methods in Algebra, Arithmetic and Topology, which is funded by the DFG

1. Introduction

Let $G$ and $A$ be finite groups such that $A$ acts on $G$ via group automorphisms. Then the action of $A$ on $G$ induces an action on the set of complex irreducible characters of $G$. Deep work of Glauberman in the 1960’s revealed an astonishing correspondence between the $A$-invariant characters of $G$, and the characters of the fixed point subgroup $\mathrm{C}_{G}(A)$, when $A$ and $G$ have coprime orders and $A$ is solvable. This work was extended by Isaacs in the 1970’s and their combined results became known as the Glauberman–Isaacs correspondence. This is a very important result in the representation theory of finite groups, and it has been used in a fundamental way in solving many problems relating to group actions, as well as in work on some of the famous open conjectures in modular representation theory such as the Alperin weight conjecture.

Theorem** (Glauberman–Isaacs Correspondence).**

Let $G$ and $A$ be finite groups. Suppose that $A$ acts on $G$ via group automorphisms and suppose that $(|A|,|G|)=1$. Then there exists a canonical bijection between the set of $A$-invariant irreducible characters of $G$ and the set of irreducible characters of the fixed point subgroup $\mathrm{C}_{G}(A)$.

We are concerned with an $\ell$-modular analogue of the Glauberman–Isaacs correspondence. Let $\ell$ be a prime and suppose that $A$ and $G$ are as in the Theorem above. Then the action of $A$ on $G$ yields a natural action on the $\ell$-Brauer characters of $G$. In [11, Problem 4], Navarro posed the following question: does there exist a bijection between the set of $A$-invariant irreducible $\ell$-Brauer characters of $G$ and the set of irreducible $\ell$-Brauer characters of $\mathrm{C}_{G}(A)$? The answer is known in some specific cases, for example the answer is yes if $G$ is $\ell$-solvable [18], but it is not known in general whether such a correspondence always exists. Späth and Vallejo Rodríguez proved a reduction theorem for this question in [16]. For this they introduced a set of conditions on finite simple groups which they called the inductive Brauer-Glauberman (IBG) condition [16, Definition 6.1], and proved the following.

Theorem** ([16, Theorem A]).**

Let $G$ and $A$ be finite groups. Suppose that $A$ acts on $G$ with $(|A|,|G|)=1$. Suppose that all finite non-abelian simple groups involved in $G$ satisfy the IBG condition. Then the number of $A$-invariant irreducible $\ell$-Brauer characters of $G$ is equal to the number of irreducible $\ell$-Brauer characters of $\mathrm{C}_{G}(A)$.

The IBG condition is known to hold for finite simple groups of Lie type defined over a field of characteristic $\ell$, and for finite simple groups not of Lie type [14]. In order to prove Navarro’s conjecture it therefore remains to show that the IBG condition holds for all finite simple groups of Lie type in non-defining characteristic. This condition has two parts. The first part is a stronger, suitably equivariant version of a modular Glauberman–Isaacs correspondence. The second part is the existence of what Späth and Vallejo Rodríguez call a fake $m$th Galois action, see Definition 2.2. Galois automorphisms act on ordinary characters and change their Clifford theory. A fake Galois action is a bijective map on the set of irreducible Brauer characters changing the Clifford theory in an analogous way. We are concerned here with this part of the condition and prove the following result.

Theorem A**.**

Let $S$ be a non-abelian simple group and let $X$ be the universal covering group of $S$. Then for all non-negative integers $m$ such that $(|X|,m)=1$, there exists a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$.

The organisation of this paper is as follows. In Section 2 we recall the most important definitions and notation. We also state some general results about projective representations which might be of independent interest. In Section 3 we prove the existence of fake Galois actions for groups of Lie type A. The main idea is to use the ordinary Galois action on complex characters and the fact that the $\ell$-decomposition matrix of $\mathrm{SL}_{n}(q)$ is unitriangular. In Section 4 we deal with all groups of Lie type not of type A. As the outer automorphism group of D${}_{4}(q)$ is more complicated than in the other cases, it often has to be dealt with using separate arguments. The construction of fake Galois actions for exceptional covering groups of simple finite groups of Lie type with non-cyclic outer automorphism groups is in Section 5. Finally, the proof of the main theorem is in Section 6.

Acknowledgements We are extremely grateful to Professor Britta Späth for her support and encouragement on this project. The first author would like to thank the Bergische Universität Wuppertal for generously hosting her for a period of six months during which much of this article was written. We also thank Professor Gunter Malle for his careful reading of an earlier manuscript and for his helpful comments.

2. Notation, definitions and preliminary results

2.1. Notation

Let $\ell$ be a prime and let ${\mathbb{F}}=\overline{\mathbb{F}}_{\ell}$ be an algebraic closure of the finite field of $\ell$ elements. Let $G$ be a finite group and let $N\lhd G$. We let ${\operatorname{Irr}}(G)$ denote the set of complex irreducible characters of $G$ and ${\operatorname{IBr}}(G)$ denote the set of $\ell$-modular Brauer characters of $G$ (see [12, Chapter 2]). For $\chi\in{\operatorname{Irr}}(G)$, $\chi_{N}$ denotes the restriction of $\chi$ to $N$ and ${\operatorname{Irr}}(N\mid\chi)$ denotes the set of irreducible constituents of $\chi_{N}$. For $\theta\in{\operatorname{Irr}}(N)$, $\theta^{G}$ denotes the induced character of $G$ and ${\operatorname{Irr}}(G\mid\theta)$ denotes the set of irreducible constituents of $\theta^{G}$. The notation for irreducible Brauer characters is analogous (see [12, Corollary 8.7]). For $\theta$ a complex or irreducible character of $G$, let $G_{\theta}$ denote the stabiliser of $\theta$ in $G$. Let $G_{\ell^{\prime}}$ denote the subset of $G$ of elements with order coprime to $\ell$. Then $d^{1}:{\mathbb{Z}}{\operatorname{Irr}}(G)\to{\mathbb{Z}}{\operatorname{IBr}}(G)$ denotes the decomposition map given by $d^{1}(\chi)=\chi\circ\nu$ where $\nu$ is the characteristic function of $G_{\ell^{\prime}}$.

2.2. Fake $m$th Galois action

Recall that a modular character triple is a triple $(G,N,\theta)$ where $N\lhd G$ are finite groups and $\theta$ is a $G$-stable Brauer character of $N$. Suppose that $\mathcal{D}:G\rightarrow{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ is a representation of $N$ affording $\theta$. Then there exists a projective representation ${\mathcal{P}}$ of $G$ such that ${\mathcal{P}}_{N}=\mathcal{D}$, and we can choose ${\mathcal{P}}$ such that its associated factor set $\alpha:G\times G\rightarrow{\mathbb{F}}^{\times}$ satisfies $\alpha(g,n)=\alpha(n,g)=1$ for all $g\in G$ and $n\in N$. We say that such a projective representation ${\mathcal{P}}$ is associated to the triple $(G,N,\theta)$, or associated to $\theta$.

Definition 2.1**.**

Let $(G,N,\theta)$ and $(G,N,\theta^{\prime})$ be modular character triples and let $m$ be a positive integer such that $(m,|N|)=1$. Then we write

[TABLE]

if there exist projective representations ${\mathcal{P}}$ and ${\mathcal{P}}^{\prime}$ of $G$ associated to $\theta$ and $\theta^{\prime}$ respectively, such that

• •

for every $c\in\mathrm{C}_{G}(N)$ the scalar matrices ${\mathcal{P}}(c)$ and ${\mathcal{P}}^{\prime}(c)$ are associated to some $m^{\prime}$ roots of unity $\xi$ and $\xi^{m}$;

• •

the factor sets $\alpha$ and $\alpha^{\prime}$ of ${\mathcal{P}}$ and ${\mathcal{P}}^{\prime}$ respectively take as values $m^{\prime}$ roots of unity and satisfy

[TABLE]

for every $g,g^{\prime}\in G$.

Definition 2.2**.**

Let $N\lhd G$. There exists a fake $m$th Galois action on ${\operatorname{IBr}}(N)$ with respect to $N\lhd G$ if there exists a $G$-equivariant bijection

[TABLE]

such that for every $\theta\in{\operatorname{IBr}}(N)$,

[TABLE]

Let ${\mathcal{P}}:G\rightarrow{\textnormal{GL}}_{n}({\mathbb{F}})$ be a projective representation with factor set $\alpha:G\times G\rightarrow{\mathbb{F}}^{\times}$. For an element $x\in G$ we let ${\mathcal{P}}^{x}$ denote the projective representation given by ${\mathcal{P}}^{x}(g)={\mathcal{P}}(g^{x^{-1}})$ for all $g\in G$.

Lemma 2.3**.**

Let $N\lhd G$ be finite groups. Let $\theta\in{\operatorname{IBr}}(N)$ and let $f:{\operatorname{IBr}}(N)\rightarrow{\operatorname{IBr}}(N)$ be a $G$-equivariant bijection. Then $(G_{\theta},N,\theta)^{(m)}\approx(G_{\theta},N,f(\theta))$ if and only if $(G_{\theta^{g}},N,\theta^{g})^{(m)}\approx(G_{\theta^{g}},N,f(\theta^{g}))$ for any $g\in G$.

Proof.

Let ${\mathcal{P}}$ and ${\mathcal{P}}^{\prime}$ be the projective representations associated to $\theta$ and $f(\theta)$ respectively such that $(G_{\theta},N,\theta)^{(m)}\approx(G_{\theta},N,f(\theta))$ as in Definition 2.1. Then for any $g\in G$, ${\mathcal{P}}^{g}$ and ${\mathcal{P}}^{\prime g}$ are projective representations associated to $\theta^{g}$ and $f(\theta)^{g}$ respectively, and they satisfy the properties of Definition 2.1 so $(G_{\theta^{g}},N,\theta^{g})^{(m)}\approx(G_{\theta^{g}},N,f(\theta^{g}))$. ∎

2.3. Results from group theory

Recall that every subgroup $U$ of $G\times H$ has the form

[TABLE]

for some $5$-tuple $(G_{1},G_{2},H_{1},H_{2},\varphi)$ where $G_{2}\lhd G_{1}\leq G$ and $H_{2}\lhd H_{1}\leq H$ and $\varphi:G_{1}/G_{2}\to H_{1}/H_{2}$ is an isomorphism. We note that $U$ is a subgroup of $G_{1}\times H_{1}$, and $G_{2}\times 1$ and $1\times H_{2}$ are normal subgroups of $U$.

Lemma 2.4**.**

Let $G$ be a finite group and let $a\in{\mathbb{N}}_{>0}$. Suppose that $U$ is a subgroup of $G\times C_{a}$ associated to a 5-tuple $(G_{1},G_{2},H_{1},H_{2},\varphi)$ such that $G_{1}$ is abelian or $G_{1}=G_{2}\rtimes K$ for some finite group $K$. Then $U\cong G_{2}\rtimes H_{1}$.

Proof.

If $G_{1}$ is abelian then $U$ is a subgroup of an abelian group so the result follows from the classification of finite abelian groups. Suppose now that $G_{1}=G_{2}\rtimes K$ for some finite group $K$.

Consider the projection map $\pi:U\to H_{1}$, given by $\pi(g,h)=h$. Since ker$(\pi)=G_{2}\times 1$ we obtain an exact sequence,

[TABLE]

Let $H_{1}=\langle h_{1}\rangle$. Consider the map $\pi^{\prime}:H_{1}\to U,\,h_{1}\mapsto(g_{1},h_{1})$ where $g_{1}\in K$ is such that $\varphi(g_{1}G_{2})=h_{1}H_{2}$. Since $K\cong G_{1}/G_{2}\cong H_{1}/H_{2}$, the order of $h_{1}$ equals the order of $(g_{1},h_{1})$ and hence this map is well defined. As $\pi\circ\pi^{\prime}=$ id, $\pi^{\prime}$ is a section of $\pi$ and hence $U\cong G_{2}\rtimes H_{1}$ as required. ∎

Corollary 2.5**.**

Let $a\in{\mathbb{N}}_{>0}$ and let $U\leq S_{4}\times C_{a}$ be associated to a 5-tuple $(G_{1},G_{2},H_{1},H_{2},\varphi)$. Then $U\cong G_{2}\rtimes H_{1}$. Moreover, if $A_{4}\leq G_{1}$, then $V_{4}\times 1\lhd U$.

Proof.

By Lemma 2.4, we need to show that for any $U\leq S_{4}\times C_{a}$ associated to a 5-tuple $(G_{1},G_{2},H_{1},H_{2},\varphi)$, either $G_{1}$ is abelian or $G_{1}=G_{2}\rtimes K$ for some finite group $K$. Clearly if $G_{2}$ is trivial, then the second condition automatically holds. Since $H_{1}\leq C_{a}$ is cyclic and $G_{1}/G_{2}\cong H_{1}/H_{2}$, it follows that $G_{1}/G_{2}$ is cyclic. It is therefore enough to consider each of the possibilities for $G_{1}$ and $G_{2}$ such that $G_{1}$ is not abelian, $G_{2}$ is not trivial, and $G_{1}/G_{2}$ is cyclic.

First let $G_{1}=S_{4}$. Then $G_{2}=A_{4}$ and $G_{1}=G_{2}\rtimes\langle(1,2)\rangle$ is a semidirect product. If $G_{1}=S_{3}$ then $G_{2}\cong A_{3}$ and hence $G_{1}=G_{2}\rtimes\langle(1,2)\rangle$. Now suppose that $G_{1}=D_{8}=\langle a,b\mid a^{4}=b^{2}=1,\,[b,a]=a^{2}\rangle$. Then $G_{2}$ is of order 2 or 4. If $|G_{2}|=2$ then $G_{2}=\langle a^{2}\rangle$ and $G_{1}/G_{2}\cong C_{2}\times C_{2}$ is not cyclic. Therefore $G_{2}$ is of order 4 and so either $G_{2}=\langle a\rangle$ and $G_{1}=G_{2}\rtimes\langle b\rangle$; $G_{2}=\langle a^{2},b\rangle$ and $G_{1}=G_{2}\rtimes\langle ab\rangle$; or $G_{2}=\langle ab,a^{2}\rangle$ and $G_{1}=G_{2}\rtimes\langle b\rangle$. Finally, suppose that $G_{1}=A_{4}$. Then $G_{2}=V_{4}$ and $G_{1}=G_{2}\rtimes\langle(1,2,3)\rangle$. We note that whenever $A_{4}\leq G_{1}$ we have $V_{4}\leq G_{2}$ and therefore $V_{4}\times 1\lhd U$. ∎

2.4. Results from ordinary and projective character theory

Lemma 2.6**.**

Let $N\lhd G$ be finite groups. Let $\theta\in{\operatorname{IBr}}(N)$ and let $\chi,\chi^{\prime}\in{\operatorname{IBr}}(G\mid\theta)$. Then if $G/N$ is abelian, or if $\chi$ and $\chi^{\prime}$ are extensions of $\theta$, there exists a linear character $\lambda\in{\operatorname{IBr}}(G/N)$ such that $\chi=\lambda\chi^{\prime}$.

Proof.

By definition, $\chi$ and $\chi^{\prime}$ are irreducible constituents of $\theta^{G}$. It then follows from [12, Corollary 8.20] that there exists an irreducible character $\lambda\in{\operatorname{IBr}}(G/N)$ such that $\chi=\lambda\chi^{\prime}$. If $G/N$ is abelian, then $\lambda$ is a linear character. If $\chi$ and $\chi^{\prime}$ are both extensions of $\theta$ then $\chi$ and $\chi^{\prime}$ have the same degree. Therefore, $\lambda$ is again a linear character. ∎

If $\lambda:G_{\ell^{\prime}}\to\mathbb{C}^{\times}$ is a linear Brauer character of $G$ we denote (by abuse of notation) $\lambda:G\to\mathbb{F}^{\times}$ the unique representation affording $\lambda$. The following is a generalisation of ideas present in the proof of [14, Proposition 5.2].

Lemma 2.7**.**

Let $G$ be a finite group. Suppose that $H_{1}\lhd G$ and $G=H_{1}H_{2}$. Let $H=H_{1}\cap H_{2}$ and let $\theta\in{\operatorname{IBr}}(H)$ be a $G$-stable Brauer character of $H$. Suppose that there exists an extension $\theta_{1}\in{\operatorname{IBr}}(H_{1})$ of $\theta$. Let ${\mathcal{D}}_{1}:H_{1}\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ be a representation of $H_{1}$ affording $\theta_{1}$ and let ${\mathcal{P}}_{2}:H_{2}\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ be a projective representation associated to $\theta$ with factor set $\alpha_{2}$, such that ${\mathcal{D}}_{1}$ and ${\mathcal{P}}_{2}$ agree on $H$. Then there exists a well defined projective representation ${\mathcal{P}}:G\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ associated to $\theta$ with factor set $\alpha$ such that for all $g=h_{1}h_{2},g^{\prime}=h_{1}^{\prime}h_{2}^{\prime}\in G$, $h_{1},h_{1}^{\prime}\in H_{1}$, $h_{2},h_{2}^{\prime}\in H_{2}$,

[TABLE]

and $\alpha$ satisfies

[TABLE]

where $\lambda_{h_{2}}$ is the unique linear Brauer character of ${\operatorname{IBr}}(H_{1}/H)$ such that $\theta_{1}^{h_{2}}{\lambda}_{h_{2}}=\theta_{1}$.

Proof.

Let $x_{1}\in H_{1}$ and $x_{2}\in H_{2}$ such that $g=h_{1}h_{2}=x_{1}x_{2}\in G$. Then $h_{1}^{-1}x_{1}=h_{2}x_{2}^{-1}:=h\in H$. Thus ${\mathcal{P}}(h_{1}h_{2})={\mathcal{P}}(h_{1}hh^{-1}h_{2})={\mathcal{P}}(x_{1}x_{2})$, so ${\mathcal{P}}$ is well defined.

Now let $g=h_{1}h_{2}$ and $g^{\prime}=h_{1}^{\prime}h_{2}^{\prime}\in G$. Then

[TABLE]

We define the following two maps for this fixed $h_{2}$.

[TABLE]

The maps $\nu_{1}$ and $\nu_{2}$ are representations of $H_{1}$ associated to $\theta_{1}$ and $\theta_{1}^{h_{2}}$ respectively. Since $\theta$ is $G$-stable, $\theta^{h_{2}}=\theta$ so $\theta_{1}$ and $\theta_{1}^{h_{2}}$ extend the same character $\theta$ of $H$. Hence there exists a linear character ${\lambda}_{h_{2}}\in{\operatorname{IBr}}(H_{1}/H)$ such that $\theta_{1}^{h_{2}}{\lambda}_{h_{2}}=\theta_{1}$, by Lemma 2.6.

Recall our convention that we also denote the representation affording ${\lambda}_{h_{2}}$ by ${\lambda}_{h_{2}}$. The representations $\nu_{2}{\lambda}_{h_{2}}$ and $\nu_{1}$ both afford $\theta_{1}$, so they are similar representations of $H_{1}$. Since $\nu_{2}{\lambda}_{h_{2}}$ and $\nu_{1}$ coincide on $H$, it follows that $\nu_{2}{\lambda}_{h_{2}}=\nu_{1}$. We therefore have the following:

[TABLE]

Thus ${\mathcal{P}}$ is a projective representation with factor set $\alpha$ satisfying $\alpha(g,g^{\prime})={\lambda}_{h_{2}}(h_{1}^{\prime})\alpha_{2}(h_{2},h_{2}^{\prime})^{-1}$ for all $g=h_{1}h_{2}$, $g^{\prime}=h_{1}^{\prime}h_{2}^{\prime}\in G$ where ${\lambda}_{h_{2}}\in{\operatorname{IBr}}(H_{1}/H)$ is the linear character determined by $\theta_{1}^{h_{2}}{\lambda}_{h_{2}}=\theta_{1}$ and $\alpha_{2}$ is the factor set of ${\mathcal{P}}_{2}$. Since ${\mathcal{P}}|_{H}$ is an ordinary representation of $G$ affording $\theta$, ${\mathcal{P}}$ is a projective representation of $G$ associated to $\theta$. ∎

2.5. Finite groups of Lie type

Let $p\neq\ell$ be a prime and let G be a simple simply connected algebraic group defined over $\overline{\mathbb{F}}_{p}$. Let $\Phi$ be a root system of G relative to a maximal torus T with base $\Delta$. For $\alpha\in\Phi$ let $x_{\alpha}$ be the associated one-parameter subgroup. Let $F_{0}:{\textbf{G}}\to{\textbf{G}}$ be the field automorphism defined by $x_{\alpha}(t)\mapsto x_{\alpha}(t^{p})$ for all $\alpha\in\Phi$ and $t\in{\mathbb{F}}$. For a symmetry $\tau$ of the Dynkin diagram associated to $\Delta$ we let $\tau:{\textbf{G}}\to{\textbf{G}}$ be the graph automorphism given by $x_{\alpha}(t)\mapsto x_{\tau(\alpha)}(t)$. Let $F:{\textbf{G}}\to{\textbf{G}}$ be a Frobenius endomorphism of the form $F=F_{0}^{f}\tau$, where $\tau$ is a graph automorphism as above and $q=p^{f}$, endowing G with an ${\mathbb{F}}_{q}$-structure. Let ${\textbf{G}}^{F}$ be the finite group of the fixed points of G under $F$. We say that ${\textbf{G}}^{F}$ is untwisted if $F=F_{0}^{f}$ and twisted otherwise.

Let ${\textbf{G}}^{*}$ be a connected reductive algebraic group with maximal torus ${\textbf{T}}^{*}$ such that $({\textbf{G}}^{*},{\textbf{T}}^{*},F^{*})$ is dual to $({\textbf{G}},{\textbf{T}},F)$ in the sense of [6, Definition 13.10] and the subsequent discussion. Since it will be clear which Frobenius we are referring to, we drop the * notation and just write $F$ for both Frobenius maps.

Let $\iota:{\textbf{G}}\hookrightarrow{\tilde{\textbf{G}}}$ be a regular embedding of algebraic groups, with ${\tilde{\textbf{G}}}$ a connected reductive algebraic group such that ${\textnormal{Z}}({\tilde{\textbf{G}}})$ is connected and $\left[{\tilde{\textbf{G}}},{\tilde{\textbf{G}}}\right]\subseteq{\textbf{G}}$ (see [1, Section 15.1]). The Frobenius endomorphism $F$ extends to $\tilde{\textbf{G}}$ and we also denote by $F:{\tilde{\textbf{G}}}\to{\tilde{\textbf{G}}}$ a fixed extension of $F$ to ${\tilde{\textbf{G}}}$. Then ${\tilde{\textbf{T}}}={\textnormal{Z}}({\tilde{\textbf{G}}}).{\textbf{T}}$ is an $F$-stable maximal torus of ${\tilde{\textbf{G}}}$ and we let ${\tilde{\textbf{G}}}^{*}$ and ${\tilde{\textbf{T}}}^{*}$ be such that $({\tilde{\textbf{G}}}^{*},{\tilde{\textbf{T}}}^{*},F)$ is dual to $({\tilde{\textbf{G}}},{\tilde{\textbf{T}}},F)$. Let $\iota^{\ast}:{\tilde{\textbf{G}}}^{*}\to{\textbf{G}}^{*}$ be a surjective morphism dual to $\iota$.

To each ${\textbf{G}}^{*F}$-conjugacy class $(s)$ of a semisimple element $s\in{\textbf{G}}^{*F}$ we have an associated rational Lusztig series $\mathcal{E}({\textbf{G}}^{F},(s))\subseteq{\operatorname{Irr}}({\textbf{G}}^{F})$, see remarks before [6, Proposition 14.41]. In addition we let $\mathcal{E}({\textbf{G}}^{F},\ell^{\prime})$ be the union of rational Lusztig series $\mathcal{E}({\textbf{G}}^{F},(s))$ where $s$ runs over all semisimple elements of ${\textbf{G}}^{*F}$ with order coprime to $\ell$.

3. Finite groups of Lie type of type A

For this section we let ${\textbf{G}}={\textnormal{SL}}_{n}(\overline{{\mathbb{F}}}_{p})$, ${\tilde{\textbf{G}}}={\textnormal{GL}}_{n}(\overline{{\mathbb{F}}}_{p})$ with $n>1$. Let $m\in{\mathbb{N}}$ be such that $(m,|{\textbf{G}}^{F}|)=1$ and therefore $(m,|{\tilde{\textbf{G}}}^{F}|)=1$. Let $D=\langle F_{0},\tau\rangle$ be the group of automorphisms of ${\tilde{\textbf{G}}^{F}}$ generated by field and graph automorphisms.

Remark 3.1**.**

Let $\Theta:{\operatorname{IBr}}({\textbf{G}}^{F})\to{\operatorname{Irr}}({\textbf{G}}^{F})$ denote the injective map given in [4, Definition 2.2]. It follows from [4, Theorem A and Lemma 2.3] that $\Theta({\operatorname{IBr}}({\textbf{G}}^{F}))$ is an Aut$({\textbf{G}}^{F})$-stable unitriangular basic set for ${\textbf{G}}^{F}$ and $\Theta$ is ${\operatorname{Aut}}({\textbf{G}}^{F})$-equivariant. Hence for all $H\leq$ Aut$({\textbf{G}}^{F})$ and all $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$, $H_{\theta}=H_{\Theta(\theta)}$.

Lemma 3.2**.**

For any $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$ there exists a $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F}\mid\chi)$ such that

1. $(i)$

$({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta}={\tilde{\textbf{G}}^{F}}_{\theta}\rtimes D_{\theta}$, and 2. $(ii)$

$\theta$* extends to $({\textbf{G}}^{F}\rtimes D)_{\theta}$.*

Proof.

Let $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$, $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F}\mid\chi)$ and $\psi=\Theta(\theta)$. By [2, Theorem 4.1], there exists a $g\in{\tilde{\textbf{G}}^{F}}$ such that $\psi^{g}$ satisfies (i) and (ii). We claim that $\theta^{g}\in{\operatorname{IBr}}({\textbf{G}}^{F}\mid\chi)$ satisfies (i) and (ii). Since $\Theta(\theta^{g})=\Theta(\theta)^{g}=\psi^{g}$, it follows from Remark 3.1 that $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta^{g}}=({\tilde{\textbf{G}}^{F}}\rtimes D)_{\psi^{g}}={\tilde{\textbf{G}}^{F}}_{\psi^{g}}\rtimes D_{\psi^{g}}={\tilde{\textbf{G}}^{F}}_{\theta^{g}}\rtimes D_{\theta^{g}}$ so $\theta^{g}$ satisfies (i). Since $\psi^{g}$ satisfies part (ii), there exists an extension $\tilde{\psi}\in{\operatorname{Irr}}({\textbf{G}}^{F}\rtimes D)_{\psi^{g}}$ of $\psi^{g}$. By Remark 3.1 and (i) we have that $({\textbf{G}}^{F}\rtimes D)_{\psi^{g}}={\textbf{G}}^{F}\rtimes D_{\theta^{g}}$. Then since $d^{1}(\tilde{\psi})_{{\textbf{G}}^{F}}=d^{1}(\tilde{\psi}_{{\textbf{G}}^{F}})=d^{1}(\psi^{g})$, $\theta^{g}$ is a constituent of $d^{1}(\tilde{\psi})_{{\textbf{G}}^{F}}$ of multiplicity 1. It follows that there exists some irreducible constituent $\eta\in{\operatorname{IBr}}({\textbf{G}}^{F}\rtimes D_{\theta^{g}})$ of $d^{1}(\tilde{\psi})$ such that $\theta^{g}$ is an irreducible constituent of $\eta_{{\textbf{G}}^{F}}$ of multiplicity 1. Since $\theta^{g}$ is stable in ${\textbf{G}}^{F}\rtimes D_{\theta^{g}}$, it follows that $\eta$ is an extension of $\theta^{g}$ to ${\textbf{G}}^{F}\rtimes D_{\theta^{g}}$, so $\theta^{g}$ satisfies (ii). ∎

Lemma 3.3**.**

There exists a $D$-equivariant bijection

[TABLE]

such that $\tilde{f}_{m}({\lambda}\chi)={\lambda}^{m}\tilde{f}_{m}(\chi)$ for all $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$ and all ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid 1_{{\textbf{G}}^{F}})$, and $\tilde{f}_{m}({\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu))={\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu^{m})$ for all $\nu\in{\operatorname{IBr}}(Z({\tilde{\textbf{G}}^{F}}))$.

Proof.

Recall that there exists a bijection ${\textnormal{Z}}(\tilde{\textbf{G}}^{*})^{F}\rightarrow{\operatorname{Irr}}(\tilde{\textbf{G}}^{F}\mid 1_{{\textbf{G}}^{F}})$, $z\mapsto\hat{z}$ , see for example [6, Proposition 13.30]. The set $\mathcal{E}({\tilde{\textbf{G}}^{F}},\ell^{\prime})$ is a basic set of ${\operatorname{Irr}}({\tilde{\textbf{G}}^{F}})$ and we can define an injective map $\tilde{\Theta}:{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})\to{\operatorname{Irr}}({\tilde{\textbf{G}}^{F}})$ with image $\mathcal{E}({\tilde{\textbf{G}}^{F}},\ell^{\prime})$ such that the decomposition matrix associated to this map is unitriangular [7, 2.4]. For every $z\in Z({\tilde{\textbf{G}}}^{*})_{\ell^{\prime}}^{F}$ we have that $\tilde{\Theta}(\psi d^{1}(\hat{z}))=\tilde{\Theta}(\psi)\hat{z}$ for all $\psi\in{\operatorname{IBr}}({\tilde{\textbf{G}}}^{F})$. Let $\sigma\in$ Gal$({\mathbb{Q}}_{|{\tilde{\textbf{G}}}^{F}|}/{\mathbb{Q}})$ be a Galois automorphism such that $\sigma(\zeta)=\zeta^{m}$, where $\zeta$ is a primitive $|{\tilde{\textbf{G}}}^{F}|$-th root of unity. Define a map

[TABLE]

Since $\mathcal{E}({\tilde{\textbf{G}}^{F}},\ell^{\prime})^{\sigma}=\mathcal{E}({\tilde{\textbf{G}}^{F}},\ell^{\prime})$ by [15, Lemma 9.1], this map is well defined.

Let ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid 1_{{\textbf{G}}^{F}})$ and let $z\in\mathrm{Z}({\tilde{\textbf{G}}}^{\ast})_{\ell^{\prime}}^{F}$ be the central element such that $d^{1}(\hat{z})=\lambda$. Then for all $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$,

[TABLE]

Now let $\nu\in{\operatorname{Irr}}(\mathrm{Z}({\tilde{\textbf{G}}^{F}}))$. Then since $\tilde{\Theta}({\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})\mid d^{1}(\nu))\subseteq{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu)$ and ${\operatorname{Irr}}({\tilde{\textbf{G}}^{F}}\mid\nu)=$ ${\operatorname{Irr}}({\tilde{\textbf{G}}^{F}}\mid\nu^{m})$, it follows that $\tilde{f}_{m}({\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu))=$ ${\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu^{m})$. ∎

Lemma 3.4**.**

Let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ and let $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\theta)$. Suppose that $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta}={\tilde{\textbf{G}}^{F}}_{\theta}\rtimes D_{\theta}$. Then

[TABLE]

Proof.

If $d\in D_{\theta}$ then $\chi$ and $\chi^{d}$ are both elements of ${\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\theta)$ so $\chi^{d}=\chi{\lambda}$ for some linear ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F})$ by Lemma 2.6. Hence $D_{\theta}\subseteq\{d\in D\mid\chi^{d}=\chi{\lambda}{\textnormal{ for some }}{\lambda}\in{\textnormal{ IBr}}({\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F})\}$. Now suppose that $d\in D$ is such that $\chi^{d}=\chi{\lambda}$ for some ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F})$. Then $\theta^{d}$ and $\theta$ are both elements of ${\operatorname{IBr}}({\textbf{G}}^{F}\mid\chi^{d})$, so $\theta^{d}=\theta^{g}$ for some $g\in{\tilde{\textbf{G}}^{F}}$. Therefore $g^{-1}d\in({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta}$. Since $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta}={\tilde{\textbf{G}}^{F}}_{\theta}\rtimes D_{\theta}$, it follows that $g\in{\tilde{\textbf{G}}^{F}}_{\theta}$ and $d\in D_{\theta}$. ∎

Let $\mathcal{G}$ be a set of representatives of the $({\tilde{\textbf{G}}^{F}}\rtimes D)$-orbits of characters in ${\operatorname{IBr}}({\textbf{G}}^{F})$ satisfying Lemma 3.2 (i) and (ii). For $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$, let $\theta_{0}$ denote the unique character in ${\mathcal{G}}$ such that $\theta=\theta_{0}^{gd}$ for some $gd\in{\tilde{\textbf{G}}^{F}}\rtimes D$, and let $\tilde{\theta}_{0}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\theta_{0})$. There exists a unique character $\chi_{0}^{\prime}\in\mathcal{G}$ and element $d^{\prime}\in D$ such that ${\chi_{0}^{\prime}}^{d^{\prime}}\in{\operatorname{IBr}}({\textbf{G}}^{F}\mid\tilde{f}_{m}(\tilde{\theta}_{0}))$. Since $({\tilde{\textbf{G}}^{F}}\rtimes D)_{{\chi_{0}^{\prime}}^{d^{\prime}}}=d^{\prime}({\tilde{\textbf{G}}^{F}}_{\chi_{0}^{\prime}})\rtimes d^{\prime}(D_{\chi_{0}^{\prime}})$, the character ${\chi_{0}^{\prime}}^{d^{\prime}}$ also satisfies the conditions of Lemma 3.2 (i) and (ii). We let $\chi_{0}:={\chi_{0}^{\prime}}^{d^{\prime}}$ and define a map $f_{m}:{\operatorname{IBr}}({\textbf{G}}^{F})\to{\operatorname{IBr}}({\textbf{G}}^{F})$ by

[TABLE]

We claim that this is a well defined $({\tilde{\textbf{G}}}^{F}\rtimes D)$-equivariant bijection. First suppose that $\tilde{\theta}_{0}$ and $\tilde{\theta}_{0}^{\prime}$ are two elements of ${\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\theta_{0})$. Then since ${\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F}$ is abelian, $\tilde{\theta}_{0}={\lambda}\tilde{\theta}_{0}^{\prime}$ for some linear character ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F})$ by Lemma 2.6. Further, $\tilde{f}_{m}(\tilde{\theta}_{0})={\lambda}^{m}\tilde{f}_{m}(\tilde{\theta}_{0}^{\prime})$ so ${\operatorname{IBr}}({\textbf{G}}^{F}\mid\tilde{f}_{m}(\tilde{\theta}_{0}))={\operatorname{IBr}}({\textbf{G}}^{F}\mid\tilde{f}_{m}(\tilde{\theta}_{0}^{\prime}))$. Hence $\chi_{0}$, and therefore $f_{m}(\theta)$, is independent of the choice of $\tilde{\theta}_{0}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\theta_{0})$. Thus the map is well defined, and $({\tilde{\textbf{G}}}^{F}\rtimes D)$-equivariant by construction.

Lemma 3.5**.**

In the notation as above we have $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta_{0}}=({\tilde{\textbf{G}}^{F}}\rtimes D)_{\chi_{0}}$.

Proof.

Since $\theta_{0}$ and $\chi_{0}$ satisfy Lemma 3.2 (i) and (ii), $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta_{0}}={\tilde{\textbf{G}}^{F}}_{\theta_{0}}\rtimes D_{\theta_{0}}$ and $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\chi_{0}}={\tilde{\textbf{G}}^{F}}_{\chi_{0}}\rtimes D_{\chi_{0}}.$ By Lemma 3.4,

[TABLE]

Therefore, since $\tilde{f}_{m}$ is a $D$-equivariant bijection and $\tilde{f}_{m}(\tilde{\theta}_{0})\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\chi_{0})$,

[TABLE]

Recall that ${\tilde{\textbf{G}}^{F}}_{\theta_{0}}=\bigcap\ker({\lambda})$ where ${\lambda}$ runs over the elements of Stab${}_{{\operatorname{IBr}}({\tilde{\textbf{G}}}^{F}/{\textbf{G}}^{F})}(\tilde{\theta}_{0})$. Since $\tilde{f}_{m}$ is a bijection and $\tilde{f}_{m}({\lambda}\tilde{\theta}_{0})={\lambda}^{m}\tilde{f}_{m}(\tilde{\theta}_{0})$, ${\lambda}$ stabilises $\tilde{\theta}_{0}$ if and only if $\tilde{f}_{m}({\lambda})={\lambda}^{m}$ stabilises $\tilde{f}_{m}(\tilde{\theta}_{0})$. Thus, since $(m,|{\tilde{\textbf{G}}^{F}}|)=1$, ${\tilde{\textbf{G}}^{F}}_{\theta_{0}}=\bigcap\ker({\lambda}^{m})$ where ${\lambda}^{m}$ now runs over the elements of Stab${}_{{\operatorname{IBr}}({\tilde{\textbf{G}}}^{F}/{\textbf{G}}^{F})}(\tilde{f}_{m}(\tilde{\theta}_{0}))$. In particular, ${\tilde{\textbf{G}}^{F}}_{\theta_{0}}={\tilde{\textbf{G}}^{F}}_{f_{m}(\theta_{0})}$ and the result follows. ∎

Since $\tilde{f}_{m}$ is a bijection such that $\tilde{f}_{m}({\lambda}\chi)={\lambda}^{m}\tilde{f}_{m}(\chi)$ for all $\chi\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$ and all ${\lambda}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid 1_{{\textbf{G}}^{F}})$, by construction the map $f_{m}|_{{\mathcal{G}}}:{\mathcal{G}}\to{\operatorname{IBr}}({\textbf{G}}^{F})$ is an injection. Hence by Lemma 3.5, $f_{m}:{\operatorname{IBr}}({\textbf{G}}^{F})\to{\operatorname{IBr}}({\textbf{G}}^{F})$ is a $({\tilde{\textbf{G}}^{F}}\rtimes D)$-equivariant bijection, proving the claim. It follows that $({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta}=({\tilde{\textbf{G}}^{F}}\rtimes D)_{f_{m}(\theta)}$ for all $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$.

Proposition 3.6**.**

The map $f_{m}:{\textnormal{ IBr}}({\textbf{G}}^{F})\to{\textnormal{ IBr}}({\textbf{G}}^{F})$ defines a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\tilde{\textbf{G}}^{F}}\rtimes D$.

Proof.

Since $f_{m}:{\textnormal{ IBr}}({\textbf{G}}^{F})\to{\textnormal{ IBr}}({\textbf{G}}^{F})$ is a $({\tilde{\textbf{G}}^{F}}\rtimes D)$-equivariant bijection, by Definition 2.2 and Lemma 2.3 it remains to show that

[TABLE]

for every $\theta\in\mathcal{G}$.

Let $\theta\in\mathcal{G}$. Since ${\tilde{\textbf{G}}}^{F}/{\textbf{G}}^{F}$ is cyclic there exists an extension $\theta_{1}$ of $\theta$ to ${\tilde{\textbf{G}}}^{F}_{\theta}$. Since $\theta$ satisfies Lemma 3.2 (i) and (ii), $({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}={\tilde{\textbf{G}}}^{F}_{\theta}\rtimes D_{\theta}$ and there exists an extension $\theta_{2}$ of $\theta$ to ${\textbf{G}}^{F}\rtimes D_{\theta}$. Let $H_{1}={\tilde{\textbf{G}}}^{F}_{\theta}$ and let $H_{2}={\textbf{G}}^{F}\rtimes D_{\theta}$. Let ${\mathcal{D}}_{1}$ be an ordinary representation of ${\tilde{\textbf{G}}}_{\theta}^{F}$ affording $\theta_{1}$. Let ${\mathcal{D}}_{2}$ be an ordinary representation of ${\textbf{G}}^{F}\rtimes D_{\theta}$ affording $\theta_{2}$ such that ${\mathcal{D}}_{1}$ and ${\mathcal{D}}_{2}$ agree on ${\textbf{G}}^{F}$. Then by Lemma 2.7 there exists a projective representation ${\mathcal{P}}:({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}\rightarrow GL_{\theta(1)}({\mathbb{F}})$ associated to $\theta$ with factor set $\alpha$ such that

[TABLE]

[TABLE]

for all $x=h_{1}h_{2},x^{\prime}=h_{1}^{\prime}h_{2}^{\prime}\in({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}$, $h_{1},h_{1}^{\prime}\in{\tilde{\textbf{G}}}^{F}_{\theta}$, $h_{2},h_{2}^{\prime}\in{\textbf{G}}^{F}\rtimes D_{\theta}$, where $\lambda_{h_{2}}$ is the linear Brauer character of ${\tilde{\textbf{G}}}^{F}_{\theta}/{\textbf{G}}^{F}$ such that $\theta_{1}^{h_{2}}{\lambda}_{h_{2}}=\theta_{1}$.

Let $\tilde{\theta}_{1}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}})$ correspond to $\theta_{1}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}_{\theta}\mid\theta)$ via Clifford correspondence with respect to $\theta$ [12, Theorem 8.9]. Let $\theta_{1}^{\prime}\in{\operatorname{IBr}}({\tilde{\textbf{G}}}^{F}_{f_{m}(\theta)}\mid f_{m}(\theta))$ denote the Clifford correspondent of $\tilde{f}_{m}(\tilde{\theta}_{1})$ with respect to $f_{m}(\theta)$. Since $f_{m}(\theta)$ satisfies Lemma 3.2 (i) and (ii), there exists an extension $\theta_{2}^{\prime}$ of $f_{m}(\theta)$ to ${\textbf{G}}^{F}\rtimes D_{\theta}$. Let ${\mathcal{D}}^{\prime}_{1}$ be an ordinary representation of ${\tilde{\textbf{G}}}_{\theta}^{F}$ associated to $\theta_{1}^{\prime}$. Let ${\mathcal{D}}^{\prime}_{2}$ be an ordinary representation of ${\textbf{G}}^{F}\rtimes D_{\theta}$ affording $\theta_{2}^{\prime}$ such that ${\mathcal{D}}^{\prime}_{1}$ and ${\mathcal{D}}^{\prime}_{2}$ agree on ${\textbf{G}}^{F}$. Then by Lemma 2.7 again, there exists a projective representation ${\mathcal{P}}^{\prime}:({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}\rightarrow GL_{\theta(1)}({\mathbb{F}})$ associated to $f_{m}(\theta)$ with factor set $\alpha^{\prime}$ such that

[TABLE]

[TABLE]

for all $x=h_{1}h_{2},x^{\prime}=h_{1}^{\prime}h_{2}^{\prime}\in({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}$, $h_{1},h_{1}^{\prime}\in{\tilde{\textbf{G}}}^{F}_{\theta}$, $h_{2},h_{2}^{\prime}\in{\textbf{G}}^{F}\rtimes D_{\theta}$, where $\lambda^{\prime}_{h_{2}}$ is the linear Brauer character of ${\operatorname{IBr}}({\tilde{\textbf{G}}}^{F}_{\theta}/{\textbf{G}}^{F})$ such that $\theta_{1}^{\prime h_{2}}{\lambda}^{\prime}_{h_{2}}=\theta_{1}^{\prime}$.

Let $\tilde{\lambda}_{h_{2}}^{\prime}\in{\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}/{\textbf{G}}^{F})$ be an extension of $\lambda_{h_{2}}^{\prime}$ to ${\tilde{\textbf{G}}^{F}}$. It follows from Lemma 3.3 that $\tilde{f}_{m}(\tilde{\theta}_{1})^{h_{2}}\tilde{\lambda}^{\prime}_{h_{2}}=\tilde{f}_{m}(\tilde{\theta}_{1})$. Since $\tilde{f}_{m}$ is $D$-equivariant, we have the following for any $h_{2}\in{\textbf{G}}^{F}\rtimes D_{\theta}$,

[TABLE]

Hence $\lambda_{h_{2}}^{\prime}={\lambda}_{h_{2}}^{m}$ for each $h_{2}\in{\textbf{G}}^{F}\rtimes D_{\theta}$, so $\alpha(x,x^{\prime})^{m}=\alpha^{\prime}(x,x^{\prime})$ for all $x,x^{\prime}\in({\tilde{\textbf{G}}}^{F}\rtimes D)_{\theta}$.

From the description of the action of automorphisms of ${\textbf{G}}^{F}$ given in [8, Theorem 2.5.1], we observe that $C_{{\tilde{\textbf{G}}^{F}}\rtimes D}({\textbf{G}}^{F})=Z({\tilde{\textbf{G}}}^{F})$. Therefore ${\mathcal{P}}|_{Z({\tilde{\textbf{G}}^{F}})}={\mathcal{D}}_{1}|_{Z({\tilde{\textbf{G}}^{F}})}=\nu Id$ where $\nu\in{\operatorname{IBr}}(Z({\tilde{\textbf{G}}^{F}})\mid\tilde{\theta}_{1})$. Similarly, ${\mathcal{P}}^{\prime}|_{Z({\tilde{\textbf{G}}^{F}})}={\mathcal{D}}_{1}^{\prime}|_{Z({\tilde{\textbf{G}}^{F}})}=\eta Id$ where $\eta\in{\operatorname{IBr}}(Z({\tilde{\textbf{G}}^{F}})\mid\tilde{f}_{m}(\tilde{\theta}_{1}))$. Since $\tilde{f}_{m}({\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu))={\operatorname{IBr}}({\tilde{\textbf{G}}^{F}}\mid\nu^{m})$ for all $\nu\in{\operatorname{IBr}}(Z({\tilde{\textbf{G}}^{F}}))$ by Lemma 3.3, it follows that $\eta=\nu^{m}$. Therefore $(({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta},{\textbf{G}}^{F},\theta)^{(m)}\approx(({\tilde{\textbf{G}}^{F}}\rtimes D)_{\theta},{\textbf{G}}^{F},f_{m}(\theta))$. ∎

Corollary 3.7**.**

There exists a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\textbf{G}}^{F}\rtimes\mathrm{Aut}({\textbf{G}}^{F})$.

Proof.

Let $H:={\tilde{\textbf{G}}}^{F}\rtimes D$. It follows from [8, Theorem 2.5.1] that there exists a surjective map $H\twoheadrightarrow{\operatorname{Aut}}({\textbf{G}}^{F})$ such that $H/C_{H}({\textbf{G}}^{F})\cong\mathrm{Out}({\textbf{G}}^{F})$. Therefore we can apply [16, Corollary 4.12] and conclude that there exists a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})$. ∎

4. Finite groups of Lie type of type B, C, D, E6 and E7

We continue with the notation of Section 2.5 with G a simple algebraic group of simply connected type and $F=F_{0}^{f}\tau$ a Frobenius endomorphism defining an $\mathbb{F}_{q}$-structure on G with $q=p^{f}$. For this section we assume that G is of type B, C, D, E6, or E7. In addition we assume that $\tau^{2}=1$, that is, we exclude the case where ${\textbf{G}}^{F}$ is of type 3D4.

Let $\mathcal{L}:{\textbf{G}}\to{\textbf{G}}$ denote the Lang map defined by $\mathcal{L}(g)=g^{-1}F(g)$ for all $g\in{\textbf{G}}$. Let Diag$({{\textbf{G}}^{F}})$ denote the group of diagonal automorphisms of ${\textbf{G}}^{F}$ in ${\operatorname{Out}}({\textbf{G}}^{F})$. By [17, 12.5], the proof of [9, Proposition 1.5] together with [5, Lemma 1.3],

[TABLE]

Hence each diagonal automorphism of ${\textbf{G}}^{F}$ can be realised as conjugation by some element of $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$. Since the Lang map is surjective on G and the restriction $\mathcal{L}|_{\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))}$ is a group homomorphism with kernel ${\textbf{G}}^{F}$, we have $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))/{\textbf{G}}^{F}\cong{\textnormal{Z}}({\textbf{G}})$. Let $\Gamma$ be the group of graph automorphisms which commute with $F$, regarded as automorphisms of $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$.

Define $A:=\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))\rtimes\langle F_{0},\Gamma\rangle$. Then

[TABLE]

and for convenience we identify $A/{\textbf{G}}^{F}$ with ${\textnormal{Z}}({\textbf{G}})\rtimes\langle F_{0},\Gamma\rangle$. By construction of $A$ there is a surjection

[TABLE]

Let $z\in{\textnormal{Z}}({\textbf{G}})$ and let $x\in\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$ be such that $\mathcal{L}(x)=z$. Then we let $\delta_{z}\in{\operatorname{Out}}({\textbf{G}}^{F})$ denote the diagonal automorphism of ${\textbf{G}}^{F}$ given by conjugation by $x$. In order to express $A/{\textbf{G}}^{F}$ in a more convenient form, we use the following lemma to define one more piece of notation.

Lemma 4.1**.**

If $F_{0}$ does not act trivially on ${\textnormal{Z}}({\textbf{G}})$ then there exists a graph automorphism $\gamma\in\Gamma$ such that $\gamma F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$.

Proof.

The result is immediate if ${\textnormal{Z}}({\textbf{G}})$ is trivial or of order 2 so there are just three cases to consider. We use the notation of [8, Section 1.12].

First suppose that G is of type E6 and $p\neq 3$. Let $1\neq\omega\in\overline{{\mathbb{F}}}_{p}^{\times}$ with $\omega^{3}=1$. By [8, Table 1.12.6], $h=h_{\alpha_{1}}(\omega)h_{\alpha_{2}}(\omega^{2})h_{\alpha_{5}}(\omega)h_{\alpha_{6}}(\omega^{2})$ is a generator of ${\textnormal{Z}}({\textbf{G}})$. If $p\equiv 1\,\mathrm{mod}(3)$ then $\omega^{p}=\omega$ so $F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$. If $p\equiv 2\,\mathrm{mod}(3)$ then $\omega^{p}=\omega^{2}$, so $F_{0}(h)=h^{2}$. Further, the non-trivial graph automorphism $\gamma$ acts on ${\textnormal{Z}}({\textbf{G}})$ by $\gamma(h)=h_{\alpha_{6}}(\omega)h_{\alpha_{5}}(\omega^{2})h_{\alpha_{2}}(\omega)h_{\alpha_{1}}(\omega^{2})=h^{2}$. Therefore $\gamma$ and $F_{0}$ have the same action of order 2 on ${\textnormal{Z}}({\textbf{G}})$, and hence $\gamma F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$.

Now suppose that G is of type D2m for some $m\in{\mathbb{N}}$, and $p\neq 2$. Then ${\textnormal{Z}}({\textbf{G}})$ has two generators, $h_{1}=h_{\alpha_{1}}(-1)h_{\alpha_{3}}(-1)\dots h_{\alpha_{2m-1}}(-1)$ and $h_{2}=h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(-1)$. Since $p\neq 2$, the field automorphism $F_{0}$ stabilises $-1$, hence $F_{0}$ acts trivially on the centre.

Finally, suppose that G is of type D2m+1 for some $m\in{\mathbb{N}}$, and $p\neq 2$. Let $1\neq\omega\in\overline{\mathbb{F}}_{p}^{\times}$ such that $\omega^{2}=-1$. Then ${\textnormal{Z}}({\textbf{G}})$ has one generator $h=h_{\alpha_{1}}(-1)h_{\alpha_{3}}(-1)\dots h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m}}(\omega)h_{\alpha_{2m+1}}(-\omega)$. Again, since $p$ is odd, $F_{0}$ stabilises $-1$. If $p\equiv 1$ mod $(4)$ then $F_{0}(\omega)=\omega$ so $F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$. If $p\equiv 3$ mod $(4)$ then $F_{0}(\omega)=\omega^{3}=-\omega$. The non-trivial graph automorphism $\gamma$ swaps $\alpha_{2m}$ and $\alpha_{2m+1}$ so $\gamma(h)=h_{\alpha_{1}}(-1)h_{\alpha_{3}}(-1)\dots h_{\alpha_{2m-1}}(-1)h_{\alpha_{2m+1}}(\omega)h_{\alpha_{2m}}(-\omega)$. In particular, $\gamma$ and $F_{0}$ have the same action on ${\textnormal{Z}}({\textbf{G}})$ and hence $\gamma F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$. ∎

Notation 4.2**.**

If $F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$ then we let $F_{0}^{\prime}:=F_{0}$. Otherwise we let $F_{0}^{\prime}:=\gamma F_{0}$ where $\gamma$ is the non-trivial graph automorphism such that $\gamma F_{0}$ acts trivially on ${\textnormal{Z}}({\textbf{G}})$ as given in Lemma 4.1.

Then we have

[TABLE]

where $\langle F_{0}^{\prime}\rangle\cong C_{a}$ for some positive integer $a$. As ${\textnormal{Z}}({\textbf{G}})$ is the $p^{\prime}$-part of $\Lambda(\Phi)$ given in [10, Table 9.2], and $\Gamma$ can be determined by the Dynkin diagram of G, we observe the following.

Corollary 4.3**.**

There exists a positive integer $a$ such that $A/{\textbf{G}}^{F}$ has the following form:

• •

If ${\textbf{G}}^{F}$ is of type D4 then $A/{\textbf{G}}^{F}$ is a subgroup of $(V_{4}\rtimes S_{3})\times C_{a}\cong S_{4}\times C_{a}$.

• •

If G is of type E6 then $A/{\textbf{G}}^{F}$ is a subgroup of $(C_{3}\rtimes C_{2})\times C_{a}\cong S_{3}\times C_{a}$.

• •

Otherwise $A/{\textbf{G}}^{F}$ is a subgroup of $(V_{4}\rtimes C_{2})\times C_{a}\cong D_{8}\times C_{a}$ or $(C_{4}\rtimes C_{2})\times C_{a}\cong D_{8}\times C_{a}$.

We conclude this section with results on the group theoretic structure of $A$. The following two lemmas are used in the proof of Proposition 4.10.

Lemma 4.4**.**

$\mathrm{C}_{A}({\textbf{G}}^{F})=\mathrm{Z}({\textbf{G}})\langle F\rangle$**

Proof.

It is clear that $\mathrm{Z}({\textbf{G}})\langle F\rangle$ acts trivially on ${\textbf{G}}^{F}$, so $\mathrm{Z}({\textbf{G}})\langle F\rangle\subseteq\mathrm{C}_{A}({\textbf{G}}^{F})$. Now suppose that $g\in\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$ and $d\in\langle F_{0},\Gamma\rangle$ are such that $gd\in\mathrm{C}_{A}({\textbf{G}}^{F})$. As $gd$ induces the identity on ${\textbf{G}}^{F}$ it follows from [8, Theorem 2.5.1] that $g$ and $d$ both induce the identity on ${\textbf{G}}^{F}$. Therefore $g\in{\textnormal{Z}}(\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))={\textnormal{Z}}({\textbf{G}})$ and $d=F^{i}$ for some $i$, so $gd\in{\textnormal{Z}}({\textbf{G}})\langle F\rangle$. ∎

Lemma 4.5**.**

If $F$ acts trivially on $\mathrm{Z}({\textbf{G}})$ then $F$ has order $\mathrm{exp}(\mathrm{Z}({\textbf{G}}))$ on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$. If $\mathrm{Z}({\textbf{G}})$ is cyclic and $F$ acts non-trivially on $\mathrm{Z}({\textbf{G}})$ then $F$ has order $2$ on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$, and if $\mathrm{Z}({\textbf{G}})$ is not cyclic and $F$ acts non-trivially on $\mathrm{Z}({\textbf{G}})$ then $F$ has order $4$ on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$.

Proof.

Let $z\in\mathrm{Z}({\textbf{G}})$ and let $x\in\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$ be such that $\mathcal{L}(x)=z$. Then $F(x)=xz$. If $F(z)=z$ then $F^{k}(x)=xz^{k}$ for all $k\in{\mathbb{N}}$, so $F^{\mathrm{o}(z)}(x)=x$ where $o(z)$ denotes the order of $z$. Therefore $F$ has order $\mathrm{exp}(\mathrm{Z}({\textbf{G}}))$ on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$.

Now suppose that $F$ is non-trivial on $\mathrm{Z}({\textbf{G}})$. If $\mathrm{Z}({\textbf{G}})=\langle z\rangle$ then $F(z)=z^{-1}$ and therefore $F^{2}(x)=xF(z)z=x$. Thus $F$ has order 2 on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$. If $\mathrm{Z}({\textbf{G}})$ is not cyclic then $\mathrm{Z}({\textbf{G}})\cong C_{2}\times C_{2}$. Then for all $z\in{\textnormal{Z}}({\textbf{G}})$, $F^{2}(z)=z$ so $F^{4}(x)=F^{3}(z)F^{2}(z)F(z)zx=F(z)^{2}z^{2}x=x$. Hence $F$ has order $4$ on $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))$. ∎

The following lemma is used in the proof of Proposition 4.12.

Lemma 4.6**.**

There exists an isomorphism $\tilde{{\textbf{G}}}^{F}/\mathrm{Z}(\tilde{{\textbf{G}}})\cong\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))/\mathrm{Z}({\textbf{G}}^{F})$ which maps ${\textbf{G}}^{F}\mathrm{Z}(\tilde{{\textbf{G}}}^{F})/\mathrm{Z}(\tilde{{\textbf{G}}}^{F})$ to ${\textbf{G}}^{F}{\textnormal{Z}}({\textbf{G}})/\mathrm{Z}({\textbf{G}}).$

Proof.

Let $\tilde{g}\in{\tilde{\textbf{G}}^{F}}$. Since ${\tilde{\textbf{G}}}={\textnormal{Z}}({\tilde{\textbf{G}}}){\textbf{G}}$ we can choose a $z_{\tilde{g}}\in\mathrm{Z}({\tilde{\textbf{G}}})$ such that $\tilde{g}z_{\tilde{g}}\in{\textbf{G}}$. Then $\mathcal{L}(\tilde{g}z_{\tilde{g}})=\mathcal{L}(z_{\tilde{g}})\in\mathrm{Z}({\textbf{G}})$ so we can define a set-theoretic map ${\tilde{\textbf{G}}^{F}}\to\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$ given by $\tilde{g}\mapsto\tilde{g}z_{\tilde{g}}$. The coset $\tilde{g}z_{\tilde{g}}{\textnormal{Z}}({\textbf{G}})$ is uniquely determined by $\tilde{g}$, so the map ${\tilde{\textbf{G}}^{F}}\rightarrow\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))/{\textnormal{Z}}({\textbf{G}})$ given by $\tilde{g}\mapsto\tilde{g}z_{\tilde{g}}{\textnormal{Z}}({\textbf{G}})$ is a group homomorphism. The kernel of this homomorphism is ${\textnormal{Z}}({\tilde{\textbf{G}}})$, therefore $\tilde{{\textbf{G}}}^{F}/\mathrm{Z}(\tilde{{\textbf{G}}})\cong\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))/\mathrm{Z}({\textbf{G}}^{F})$. The second claim follows directly from the construction of the isomorphism. ∎

4.1. Constructing projective representations

In this section we construct projective representations of $A_{\theta}$ associated to $\theta$ for $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$. In Section 4.2 we will then use these projective representations to define a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ first with respect to ${\textbf{G}}^{F}\lhd A$, and hence with respect to ${\textbf{G}}^{F}\lhd({\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F}))$, for every $m$ such that $(m,|{\textbf{G}}^{F}|)=1$.

The following technical lemma is central to the constructions which follow. Thanks to [14, Lemma 4.5 (b)], we do not need to construct projective representations for $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ where $\mathrm{Out}({\textbf{G}}^{F})_{\theta}$ is cyclic (see the proof of Proposition 4.13). Because of the graph automorphism of order 3, when ${\textbf{G}}^{F}$ is of type ${\textnormal{D}}_{4}$ some cases require special methods. It is therefore often convenient to make one or both of the following assumptions.

Assumption 4.7**.**

Let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$.

• (i)

$\mathrm{Out}({\textbf{G}}^{F})_{\theta}$* is non-cyclic.*

• (ii)

If ${\textbf{G}}^{F}$ is of type D4, then $A_{\theta}\leq\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\rtimes\langle F_{0},\gamma_{\theta}\rangle$ where $\gamma_{\theta}\in\Gamma$ is such that $\gamma_{\theta}^{2}=1$.

Notation 4.8**.**

In Lemma 4.9, Proposition 4.10 and Proposition 4.11, if ${\textbf{G}}^{F}$ is of type D4 then we work under Assumption 4.7 (ii). We let $\Gamma^{\prime}:=\langle\gamma_{\theta}\rangle$ if ${\textbf{G}}^{F}$ is of type D4, let $\Gamma^{\prime}:=\Gamma$ otherwise. Then $A_{\theta}\leq\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\rtimes\langle F_{0},\Gamma^{\prime}\rangle$ and $A_{\theta}/{\textbf{G}}^{F}$ is a subgroup of $({\textnormal{Z}}({\textbf{G}})\rtimes\Gamma^{\prime})\times\langle F_{0}^{\prime}\rangle$ with $\Gamma^{\prime}=\langle\gamma^{\prime}\rangle\cong C_{2}$.

Lemma 4.9**.**

Let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ be a character satisfying Assumption 4.7 (i) and (ii), and let $\gamma^{\prime}$ be as in Notation 4.8. Then there exist finite groups $U_{1}\lhd A_{\theta}$ and $U_{2}\leq A_{\theta}$ such that $A_{\theta}=U_{1}U_{2}$, $U_{1}\cap U_{2}={\textbf{G}}^{F}$, $\mathrm{Z}({\textbf{G}})\leq U_{1}$, and either $F\in U_{2}$ or $\gamma^{\prime}F\in U_{2}$ and $\gamma^{\prime}\in U_{1}$. Furthermore, $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $D_{8}$ if G is of type D, $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $S_{3}$ otherwise, and $U_{2}/{\textbf{G}}^{F}$ is cyclic.

Proof.

Let $G:=\mathrm{Z}({\textbf{G}})\rtimes\Gamma^{\prime}$ and let $H:=\langle F_{0}^{\prime}\rangle$. Let $U$ be a subgroup of $G\times H$ such that $A_{\theta}/{\textbf{G}}^{F}\cong U$. Recall that (as discussed in Section 2.3) $U$ has the form

[TABLE]

for some $5$-tuple $(G_{1},G_{2},H_{1},H_{2},\varphi)$ where $G_{2}\lhd G_{1}\leq G$, $H_{2}\lhd H_{1}\leq H$, and $\varphi:G_{1}/G_{2}\to H_{1}/H_{2}$ is an isomorphism. If G is of type D then by Assumption 4.7 (ii) and the definition of $\Gamma^{\prime}$, $G$ is a subgroup of $D_{8}$. If G is not of type D then $G$ is a subgroup of $S_{3}$. Thus there are only a small number of possibilities for $(G_{1},G_{2},H_{1},H_{2},\varphi)$, so we prove the claim by considering each of them individually. Note that $U\cap(G\times 1)=G_{2}\times 1$ and $U\cap(1\times H)=1\times H_{2}$. Since $F$ and ${\textnormal{Z}}({\textbf{G}})$ act trivially on ${\operatorname{IBr}}({\textbf{G}}^{F})$, both $F$ and ${\textnormal{Z}}({\textbf{G}})$ are contained in $A_{\theta}$ and hence $F$ and ${\textnormal{Z}}({\textbf{G}}){\textbf{G}}^{F}/{\textbf{G}}^{F}$ are contained in $A_{\theta}/{\textbf{G}}^{F}$. It follows from the definition of $G$ that ${\textnormal{Z}}({\textbf{G}}){\textbf{G}}^{F}/{\textbf{G}}^{F}\leq G\times 1\cap U=G_{2}\times 1$. Moreover, either $F\in H$ (in which case $F\in 1\times H_{2}$), or $\gamma^{\prime}F\in H$.

First suppose that $G_{2}=1$. It then follows from Lemma 2.4 that $U\cong H_{1}$, which is cyclic. Since $A_{\theta}/{\textbf{G}}^{F}$ surjects onto ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$, therefore ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$ is cyclic, contradicting Assumption 4.7 (i).

Now suppose that $G_{1}=G_{2}$. Then since $G_{1}/G_{2}\cong H_{1}/H_{2}$, it follows that $H_{1}=H_{2}$ and $\varphi(gG_{2})=hH_{2}$ for all $g\in G_{1}$, $h\in H_{1}$. Therefore $U=G_{1}\times H_{1}$. Let $U_{1},U_{2}\leq A_{\theta}$ be such that $U_{1}/{\textbf{G}}^{F}\cong G_{1}\times 1$ and $U_{2}/{\textbf{G}}^{F}\cong 1\times H_{1}$. Then $U_{2}/{\textbf{G}}^{F}$ is cyclic, and $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $D_{8}$ if G is of type D, and a subgroup of $S_{3}$ if G is not of type D. If $F\in H$ then $F\in 1\times H_{2}=1\times H_{1}$, as mentioned above, so $F\in U_{2}$. If $\gamma^{\prime}F\in H$ then since $\gamma^{\prime}\in G$, it follows that $\gamma^{\prime}\in G_{1}$ and $\gamma^{\prime}F\in H_{1}$ and hence $\gamma^{\prime}F\in U_{2}$ and $\gamma^{\prime}\in U_{1}$. As ${\textnormal{Z}}({\textbf{G}}){\textbf{G}}^{F}/{\textbf{G}}^{F}\leq G_{2}\times 1$, clearly ${\textnormal{Z}}({\textbf{G}})\leq U_{1}$.

By examination of the possibilities for $(G_{1},G_{2},H_{1},H_{2},\varphi)$, it only remains to consider the case when $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ is such that $G_{1}/G_{2}\cong H_{1}/H_{2}\cong C_{2}$.

First suppose that $G_{2}$ has a direct complement in $G_{1}$, so $G_{1}=G_{2}\rtimes\langle\nu\rangle$ where $\nu^{2}=1$. Let $i\in\mathbb{N}$ such that $H_{1}=\langle(F_{0}^{\prime})^{i}\rangle$ and $H_{2}=\langle(F_{0}^{\prime})^{2i}\rangle$. We claim that $U=G_{2}\rtimes\langle\nu(F^{\prime}_{0})^{i}\rangle$. Since $\langle\nu\rangle\cong G_{1}/G_{2}\cong H_{1}/H_{2}\cong\langle(F_{0}^{\prime})^{i}H_{2}\rangle$, the isomorphism $\varphi$ sends $\nu G_{2}$ to $(F_{0}^{\prime})^{i}H_{2}$. Any $(g,h)\in U$ can be written as $(g_{2},1)(\nu^{j},(F_{0}^{\prime})^{ik})$ for some $g_{2}\in G_{2}$ and some positive integers $j$ and $k$. Then $\varphi(\nu^{j}G_{2})=(F_{0}^{\prime})^{ik}H_{2}$ and $\varphi(\nu^{j}G_{2})=\varphi(\nu G_{2})^{j}=(F_{0}^{\prime})^{ij}H_{2}$ so $(F_{0}^{\prime})^{i(k-j)}\in H_{2}$. Hence $\varphi(\nu^{k-j})\in G_{2}\cap\langle\nu\rangle=\{1\}$ since $\langle\nu\rangle$ is a direct complement of $G_{2}$ in $G_{1}$. Therefore $\nu^{j}=\nu^{k}$ and $(g,h)=(g_{2},1)(\nu,(F_{0}^{\prime})^{i})^{j}\in G_{2}\rtimes\langle\nu(F_{0}^{\prime})^{i}\rangle$, proving the claim.

Now we let $U_{1},U_{2}\leq A_{\theta}$ such that $U_{1}/{\textbf{G}}^{F}=G_{2}\times 1$ and $U_{2}/{\textbf{G}}^{F}=1\times\langle\nu(F^{\prime}_{0})^{i}\rangle$. As in the previous case, it is immediate that ${\textnormal{Z}}({\textbf{G}})\leq U_{1}$, $U_{2}/{\textbf{G}}^{F}$ is cyclic, and $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $D_{8}$ if G is of type D, and a subgroup of $S_{3}$ otherwise. If $F\in H$ then $F\in 1\times H_{2}\leq 1\times\langle\nu(F_{0}^{\prime})^{i}\rangle$ since $\nu^{2}=1$, and hence $F\in U_{2}$. Otherwise $\gamma^{\prime}F\in H$ so $\gamma^{\prime}F=(F_{0}^{\prime})^{r}$ for some positive integer $r$. Then since $F\in U$, we have $\gamma^{\prime}(F_{0}^{\prime})^{r}\in G_{2}\rtimes\langle\nu(F_{0}^{\prime})^{i}\rangle$. If $\gamma^{\prime}\in G_{2}$ then $\gamma^{\prime}\neq\nu$ so $r$ is an even multiple of $i$, and hence $\gamma^{\prime}\in U_{1}$ and $\gamma^{\prime}F\in U_{2}$. If $\gamma^{\prime}=\nu$ then $F=\nu(F_{0}^{\prime})^{r}\in\langle\nu(F_{0}^{\prime})^{i}\rangle$, so $F\in U_{2}$. Finally, if $\gamma^{\prime}=g_{2}\nu$ for some nontrivial $g_{2}\in G_{2}$, then note that $G_{1}=G_{2}\rtimes\langle g_{2}\nu\rangle$ and by defining $U_{1}$ and $U_{2}$ now in relation to this new decomposition of $G_{1}$, again we see that $F\in U_{2}$.

The final case to consider is when $G_{1}/G_{2}\cong H_{1}/H_{2}\cong C_{2}$ and $G_{2}$ does not have a direct complement in $G_{1}$. This occurs only if G is of type Dn with $G\cong D_{8}$ and $G_{1}\cong C_{4}$, the unique cyclic subgroup of $D_{8}$ of order 4. If $n$ is even then let $z_{1},z_{2}\in{\textbf{G}}$ be such that ${\textnormal{Z}}({\textbf{G}})=\langle z_{1},z_{2}\rangle$. Then $G=(\langle z_{1}\rangle\times\langle z_{2}\rangle)\rtimes\langle\gamma^{\prime}\rangle$. In this case fix $z_{3}:=z_{1}z_{2}$. Then $G_{1}=\langle z_{1}\gamma^{\prime}\rangle$ or $\langle z_{2}\gamma^{\prime}\rangle$, and $G_{2}=\langle z_{3}\rangle$. If $n$ is odd then then let $z_{1}\in{\textbf{G}}$ be such that ${\textnormal{Z}}({\textbf{G}})=\langle z_{1}\rangle$. Then $G=\langle z_{1}\rangle\rtimes\langle\gamma^{\prime}\rangle$ and $G_{1}=\langle z_{1}\rangle$. In this case we fix $z_{3}:=z_{1}^{2}$, so again we get $G_{2}=\langle z_{3}\rangle$. In both cases we can show that $U=\langle z_{3}\rangle\times\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$, using arguments similar to the case above where $G_{2}$ has a direct complement in $G_{1}$. As usual, we let $U_{1},U_{2}\leq A_{\theta}$ such that $U_{1}/{\textbf{G}}^{F}\cong\langle z_{3}\rangle\times 1$ and $U_{2}/{\textbf{G}}^{F}\cong 1\times\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$. Clearly ${\textnormal{Z}}({\textbf{G}})\leq U_{1}$, $U_{2}/{\textbf{G}}^{F}$ is cyclic, and $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $D_{8}$. Since $F\in U$, either $F\in\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$ or $z_{3}F\in\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$. In the first case then $F\in U_{2}$ so we are done. Suppose that $z_{3}F\in\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$. The map $\delta:A/{\textbf{G}}^{F}\twoheadrightarrow{\operatorname{Out}}({\textbf{G}}^{F})$ introduced at the beginning of Section 4 induces a surjection $U\twoheadrightarrow{\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$, so ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}\leq\langle\delta_{z_{2}},\delta_{z_{1}}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$. Now since $z_{3}F\in\langle z_{1}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$, we have $\delta_{z_{3}}F\in\langle\delta_{z_{1}}\gamma^{\prime}(F_{0}^{\prime})^{i}\rangle$ and therefore ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$ is cyclic, contradicting Assumption 4.7 (i). ∎

Proposition 4.10**.**

Suppose that G is of type B, C, E6 or E7 and let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ satisfying Assumption 4.7 (i). There exists a projective representation ${\mathcal{P}}$ of $A_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{6}=1$ and for every $c\in C_{A}({\textbf{G}}^{F})$, ${\mathcal{P}}(c)=\xi Id$ for a sixth root of unity $\xi$.

Proof.

By Lemma 4.9, there exist finite groups $U_{1}$ and $U_{2}$ such that $\mathrm{Z}({\textbf{G}})\leq U_{1}\lhd A_{\theta}$, $U_{2}\leq A_{\theta}$, $A_{\theta}=U_{1}U_{2}$, $U_{1}\cap U_{2}={\textbf{G}}^{F}$, $U_{2}/{\textbf{G}}^{F}$ is cyclic and $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $S_{3}$, and either $F\in U_{2}$, or $F\gamma^{\prime}\in U_{2}$ and $\gamma^{\prime}\in U_{1}$.

Since $U_{1}/{\textbf{G}}^{F}$ has cyclic Sylow subgroups, $\theta$ extends to a character $\theta_{1}$ of $U_{1}$ by [12, Theorem 8.29], and since $U_{2}/{\textbf{G}}^{F}$ is cyclic $\theta$ also extends to a character $\theta_{2}$ of $U_{2}$. Let ${\mathcal{D}}_{1}:U_{1}\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ be a representation of $U_{1}$ affording $\theta_{1}$ and let ${\mathcal{D}}_{2}:U_{2}\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ be a representation of $U_{2}$ affording $\theta_{2}$ which agrees with ${\mathcal{D}}_{1}$ on ${\textbf{G}}^{F}$. It then follows from Lemma 2.7 that there exists a projective representation ${\mathcal{P}}:A_{\theta}\to{\textnormal{GL}}_{\theta(1)}({\mathbb{F}})$ associated to $\theta$ with factor set $\alpha$ such that for all $g=u_{1}u_{2},g^{\prime}=u_{1}^{\prime}u_{2}^{\prime}\in A_{\theta}$, $u_{1},u_{1}^{\prime}\in U_{1}$, $u_{2},u_{2}^{\prime}\in U_{2}$,

[TABLE]

where $\lambda_{u_{2}}$ denotes the linear Brauer character of $U_{1}/{\textbf{G}}^{F}$ such that $\theta_{1}^{u_{2}}{\lambda}_{u_{2}}=\theta_{1}$. Since $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $S_{3}$, $\lambda_{u_{2}}$ takes as values 2nd or 3rd roots of unity. In particular, $\alpha^{6}=1$.

Recall that $\mathrm{C}_{A}({\textbf{G}}^{F})\cong\mathrm{Z}({\textbf{G}})\langle F\rangle$ by Lemma 4.4. For $c=zF^{i}$ an arbitrary element of $\mathrm{C}_{A}({\textbf{G}}^{F})$, we have

[TABLE]

where the second equality holds because ${\textnormal{Z}}({\textbf{G}})\leq U_{1}$. Since G is of type B, C, E6 or E7, $|\mathrm{Z}({\textbf{G}})|\leq 3$ and by Lemma 4.5, $F$ has order less than or equal to $3$ on ${\mathcal{L}}^{-1}(\mathrm{Z}({\textbf{G}}))$.

If $F\in U_{2}$ then ${\mathcal{P}}(F^{i})={\mathcal{D}}_{2}(F^{i})$ and since ${\mathcal{D}}_{1}(z)$ and ${\mathcal{D}}_{2}(F^{i})$ are matrices containing roots of unity of order less or equal than $3$, it follows that ${\mathcal{P}}(c)=\xi Id$ for some $\xi$ such that $\xi^{6}=1$. If $\gamma^{\prime}F\in U_{2}$ and $\gamma^{\prime}\in U_{1}$ then ${\mathcal{P}}(F)={\mathcal{D}}_{1}(\gamma^{\prime}){\mathcal{D}}_{2}(\gamma^{\prime}F)$. Since $F\in\mathrm{C}_{A}({\textbf{G}}^{F})$, ${\mathcal{P}}(F)$ is a scalar matrix and hence by a base change we can also assume that ${\mathcal{D}}_{2}(\gamma^{\prime}F)$ is a diagonal matrix. Therefore ${\mathcal{D}}_{1}(\gamma^{\prime})={\mathcal{P}}(F){\mathcal{D}}_{2}(\gamma^{\prime}F)^{-1}$ is also a scalar matrix. Since the orders of $\gamma^{\prime}$ and $\gamma^{\prime}F$ are divisors of $6$, the entries of ${\mathcal{D}}_{1}(\gamma^{\prime})$ and ${\mathcal{D}}_{2}(\gamma^{\prime}F)$ are $6$th roots of unity. Hence the scalar associated to ${\mathcal{P}}(F)$, and therefore the scalar associated to ${\mathcal{P}}(F^{i})$, is a $6$th root of unity and so ${\mathcal{P}}(c)=\xi Id$ for some sixth root of unity $\xi$. ∎

Proposition 4.11**.**

Suppose that G is of type D and suppose that $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ satisfies Assumption 4.7 (i) and (ii). Then there exists a projective representation ${\mathcal{P}}$ of $A_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{4}=1$ and for every $c\in C_{A}({\textbf{G}}^{F})$, ${\mathcal{P}}(c)=\xi Id$ for a fourth root of unity $\xi$.

Proof.

Lemma 4.9 shows that there exist finite groups $U_{1}$ and $U_{2}$ such that $\mathrm{Z}({\textbf{G}})\leq U_{1}\lhd A_{\theta}$, $U_{2}\leq A_{\theta}$, $A_{\theta}=U_{1}U_{2}$, $U_{1}\cap U_{2}={\textbf{G}}^{F}$, $U_{1}/{\textbf{G}}^{F}$ is a subgroup of $D_{8}$, $U_{2}/{\textbf{G}}^{F}$ is cyclic and either $F\in U_{2}$, or $F\gamma^{\prime}\in U_{2}$ and $\gamma^{\prime}\in U_{1}$. Unlike Proposition 4.10, however, when G is of type D we cannot assume that $\theta$ extends to $U_{1}$ as it is possible that $U_{1}/{\textbf{G}}^{F}$ has non-cyclic Sylow subgroups. We therefore need to construct $\mathcal{{\mathcal{P}}}$ in two steps.

We claim that there exist finite groups $K_{1}$ and $K_{2}$ such that $U_{1}=K_{1}K_{2}$, $K_{1}\lhd U_{1}$, $K_{1}\cap K_{2}={\textbf{G}}^{F}$, $K_{2}$ is normalised by $U_{2}$, and such that $K_{1}/{\textbf{G}}^{F}$ is a subgroup of $C_{4}$ and $K_{2}/{\textbf{G}}^{F}$ is a subgroup of $C_{2}$. In addition, $\mathrm{Z}({\textbf{G}})$ is contained in either $K_{1}$ or $K_{2}$ and whenever $\gamma^{\prime}\in U_{1}$, either $\gamma^{\prime}\in K_{1}$ or $\gamma^{\prime}\in K_{2}$. If $U_{1}/{\textbf{G}}^{F}$ is cyclic then the claim holds trivially with $K_{1}=U_{1}$ and $K_{2}={\textbf{G}}^{F}$.

Suppose that $U_{1}/{\textbf{G}}^{F}$ is not cyclic. If $U_{1}/{\textbf{G}}^{F}$ is isomorphic to $D_{8}$ then $\gamma^{\prime}$ is not a power of the field automorphism $F_{0}$, and hence ${\textbf{G}}^{F}$ is untwisted. In this case we let $K_{1}$ and $K_{2}$ be subgroups of $U_{1}$ such that $K_{1}/{\textbf{G}}^{F}$ is the unique cyclic subgroup of $U_{1}/{\textbf{G}}^{F}$ of order $4$ and $K_{2}/{\textbf{G}}^{F}=\langle\gamma^{\prime}\rangle$. If $n$ is even then since ${\textbf{G}}^{F}$ is untwisted, $\mathrm{Z}({\textbf{G}})=\mathrm{Z}({\textbf{G}})^{F}\leq{\textbf{G}}^{F}\leq K_{1}$. If $n$ is odd then $K_{1}/{\textbf{G}}^{F}\cong\mathrm{Z}({\textbf{G}})$ and therefore $\mathrm{Z}({\textbf{G}})\leq K_{1}$. Hence the properties for ${\textnormal{Z}}({\textbf{G}})$ and $\gamma^{\prime}$ are satisfied. Now suppose that $U_{1}/{\textbf{G}}^{F}$ is isomorphic to $C_{2}\times C_{2}$. If ${\textnormal{Z}}({\textbf{G}})\neq{\textnormal{Z}}({\textbf{G}})^{F}$ then we let $K_{2}=\mathrm{Z}({\textbf{G}}){\textbf{G}}^{F}$ and if ${\textnormal{Z}}({\textbf{G}})={\textnormal{Z}}({\textbf{G}})^{F}$ then we let $K_{2}$ be such that $K_{2}/{\textbf{G}}^{F}$ is a subgroup of $U_{1}/{\textbf{G}}^{F}$ of order $2$ stabilised by the action of $U_{2}/{\textbf{G}}^{F}$. If $\gamma^{\prime}\in U_{1}$ but $\gamma^{\prime}\notin K_{2}$ we let $K_{1}/{\textbf{G}}^{F}=\langle\gamma^{\prime}\rangle$. Otherwise, we let $K_{1}/{\textbf{G}}^{F}$ be a direct complement of $K_{2}/{\textbf{G}}^{F}$ in $U_{1}/{\textbf{G}}^{F}$. Again, the properties for ${\textnormal{Z}}({\textbf{G}})$ and $\gamma^{\prime}$ are clearly satisfied. Finally, since $U_{1}/{\textbf{G}}^{F}$ is not cyclic, it follows from the proof of Lemma 4.9 that (in the notation of Lemma 4.9) either $U_{1}/{\textbf{G}}^{F}\cong G_{1}=G_{2}$, $U_{2}/{\textbf{G}}^{F}\cong H_{1}$ and $A_{\theta}/{\textbf{G}}^{F}\cong G_{1}\times H_{1}$; or $G_{1}/G_{2}\cong C_{2}$, $G_{2}$ has a direct complement in $G_{1}$, $U_{1}/{\textbf{G}}^{F}\cong G_{2}$, $U_{2}/{\textbf{G}}^{F}\cong\langle\nu(F_{0}^{\prime})^{i}\rangle$ and $A_{\theta}/{\textbf{G}}^{F}\cong G_{2}\times\langle\nu(F_{0}^{\prime})^{i}\rangle$. In particular, $U_{2}/{\textbf{G}}^{F}$ centralises $U_{1}/{\textbf{G}}^{F}$ and hence $U_{2}$ centralises $K_{2}$ and the claim is proved.

We now construct ${\mathcal{P}}$. Let ${\mathcal{D}}$ be a representation of ${\textbf{G}}^{F}$ affording $\theta$. As $K_{1}/{\textbf{G}}^{F}$, $K_{2}/{\textbf{G}}^{F}$ and $U_{2}/{\textbf{G}}^{F}$ are cyclic, ${\mathcal{D}}$ extends to representations ${\mathcal{D}}_{K_{1}}$ of $K_{1}$, ${\mathcal{D}}_{K_{2}}$ of $K_{2}$ and ${\mathcal{D}}_{U_{2}}$ of $U_{2}$. As $K_{2}$ is normalised by $U_{2}$, it follows from Lemma 2.7 that there exists a projective representation ${\mathcal{P}}_{K_{2}U_{2}}$ of $K_{2}U_{2}$ associated to $\theta$ with factor set $\alpha_{K_{2}U_{2}}$ such that

[TABLE]

[TABLE]

for any $ku,k^{\prime}u^{\prime}\in K_{2}U_{2}$, where ${\lambda}_{u}\in{\operatorname{IBr}}(K_{2}/{\textbf{G}}^{F})$. Applying Lemma 2.7 again yields a projective representation ${\mathcal{P}}$ of $A_{\theta}=K_{1}K_{2}U_{2}$ associated to $\theta$ with factor set $\alpha$ such that

[TABLE]

[TABLE]

for any $k_{1}k_{2}u,k_{1}^{\prime}k_{2}^{\prime}u^{\prime}\in K_{1}K_{2}U_{2}$, where $\lambda_{k_{2}u}\in{\operatorname{IBr}}(K_{1}/{\textbf{G}}^{F})$. Since $K_{1}/{\textbf{G}}^{F}$ is a subgroup of $C_{4}$ and $K_{2}/{\textbf{G}}^{F}$ is a subgroup of $C_{2}$, it follows that $\lambda_{k_{2}u}(k_{1}^{\prime})\lambda_{u}(k_{2}^{\prime})$ has order dividing 4 and hence $\alpha^{4}=1$.

Finally, let $c=zF^{i}$ be an arbitrary element of $\mathrm{C}_{A}({\textbf{G}}^{F})$. Note that ${\mathcal{P}}(c)=\alpha(z,F^{i})^{-1}{\mathcal{P}}(z){\mathcal{P}}(F^{i})$. By construction, ${\textnormal{Z}}({\textbf{G}})$ is contained in either $K_{1}$ or $K_{2}$ so ${\mathcal{P}}(z)={\mathcal{D}}_{K_{1}}(z)$ or ${\mathcal{P}}(z)={\mathcal{D}}_{K_{2}}(z)$. Because G is of type D, the central element $z$ has order dividing 4 and hence the scalar associated to the scalar matrix ${\mathcal{P}}(z)$ is a fourth root of unity. It follows from Lemma 4.1 that the order of $F$ on ${\mathcal{L}}^{-1}(\mathrm{Z}({\textbf{G}}))$ divides 4. If $F\in U_{2}$ then ${\mathcal{P}}(F^{i})={\mathcal{D}}_{U_{2}}(F^{i})$, a scalar matrix associated to a fourth root of unity. Otherwise, $\gamma^{\prime}F\in U_{2}$ and $\gamma^{\prime}\in U_{1}$ and hence, as in Proposition 4.10, ${\mathcal{P}}(F)={\mathcal{D}}_{K_{1}}(\gamma^{\prime}){\mathcal{D}}_{U_{2}}(\gamma^{\prime}F)$ or ${\mathcal{D}}_{K_{2}}(\gamma^{\prime}){\mathcal{D}}_{U_{2}}(\gamma^{\prime}F)$. As in Proposition 4.10, we can assume that ${\mathcal{D}}_{U_{2}}(\gamma^{\prime}F)$ is a diagonal matrix and since the orders of $\gamma^{\prime}$ and $\gamma^{\prime}F$ divide 4, the scalar associated to ${\mathcal{P}}(F)$, and therefore the scalar associated to ${\mathcal{P}}(F^{i})$, is a fourth root of unity. In particular, ${\mathcal{P}}(c)=\xi Id$ for some fourth root of unity $\xi$. ∎

Proposition 4.12**.**

Suppose that ${\textbf{G}}^{F}$ is of type ${\textnormal{D}}_{4}$ and suppose that $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ satisfies Assumption 4.7 (i) but not Assumption 4.7 (ii). Then there exists an integer $r\in\{4,6\}$ and a projective representation ${\mathcal{P}}$ of $A_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{r}=1$ and for every $c\in C_{A}({\textbf{G}}^{F})$, ${\mathcal{P}}(c)=\xi Id$ for an $r$th root of unity $\xi$.

Proof.

Recall that by Corollary 4.3, $A_{\theta}/{\textbf{G}}^{F}$ is isomorphic to a subgroup of $S_{4}\times C_{a}$ for some positive integer $a$. As ${\textbf{G}}^{F}$ is of type D4 the automorphism $F$ acts trivially on $\mathrm{Z}({\textbf{G}})$, see for instance the proof of Lemma 4.1. Therefore, $\mathrm{Z}({\textbf{G}})=\mathrm{Z}({\textbf{G}})^{F}=\mathrm{Z}({\textbf{G}}^{F})$ by [3, Proposition 3.6.8].

Let $\nu\in{\operatorname{IBr}}({\textnormal{Z}}({\textbf{G}}))$ be such that $\theta_{\mathrm{Z}(\mathbf{G})_{\ell^{\prime}}}=\theta(1)\nu$. Note that $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$ and $F_{0}$ act trivially on $\mathrm{Z}({\textbf{G}})$, so in fact only $\Gamma$ acts non-trivially on $\mathrm{Z}({\textbf{G}})$. Using the parametrisation of generators of $\mathrm{Z}({\textbf{G}})$ given in [8, Table 1.12.6] we see that any graph automorphism of order 3 acts non-trivially on $\mathrm{Z}({\textbf{G}})$. However, the only element of ${\operatorname{IBr}}(\mathrm{Z}({\textbf{G}}))$ stabilised by an action of order 3 is the trivial character. Thus, if $\nu$ is nontrivial then $A_{\theta}\leq\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\rtimes\langle F_{0},\gamma\rangle$ for some suitable graph automorphism $\gamma\in\Gamma$ of order $2$. In other words, then $\theta$ satisfies Assumption 4.7 (ii). We can therefore assume that $\theta$ is trivial on ${\textnormal{Z}}({\textbf{G}})$.

Note that ${\textbf{G}}^{F}\langle F\rangle\lhd A_{\theta}$. We can extend $\theta$ to a character $\theta_{0}=\theta\times 1_{\langle F\rangle}\in{\operatorname{IBr}}({\textbf{G}}^{F}\langle F\rangle)$, and since $\theta$ is trivial on $\mathrm{Z}({\textbf{G}})$ we have $A_{\theta}=A_{\theta_{0}}$. Since $A_{\theta}/({\textbf{G}}^{F}\langle F\rangle)$ is isomorphic to a subgroup of $S_{4}\times C_{a}$, Corollary 2.5 shows that $A_{\theta}/({\textbf{G}}^{F}\langle F\rangle)\cong G_{2}\rtimes H_{1}$ for some subgroup $G_{2}$ of $S_{4}$ and some cyclic group $H_{1}$. If $G_{2}$ is isomorphic to a subgroup of $S_{3}$ or $D_{8}$ then the existence of a projective representation ${\mathcal{P}}$ associated to $\theta$ with factor set $\alpha$ satisfying $\alpha^{6}=1$ or $\alpha^{4}=1$ (respectively) follows from the arguments in Propositions 4.10 and 4.11.

It remains to consider the case that $A_{\theta}/({\textbf{G}}^{F}\langle F\rangle)\cong G_{2}\rtimes H_{1}$ and $G_{2}\cong A_{4}$ or $S_{4}$. As shown in Corollary 2.5, then $V_{4}\times 1\lhd A_{\theta}/({\textbf{G}}^{F}\langle F\rangle)$. Thus $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\leq A_{\theta}$, so $\theta$ is stable in $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$. We claim that $\theta$ extends to $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$ and hence to a character of $\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))\langle F\rangle$ lying over $\theta_{0}$. If $\ell=2$ then the quotient $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))/{\textbf{G}}^{F}$ is an $\ell$-group and hence $\theta$ extends to a character $\hat{\theta}$ of $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$ by Green’s indecomposability theorem [12, Theorem 8.11]. Suppose now that $\ell\neq 2$. As $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\leq A_{\theta}$ it follows that the character $\theta$ is stable under diagonal automorphisms, and hence $\theta$ is ${\tilde{\textbf{G}}^{F}}\!$-stable. In particular, it follows from [7, Theorem B] that $\theta$ extends to a character $\tilde{\theta}$ of $\tilde{{\textbf{G}}}^{F}$ such that $\tilde{\theta}$ also lies over the trivial character of $\mathrm{Z}(\tilde{{\textbf{G}}}^{F})$. But we saw in Lemma 4.6 that $\tilde{{\textbf{G}}}^{F}/\mathrm{Z}(\tilde{{\textbf{G}}}^{F})\cong\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))/\mathrm{Z}({\textbf{G}}^{F})$, therefore $\theta$ extends to a character $\hat{\theta}$ of $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))$. Note that $F$ acts trivially on the quotient $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}})/\mathrm{Z}({\textbf{G}})$. Therefore, the extension $\hat{\theta}\in{\operatorname{IBr}}(\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}})))/\mathrm{Z}({\textbf{G}}))$ is $F$-stable and hence can be extended to a character $\hat{\theta}_{0}$ of $\mathcal{L}^{-1}(\mathrm{Z}({\textbf{G}}))\langle F\rangle$ lying above $\theta_{0}$, as claimed.

We can now construct a suitable projective representation ${\mathcal{P}}$ of $A_{\theta}=A_{\theta_{0}}$ which is associated to $\theta_{0}$, and therefore to $\theta$. Since $A_{\theta}/{\textbf{G}}^{F}\langle F\rangle$ is isomorphic to a subgroup of $V_{4}\rtimes(S_{3}\rtimes C_{a})$, there exist groups $H_{1}=\mathcal{L}^{-1}({\textnormal{Z}}({\textbf{G}}))\langle F\rangle\lhd A_{\theta}$ and $H_{2}\leq A_{\theta}$ such that $H_{1}H_{2}=A_{\theta}$ and $H_{1}\cap H_{2}={\textbf{G}}^{F}\langle F\rangle$ with $H_{1}/{\textbf{G}}^{F}\langle F\rangle\cong V_{4}$ and $H_{2}/{\textbf{G}}^{F}\langle F\rangle$ isomorphic to subgroup of $S_{3}\times C_{a}$. Furthermore, there exist groups $K_{1}\lhd H_{2}$ and $K_{2}\leq H_{2}$ such that $K_{1}K_{2}=H_{2}$ and $K_{1}\cap K_{2}={\textbf{G}}^{F}\langle F\rangle$ with $K_{1}/{\textbf{G}}^{F}\langle F\rangle$ isomorphic to a subgroup of $S_{3}$ and $K_{2}/{\textbf{G}}^{F}\langle F\rangle$ cyclic. As $K_{1}/{\textbf{G}}^{F}\langle F\rangle$ has cyclic Sylow subgroups, $\theta_{0}$ extends to $K_{1}$, so by Lemma 2.7 there exists a projective representation $\mathcal{P}_{2}$ of $H_{2}$ associated to $\theta$ with factor set $\alpha_{2}$ such that $\alpha_{2}^{r}=1$ for some $r\in\{2,3\}$. By the claim above, $\theta_{0}$ extends to a character $\hat{\theta}_{0}$ of $H_{1}$. Let $\mathcal{D}_{1}$ be an ordinary representation of $H_{1}$ affording $\hat{\theta}_{0}$ such that $\mathcal{D}_{1}$ and $\mathcal{P}_{2}$ agree on ${\textbf{G}}^{F}\langle F\rangle$. Then by Lemma 2.7 again, there exists a projective representation $\mathcal{P}$ associated to $\theta_{0}$ with factor set $\alpha$ such that $\alpha^{6}=1$. Since $\theta_{0}$ extends $\theta$, ${\mathcal{P}}$ is also associated to $\theta$.

Finally, since ${\textnormal{Z}}({\textbf{G}})={\textnormal{Z}}({\textbf{G}})^{F}$ Lemma 4.4 shows that $\mathrm{C}_{A}({\textbf{G}}^{F})={\textnormal{Z}}({\textbf{G}})\langle F\rangle\leq{\textbf{G}}^{F}\langle F\rangle$. Hence for any $c=zF^{i}\in\mathrm{C}_{A}({\textbf{G}}^{F})$ and any projective representation ${\mathcal{P}}^{\prime}$ of $A_{\theta}$ associated to $\theta_{0}$ (in particular, for ${\mathcal{P}}$), we have

[TABLE]

so ${\mathcal{P}}^{\prime}(c)=\xi Id$ for a root of unity $\xi$ of order at most 2. ∎

4.2. Fake Galois actions

For $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$, we let $\overline{\theta}\in{\operatorname{IBr}}({\textbf{G}}^{F})$ be the character given by $\overline{\theta}(g)=\theta(g^{-1})$ for all $g\in{\textbf{G}}^{F}$.

Proposition 4.13**.**

Let G be a simple, simply connected algebraic group of type B, C, D, E6 or E7. For any positive integer $m$ such that $(m,|{\textbf{G}}^{F}|)=1$, there exists a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})$.

Proof.

Let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$. We fix a positive integer $r_{\theta}$ as follows. If ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$ is cyclic then $r_{\theta}:=\exp({\textnormal{Z}}({\textbf{G}}))$. If ${\operatorname{Out}}({\textbf{G}}^{F})_{\theta}$ is not cyclic then it follows from Propositions 4.10, 4.11 and 4.12 that there exists a projective representation ${\mathcal{P}}$ of $A_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{k}=1$ for some $k\in\{4,6\}$. In this case we fix $r_{\theta}:=k$.

Since $(m,|{\textbf{G}}^{F}|)=1$ and $|{\textbf{G}}^{F}|$ is divisible by $2$ and $3$ by [10, Table 24.1], $m$ and $r_{\theta}$ are coprime and hence $m\equiv\pm 1\mathrm{\,mod\,}(r_{\theta})$. We can therefore define a map on ${\operatorname{IBr}}({\textbf{G}}^{F})$ by

[TABLE]

Since $A_{\bar{\theta}}=A_{\theta}$ and $f_{m}(\bar{\theta})=\overline{f_{m}(\theta)}$, the map $f_{m}:{\operatorname{IBr}}({\textbf{G}}^{F})\rightarrow{\operatorname{IBr}}({\textbf{G}}^{F})$ is a well defined bijection. We claim that this bijection defines a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})$.

First suppose that $\mathrm{Out}({\textbf{G}}^{F})_{\theta}$ is cyclic and let $\nu\in\mathrm{IBr}(\mathrm{Z}({\textbf{G}}^{F})\mid\theta)$. Then

[TABLE]

so $\nu^{m}\in\mathrm{IBr}(\mathrm{Z}({\textbf{G}}^{F})\mid f_{m}(\theta))$. Thus it follows from [14, Lemma 4.5(b)] that

[TABLE]

Now suppose that $\mathrm{Out}({\textbf{G}}^{F})_{\theta}$ is not cyclic. By definition of $r_{\theta}$, there exists a projective representation ${\mathcal{P}}$ of $A_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{r_{\theta}}=1$. Moreover, by Propositions 4.10, 4.11 and 4.12, for every $c\in C_{A}({\textbf{G}}^{F})$, we have ${\mathcal{P}}(c)=\xi Id$ where $\xi$ is an $r_{\theta}$th root of unity. It then follows from [14, Proposition 4.7] that

[TABLE]

By construction of $A$, there exists a surjective map $A\twoheadrightarrow{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})$ such that $A/C_{A}({\textbf{G}}^{F})\cong\mathrm{Out}({\textbf{G}}^{F})$, and hence there exists a surjective map $A_{\theta}\twoheadrightarrow{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})_{\theta}$ such that $A_{\theta}/C_{A}({\textbf{G}}^{F})\cong\mathrm{Out}({\textbf{G}}^{F})_{\theta}$. Thus by [16, Corollary 4.12],

[TABLE]

Consequently, by Definition 2.2, there exists a fake $m$th Galois action on ${\operatorname{IBr}}({\textbf{G}}^{F})$ with respect to ${\textbf{G}}^{F}\lhd{\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})$. ∎

5. Exceptional Covering Groups

In this section we deal with the simple finite groups of Lie type with non-cyclic outer automorphism group whose Schur multiplier has non-trivial exceptional part. We continue to use the notation of Section 2.5 with G a simple simply connected algebraic group defined over $\overline{\mathbb{F}}_{p}$ where $p\neq\ell$, $F:{\textbf{G}}\to{\textbf{G}}$ a Frobenius endomorphism, and ${\textbf{G}}^{F}$ the finite group of the fixed points of G under $F$. Let $S\cong{\textbf{G}}^{F}/{\textnormal{Z}}({\textbf{G}}^{F})$ be a finite simple group of Lie type, and let $X$ be the universal cover of $S$. Let $M(S)$ denote the Schur multiplier of $S$ and recall that ${\textnormal{Z}}(X)\cong M(S)\cong M_{c}(S)\times M_{e}(S)$, where $M_{c}(S)$ denotes the canonical part of the Schur multiplier and $M_{e}(S)$ denotes the exceptional part.

Lemma 5.1**.**

Suppose that $S\in\{O^{+}_{8}(2),U_{6}(2),{}^{2}E_{6}(2),U_{4}(3)\}$. Let $\nu\in\mathrm{IBr}(\mathrm{Z}(X))$. Then either $\nu$ is trivial on $M_{e}(S)$ or $\mathrm{Out}(S)_{\nu}$ is cyclic.

Proof.

Write $\nu=\nu_{1}\times\nu_{2}$ with $\nu_{1}\in\mathrm{IBr}(M_{c}(S))$ and $\nu_{2}\in\mathrm{IBr}(M_{e}(S))$ and suppose that $\nu_{2}$ is non-trivial. If $S\in\{O^{+}_{8}(2),U_{6}(2),{}^{2}E_{6}(2)\}$ then by [8, Table 6.3.1], the outer automorphism group $\mathrm{Out}(S)\cong S_{3}$ acts faithfully on $M_{e}(S)\cong C_{2}\times C_{2}$. However, no non-trivial character of $C_{2}\times C_{2}$ is stabilized by a non-trivial action of an automorphism of order $3$, therefore ${\operatorname{Out}}(S)_{\nu}\cong C_{2}$ is cyclic. If $S=U_{4}(3)$ then $\mathrm{Out}(S)\cong D_{8}$ acts faithfully on $M_{e}(S)\cong C_{3}\times C_{3}$. Again, no non-trivial character of $C_{3}\times C_{3}$ is stabilized by $C_{2}\times C_{2}$ acting faithfully on $C_{3}\times C_{3}$, hence ${\operatorname{Out}}(S)_{\nu}\cong C_{4}$ is cyclic. ∎

Remark 5.2**.**

Let $\pi:X\rightarrow{\textbf{G}}^{F}$ be the quotient map. Then there is a natural bijection

[TABLE]

so we identify any $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ with a character of $X$ which is trivial on $M_{e}(S)$, which we also denote by $\theta$. By [13, Corollary B.8] there exists an isomorphism

[TABLE]

sending $\phi\mapsto\hat{\phi}$ where $\phi\pi=\pi\hat{\phi}$. This induces a quotient map

[TABLE]

such that $\pi(X)={\textbf{G}}^{F}$. Let $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$ and let ${\mathcal{P}}$ be a projective representation of ${\textbf{G}}^{F}\rtimes{\operatorname{Aut}}({\textbf{G}}^{F})_{\theta}$ associated to $\theta$ with factor set $\alpha$. Let $\hat{\mathcal{P}}$ denote the projective representation of $X\rtimes{\operatorname{Aut}}(X)_{\theta}$ given by

[TABLE]

for all $x\in X\rtimes{\operatorname{Aut}}(X)_{\theta}$. Then the factor set $\hat{\alpha}$ of $\hat{\mathcal{P}}$ satisfies $\hat{\alpha}(x,y)=\alpha(\pi(x),\pi(y))$ for all $x,y\in X\rtimes{\operatorname{Aut}}(X)_{\theta}$.

We set up some final pieces of notation. Let $G$ be a finite group and let ${\mathcal{P}}:G\rightarrow{\textnormal{GL}}_{n}({\mathbb{F}})$ be a projective representation with factor set $\alpha:G\times G\rightarrow{\mathbb{F}}^{\times}$. We denote by $\overline{\mathcal{P}}$ the projective representation given by $\overline{\mathcal{P}}(g)=({\mathcal{P}}(g)^{T})^{-1}$ for all $g\in G$. Its factor set $\overline{\alpha}$ satisfies $\overline{\alpha}(g,g^{\prime})=\alpha(g,g^{\prime})^{-1}$ for all $g,g^{\prime}\in G$. Let $\sigma:{\mathbb{F}}\to{\mathbb{F}}$ denote the Galois automorphism given by $x\mapsto x^{\ell}$ for $x\in{\mathbb{F}}$, and recall that if $\theta\in{\operatorname{IBr}}(G)$ then $\theta^{\sigma}\in{\operatorname{IBr}}(G)$, see [12, Problem 2.10]. We denote by ${\mathcal{P}}^{\sigma}$ the projective representation given by ${\mathcal{P}}^{\sigma}(g)=\sigma({\mathcal{P}}(g))$ for all $g\in G$. The associated factor set $\alpha^{\sigma}$ satisfies $\alpha^{\sigma}(g,g^{\prime})=\sigma(\alpha(g,g^{\prime}))$ for all $g,g^{\prime}\in G$.

Proposition 5.3**.**

Suppose that $S\in\{O^{+}_{8}(2),U_{6}(2),{}^{2}E_{6}(2),U_{4}(3)\}$ and let $X$ be the universal cover of $S$. Then for any positive integer $m$ such that $(m,|X|)=1$, there exists a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$.

Proof.

Let $\mathfrak{X}:=\mathrm{IBr}(X)\setminus\mathrm{IBr}(X\mid 1_{M_{e}(S)})$ and fix a positive integer $m$ such that $(m,|X|)=1$. We claim that there exist suitable bijections $f_{m}:{\operatorname{IBr}}(X\mid 1_{M_{e}(S)})\rightarrow{\operatorname{IBr}}(X\mid 1_{M_{e}(S)})$ and $g_{m}:\mathfrak{X}\rightarrow\mathfrak{X}$, such that the bijection $h_{m}:{\operatorname{IBr}}(X)\rightarrow{\operatorname{IBr}}(X)$ given by

[TABLE]

defines a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$.

Let $f_{m}:\mathrm{IBr}({\textbf{G}}^{F})\to\mathrm{IBr}({\textbf{G}}^{F})$ be the bijection given in Corollary 3.7 (if $S=U_{6}(2)$ or $U_{4}(3)$) or Proposition 4.13 (if $S=O^{+}_{8}(2)$ or ${}^{2}\!E_{6}(2)$) such that

[TABLE]

for every $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$. By Remark 5.2, it follows that

[TABLE]

for all $\theta\in{\operatorname{IBr}}({\textbf{G}}^{F})$.

We will now construct $g_{m}$. Suppose first that $S\in\{O^{+}_{8}(2),U_{6}(2),\,^{2}\!E_{6}(2)\}$ or $S=U_{4}(3)$ and $\ell=2$. Define a map $g_{m}:\mathfrak{X}\rightarrow\mathfrak{X}$ by

[TABLE]

This is an $\mathrm{Aut}(X)$-equivariant bijection. Let $\nu\in\mathrm{IBr}(\mathrm{Z}(X)\mid\theta)$. Note that if $S\in\{O^{+}_{8}(2),U_{6}(2),\,^{2}\!E_{6}(2)\}$ then $\mathrm{Z}(X)$ is a subgroup of $C_{2}\times C_{2}\times C_{3}$ and if $S=U_{4}(3)$ and $\ell=2$, then ${\textnormal{Z}}(X)_{\ell^{\prime}}=C_{3}^{2}$. Therefore

[TABLE]

and hence $\nu^{m}\in\mathrm{IBr}(\mathrm{Z}(X)\mid g_{m}(\theta))$. Since $\ell$ divides the order of $S$ and $p\neq\ell$, it only remains to define $g_{m}$ in the case where $S=U_{4}(3)$ and $\ell\in\{5,7\}$. If $\ell=5$ we fix a pair of integers $(k_{1},k_{2}):=(3,4)$ and if $\ell=7$ we fix $(k_{1},k_{2}):=(4,3)$. We then define a bijection $g_{m}:\mathfrak{X}\rightarrow\mathfrak{X}$ by

[TABLE]

Again, clearly $g_{m}$ is an $\mathrm{Aut}(X)$-equivariant bijection, and if $\nu\in\mathrm{IBr}(\mathrm{Z}(X)\mid\theta)$ then $\nu^{m}\in\mathrm{IBr}(\mathrm{Z}(X)\mid g_{m}(\theta))$. Now since ${\operatorname{Out}}(S)_{\theta}$ is cyclic for all $\theta\in\mathfrak{X}$ by Lemma 5.1, it follows from [14, Lemma 4.5(b)] and [16, Corollary 4.12] that

[TABLE]

for all $\theta\in\mathfrak{X}$.

Finally, with $h_{m}:{\operatorname{IBr}}(X)\rightarrow{\operatorname{IBr}}(X)$ defined as above, we have

[TABLE]

for all $\theta\in{\operatorname{IBr}}(X)$, and hence $h_{m}$ defines a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\times{\operatorname{Aut}}(X)$. ∎

We now set up some notation for the following proposition. Let ${\textbf{G}}={\textnormal{SL}}_{3}(\overline{{\mathbb{F}}}_{4})$, $\tilde{{\textbf{G}}}=\mathrm{GL}_{3}(\overline{\mathbb{F}}_{4})$, and let $F_{0}:\tilde{{\textbf{G}}}\to\tilde{{\textbf{G}}}$ be the field automorphism given by squaring all matrix entries. Fix a Frobenius automorphism $F:\tilde{\textbf{G}}\rightarrow\tilde{\textbf{G}}$ given by $F:=F_{0}^{2}$. Then ${\textbf{G}}^{F}/\mathrm{Z}({\textbf{G}}^{F})=L_{3}(4)$ which we denote by $S$. Let $X$ be the universal cover of $S$. We denote the graph automorphism by $\gamma:{\textbf{G}}\to{\textbf{G}},\,(a_{ij})\mapsto(a_{ij})^{-T}$. Let $\tilde{T}$ be the set of diagonal matrices in $\tilde{\textbf{G}}^{F}$ and note that $\tilde{T}\cong C_{3}^{3}$ is an elementary abelian group of exponent $3$. Let $t\in\tilde{T}$. It follows from [9, Proposition 1.5] and [5, Lemma 1.3] that the map $\delta:{\textbf{G}}^{F}\to{\textbf{G}}^{F}$ given by $x\mapsto{}^{t}\!x$ realizes a diagonal automorphism of order $3$. Thus the order of $\delta\in\mathrm{Aut}({\textbf{G}}^{F})$ coincides with the order of $\delta\mathrm{Inn}({\textbf{G}}^{F})$ in $\mathrm{Out}({\textbf{G}}^{F})$.

Let $A:=\langle F_{0},\gamma,\delta\rangle$ considered as automorphisms of $S$ and let $\tilde{A}$ be the preimage of $A$ under the isomorphism $\mathrm{Aut}(X)\to\mathrm{Aut}(S)$ given in [13, Corollary B.8]. Since $F_{0}(\delta)=\gamma(\delta)=\delta^{-1}$, it follows that $A\cong{\operatorname{Aut}}(S)/{\operatorname{Inn}}(S)$. Therefore $\tilde{A}\cong{\operatorname{Aut}}(X)/{\operatorname{Inn}}(X)$ and hence $\mathrm{C}_{X\rtimes\tilde{A}}(X)=\mathrm{Z}(X)$.

Proposition 5.4**.**

Let $S=L_{3}(4)$ and let $X$ be the universal cover of $S$. Then for any positive integer $m$ such that $(m,|X|)=1$, there exists a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$.

Proof.

Since $\ell$ divides the order of $S$ and $p\neq\ell$, we can assume that $\ell\in\{3,5,7\}$. If $\ell\in\{3,5\}$ we fix a pair of integers $(k_{1},k_{2}):=(3,4)$ and if $\ell=7$ we fix $(k_{1},k_{2}):=(4,3)$. Define a map $f_{m}:{\operatorname{IBr}}(X)\rightarrow{\operatorname{IBr}}(X)$ by

[TABLE]

This is clearly an ${\operatorname{Aut}}(X)$-invariant bijection.

Let $\theta\in{\operatorname{IBr}}(X)$. Since $(X\rtimes\tilde{A})/\tilde{A}\cong\tilde{A}\cong\mathrm{Out}(S)\cong S_{3}\times C_{2}$, it follows from Lemma 2.7 that there exists a projective representation $\mathcal{P}$ of $(X\rtimes\tilde{A})_{\theta}$ associated to $\theta$ with factor set $\alpha$ such that $\alpha^{2}=1$. The projective representations $\mathcal{P}^{\sigma}$, $\overline{\mathcal{P}}$ and $\overline{\mathcal{P}^{\sigma}}$ associated to $\theta^{\sigma}$, $\overline{\theta}$ and $\overline{\theta^{\sigma}}$ respectively, then also have factor set $\alpha$, and hence the second condition of Definition 2.1 is satisfied. Now since $\mathrm{C}_{X\rtimes\tilde{A}}(X)=\mathrm{Z}(X)\cong C_{3}\times C_{4}^{2}$, one can check, as in the previous proposition, that if $\nu\in\mathrm{IBr}(\mathrm{Z}(X)\mid\theta)$ then $\nu^{m}\in\mathrm{IBr}(\mathrm{Z}(X)\mid f_{m}(\theta))$ and hence the first condition of Definition 2.1 is also satisfied. Thus, for every $\theta\in{\operatorname{IBr}}(X)$,

[TABLE]

Since there exists a surjective map $X\rtimes\tilde{A}\twoheadrightarrow X\rtimes{\operatorname{Aut}}(X)$ such that $(X\rtimes\tilde{A})/C_{X\rtimes\tilde{A}}(X)\cong\mathrm{Out}(X)$, [16, Corollary 4.12] implies that

[TABLE]

for all $\theta\in{\operatorname{IBr}}(X)$. Hence there exists a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$. ∎

6. Proof of Theorem A

Theorem A**.**

Let $S$ be a non-abelian simple group and let $X$ be the universal covering group of $S$. Then for all non-negative integers $m$ such that $(|X|,m)=1$, there exists a fake $m$th Galois action on ${\operatorname{IBr}}(X)$ with respect to $X\lhd X\rtimes{\operatorname{Aut}}(X)$.

Proof.

First suppose that $S$ has cyclic outer automorphism group. Then the result follows from [14, Theorem 4.4]. Now suppose that $S:={\textbf{G}}^{F}/{\textnormal{Z}}({\textbf{G}}^{F})$ is a simple group for some simple simply connected algebraic group G and a Frobenius endomorphism $F$ as defined in Section 2.5. Then if $p=\ell$, the result follows from [14, Theorem 5.1]. If $p\neq\ell$ and $X={\textbf{G}}^{F}$, then the result follows from Corollary 3.7 and Proposition 4.13. If $p\neq\ell$ and $X\neq{\textbf{G}}^{F}$, then the result follows from Propositions 5.3 and 5.4. ∎