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Uniqueness of factorization for fusion-invariant representations

José Cantarero Centro de Investigación en Matemáticas, A.C. Unidad Mérida
Parque Científico y Tecnológico de Yucatán
Carretera Sierra Papacal–Chuburná Puerto Km 5.5
Sierra Papacal, Mérida, YUC 97302
Mexico.
cantarero@cimat.mx
 and  Germán Combariza Universidad Externado de Colombia
Departamento de Matemáticas
Calle 12 No. 1-17 Este.
C.P. 111711, Bogotá
Colombia.
german.combariza@uexternado.edu.co
Abstract.

Given a saturated fusion system {\mathcal{F}} over a finite pp-group SS, we provide criteria to determine when uniqueness of factorization into irreducible {\mathcal{F}}–invariant representations holds. We use them to prove uniqueness of factorization when the order of SS is at most p3p^{3}. We also describe an example where the monoid of fusion-invariant representations is not even half-factorial. Finally, we find other examples of fusion systems where this monoid is not factorial using GAP.

Key words and phrases:
Factorial monoids, Fusion systems, Monoids of representations
1991 Mathematics Subject Classification:
Primary 20D20; Secondary 20C15, 20M14

Introduction

The KK-theory of the classifying space of a finite group GG is determined by the Atiyah-Segal completion theorem [2], [3] which established it to be the completion of the representation ring of GG with respect to the augmentation ideal. This is generally different from the Grothendieck group of complex vector bundles over BGBG, which is described in [18].

These results were generalized recently to classifying spaces of pp-local finite groups. Given a pp-local finite group (S,,)(S,{\mathcal{F}},{\mathcal{L}}), the Grothendieck group of complex vector bundles over ||p|{\mathcal{L}}|^{\wedge}_{p} was found in [8] to be isomorphic to the representation ring R()R({\mathcal{F}}), which can expressed as an inverse limit of representation rings over the orbit category of {\mathcal{F}}. On the other hand, the KK-theory of ||p|{\mathcal{L}}|^{\wedge}_{p} is isomorphic to the completion of R()R({\mathcal{F}}) with respect to the augmentation ideal (see [4]).

The representation ring R()R({\mathcal{F}}) can also be regarded as the Grothendieck group associated to the monoid of equivalence classes of representations of SS which are {\mathcal{F}}–invariant. For instance, when {\mathcal{F}} is the fusion system associated to a finite group GG with pp-Sylow subgroup SS, they are precisely the representations of SS whose characters are constant on GG–conjugacy classes of elements of SS. We begin the paper by giving a brief review of fusion-invariant representations and their main properties in Section 1. Even though we prove our results in the full generality of saturated fusion systems, our main examples come from finite groups, hence this paper should be of interest to people who are only interested in the representation theory of finite groups.

The group R()R({\mathcal{F}}) is free abelian and its rank is the number of {\mathcal{F}}–conjugacy classes of elements of SS. However, there are examples in [12] and [23] where an {\mathcal{F}}–invariant representation does not decompose as sum of irreducible {\mathcal{F}}–invariant representations in a unique way. This represents a significant departure from the behaviour of the representation ring for finite groups, and this article aims to provide a better understanding of this phenomenon.

This lack of uniqueness is better understood from the point of view of the monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) of isomorphism classes of {\mathcal{F}}–invariant representations. Using the language of reduced commutative monoids, this is a submonoid of the free monoid Rep(S)\operatorname{Rep}\nolimits(S), and these examples show that it is not necessarily free. Free monoids are also called factorial to stress the uniqueness of factorization as a sum of atoms. We find several criteria in Section 2 that guarantee the uniqueness of factorization, some of them characterizations in fact. The following theorem outlines the main results in this section.

Theorem.

Let {\mathcal{F}} be a saturated fusion system over a finite pp-group SS. If one of the following conditions hold, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial.

  • The number of {\mathcal{F}}–conjugacy classes of elements of SS equals the number of irreducible {\mathcal{F}}–invariant representations of SS.

  • As a free abelian group, R()R({\mathcal{F}}) has a basis {ρ1,,ρk}\{\rho_{1},\ldots,\rho_{k}\} of representations which satisfies

    Irr(ρj)ijIrr(ρi)\operatorname{Irr}\nolimits(\rho_{j})-\bigcup_{i\neq j}\operatorname{Irr}\nolimits(\rho_{i})\neq\emptyset

    for all jj.

  • As a free abelian group, R()R({\mathcal{F}}) has a basis of disjoint virtual representations.

  • The fusion system {\mathcal{F}} is transitive.

  • The essential rank of {\mathcal{F}} is zero.

Moreover, the first two conditions are characterizations of the factoriality of Rep()\operatorname{Rep}\nolimits({\mathcal{F}}).

The essential rank of any saturated fusion system over an abelian pp-group SS is zero. In particular, factoriality is guaranteed when the order of SS is at most p2p^{2}. It is natural to wonder if this holds when the order is p3p^{3}. Saturated fusion systems over groups of order eight are few and easy to handle, and over groups of order p3p^{3} with pp odd, there is a classification in [24]. Using a case-by-case analysis and the criteria from Section 2, we determine the factoriality of Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) for all these possibilities. The following theorem corresponds to Theorem 3.2, which is the main result in Section 3.

Theorem.

If |S|p3|S|\leq p^{3} and {\mathcal{F}} is a saturated fusion system over SS, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is a factorial monoid.

This result can not be taken further, since the examples without uniqueness of factorization in [12] and [23] correspond to fusion systems over groups of order 343^{4} and 242^{4}, respectively. They are in fact the fusion systems of Σ9\Sigma_{9} and PSL3(𝔽3)PSL_{3}({\mathbb{F}}_{3}) at the primes 33 and 22. One may wonder if weaker factorization properties hold in Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) in general. One such property is half-factoriality, where we do not ask for uniqueness of factorization, but only for uniqueness of the length of factorization of each element. In Section 4, we show that Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is not half-factorial when {\mathcal{F}} is the fusion system of A6A_{6} at the prime 22.

Even though the example of A6A_{6} is completely described mathematically, we found this example thanks to algorithms that we programmed and implemented in GAP [11]. Namely, for a fusion system {\mathcal{F}} associated to a finite group GG with pp-Sylow subgroup SS, we built functions which can determine the number of {\mathcal{F}}–conjugacy classes, the complexification of the representation ring, and the number of irreducible {\mathcal{F}}–invariant representations which are subrepresentations of the regular representations. In some cases, we noticed that these pieces of information can be used to determine all the irreducible {\mathcal{F}}–invariant representations. Apart from the examples from [12] and [23], we found new examples of fusion systems without uniqueness of factorization into irreducible fusion-invariant representations. Namely, they are the fusion systems of the groups:

  • Σ6\Sigma_{6}, Σ8\Sigma_{8}, PSL2(𝔽17)PSL_{2}({\mathbb{F}}_{17}), PSU3(𝔽5)PSU_{3}({\mathbb{F}}_{5}), M10M_{10}, PSL3(𝔽5)PSL_{3}({\mathbb{F}}_{5}), PSL2(𝔽31)PSL_{2}({\mathbb{F}}_{31}), PSU3(𝔽9)PSU_{3}({\mathbb{F}}_{9}) and GL3(𝔽3)GL_{3}({\mathbb{F}}_{3}) at the prime 22.

  • A9A_{9} and PSL3(𝔽19)PSL_{3}({\mathbb{F}}_{19}) at the prime 33.

Apart from the non-factorial examples above, we found that factoriality holds for A8A_{8} at the prime 22 and for PSp4(𝔽3)PSp_{4}({\mathbb{F}}_{3}) and PSL4(𝔽7)PSL_{4}({\mathbb{F}}_{7}) at the prime 33. Even though we have characterizations of the factoriality of Rep()\operatorname{Rep}\nolimits({\mathcal{F}}), they are not completely satisfactory because they require determining the number of irreducible {\mathcal{F}}–invariant representations or studying the inclusion of R()R({\mathcal{F}}) into R(S)R(S) to find a particular basis, but we do not have a direct criterion in terms of {\mathcal{F}}.

There is another interesting aspect of the irreducible {\mathcal{F}}–invariant representations which we could not settle completely in this paper. In Section 2 we show that if Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial, these representations must be subrepresentations of the regular representation of SS. We conjecture that this is always the case, even when Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is not factorial, and indeed it holds for all the examples where we managed to determine all the irreducible GG–invariant representations.

1. Fusion-invariant representations

In this section we provide an introduction to fusion-invariant representations. Note that all the representations in this paper are finite-dimensional unitary representations and we are only interested in representations up to equivalence. We begin with a particular case which covers most of the examples found in this paper. However, the main results apply to the general definition.

Definition 1.1.

Let GG be a finite group and SS a pp-Sylow subgroup of GG. We say that a representation ρ:SUn\rho\colon S\to U_{n} is GG–invariant if the representations ρ|P\rho_{|P} and ρ|gPg1cg\rho_{|gPg^{-1}}\circ c_{g} are equivalent for any PSP\leq S and any gGg\in G that satisfies gPg1SgPg^{-1}\leq S.

The category of subgroups of SS with GG–subconjugations as morphisms appears as the indexing category in the Theorem of Cartan-Eilenberg on the pp-local cohomology of finite groups (see for instance Theorem III.10.3 in [6]). This theorem can be improved to a homotopical decomposition of BGpBG^{\wedge}_{p} and C. Broto, R. Levi and B. Oliver [5] created a suitable language and techniques to study these decompositions and other spaces with similar properties. In this paper the authors define the notion of saturated fusion system over a finite pp-group SS. This is a category whose objects are the subgroups of SS and whose morphisms are group homomorphisms which are required to satisfy some axioms.

The interested reader can see Definitions 1.1 and 1.2 in [5] for the formal definition, but the intuitive idea is that these morphisms behave like conjugations by elements of a bigger group. In fact, the main source of examples of saturated fusion systems comes from finite groups. If GG is a finite group and SS is a pp-Sylow subgroup of GG, there is a saturated fusion system called S(G){\mathcal{F}}_{S}(G) whose morphism sets are given by

HomS(G)(P,Q)=HomG(P,Q)={f:PQf=cg for some gG}\operatorname{Hom}\nolimits_{{\mathcal{F}}_{S}(G)}(P,Q)=\operatorname{Hom}\nolimits_{G}(P,Q)=\{f\colon P\to Q\mid f=c_{g}\text{ for some }g\in G\}

But there are also exotic fusion systems, meaning that they are not of this form. At this moment, there are many constructions of exotic fusion systems in the literature, although we will not encounter them in this paper. We direct the interested reader to [19] and [24] for some examples. Next we extend Definition 1.1 to this general setting.

Definition 1.2.

Let {\mathcal{F}} be a saturated fusion system over a finite pp-group SS. We say that a representation ρ:SUn\rho\colon S\to U_{n} is {\mathcal{F}}–invariant if the representations ρ|P\rho_{|P} and ρ|f(P)f\rho_{|f(P)}\circ f are equivalent for any PSP\leq S and any fHom(P,S)f\in\operatorname{Hom}\nolimits_{{\mathcal{F}}}(P,S).

In the particular case of S(G){\mathcal{F}}_{S}(G), this corresponds to GG–invariant representations. When {\mathcal{F}} is clear, we also call them fusion-invariant representations. These representations have been studied in [4], [8], [12] and [23], but also in [7], [22], [27] and [28] in a more general context.

In order to determine whether a representation is fusion-invariant, we can use character theory and certain families of subgroups which generate the fusion system in a certain sense. For the following lemma, we say that two elements xx, yy of SS are {\mathcal{F}}–conjugate if there exists a morphism ff in {\mathcal{F}} such that f(x)=yf(x)=y. The definitions of centric, radical and essential subgroups can be found in Definition 1.6 of [5], Definition A.9 of [5] and in the second paragraph of page 2 in [16], respectively.

Lemma 1.3.

Let ρ:SUn\rho\colon S\to U_{n} be a representation. The following conditions are equivalent.

  1. (1)

    ρ\rho is {\mathcal{F}}–invariant.

  2. (2)

    ρ|P\rho_{|P} and ρ|Pf\rho_{|P}\circ f are isomorphic for any PSP\leq S and any fAut(P)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(P).

  3. (3)

    ρ|P\rho_{|P} and ρ|Pf\rho_{|P}\circ f are isomorphic for any {\mathcal{F}}–centric radical subgroup PP of SS and any fAut(P)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(P).

  4. (4)

    ρ|P\rho_{|P} and ρ|Pf\rho_{|P}\circ f are isomorphic for any {\mathcal{F}}–essential subgroup PP of SS and any fAut(P)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(P) and for P=SP=S with any fAut(S)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S).

  5. (5)

    χρ(x)=χρ(y)\chi_{\rho}(x)=\chi_{\rho}(y) for any pair of {\mathcal{F}}–conjugate elements xx, yy.

In the case of S(G){\mathcal{F}}_{S}(G), the last condition states that a representation ρ\rho is GG–invariant if and only χρ(x)=χρ(y)\chi_{\rho}(x)=\chi_{\rho}(y) for any pair of GG–conjugate elements of SS. We now enumerate some immediate properties of {\mathcal{F}}–invariant representations.

  • If α\alpha is equivalent to β\beta and α\alpha is {\mathcal{F}}–invariant, so is β\beta.

  • If ρ\rho and α\alpha are {\mathcal{F}}–invariant, so are ρα\rho\oplus\alpha and ρα\rho\otimes\alpha.

  • The trivial representation and the reduced regular representation are {\mathcal{F}}–invariant for any {\mathcal{F}}.

  • If ρ\rho and ρα\rho\oplus\alpha are {\mathcal{F}}–invariant, so is α\alpha.

  • If nρn\rho is {\mathcal{F}}–invariant for some n0n\neq 0, so is ρ\rho.

  • If ρα\rho\oplus\alpha is isomorphic to ρβ\rho\oplus\beta, then α\alpha is isomorphic to β\beta.

Definition 1.4.

An {\mathcal{F}}–invariant representation is called irreducible if it can not be decomposed as the direct sum of two nontrivial {\mathcal{F}}–invariant representations.

It should be clear that any {\mathcal{F}}–invariant representation can be decomposed as a direct sum of irreducible {\mathcal{F}}–invariant representations. Note that an irreducible {\mathcal{F}}–invariant representation is not necessarily irreducible if we regard it as a representation of SS. The following example illustrates this.

Example 1.5.

Consider the 33-Sylow subgroup S={1,(1,2,3),(1,3,2)}S=\{1,(1,2,3),(1,3,2)\} of Σ3\Sigma_{3}. The irreducible representations of SS are the 11-dimensional representations ρj\rho_{j} determined by

ρj(1,2,3)=e2πij/3\rho_{j}(1,2,3)=e^{2\pi ij/3}

for j=0,1,2j=0,1,2. Up to equivalence, the representations of SS have the form ρ=m0ρ0+m1ρ1+m2ρ2\rho=m_{0}\rho_{0}+m_{1}\rho_{1}+m_{2}\rho_{2} with mim_{i}\in{\mathbb{N}}. Since SS does not have any nontrivial proper subgroup and AutΣ3(S)={1,c(1,2)}\operatorname{Aut}\nolimits_{\Sigma_{3}}(S)=\{1,c_{(1,2)}\}, we have that ρ\rho is Σ3\Sigma_{3}–invariant if and only if ρc(1,2)\rho\circ c_{(1,2)} is equivalent to ρ\rho. But

ρc(1,2)=m0ρ0+m1ρ2+m2ρ1\rho\circ c_{(1,2)}=m_{0}\rho_{0}+m_{1}\rho_{2}+m_{2}\rho_{1}

and so this holds if and only if m1=m2m_{1}=m_{2}. Hence any Σ3\Sigma_{3}–invariant representation is a nonnegative integral combination of ρ0\rho_{0} and ρ1+ρ2\rho_{1}+\rho_{2}. These are irreducible Σ3\Sigma_{3}–invariant representations, since they can not be decomposed as the sum of two nontrivial Σ3\Sigma_{3}–invariant subrepresentations. Note that ρ1+ρ2\rho_{1}+\rho_{2} is not irreducible as a representation of SS.

The isomorphism classes of {\mathcal{F}}–invariant representations of SS form a semiring with the operations of direct sum and tensor product. We denote it by Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) and when =S(G){\mathcal{F}}={\mathcal{F}}_{S}(G), we denote it by RepG(S)\operatorname{Rep}\nolimits_{G}(S) instead. By definition, this is a subsemiring of the semiring Rep(S)\operatorname{Rep}\nolimits(S). For the purpose of studying the uniqueness of decomposition as a sum of irreducibles, we can ignore their multiplicative structures and regard them simply as commutative monoids.

Remark 1.6.

We chose to write RepG(S)\operatorname{Rep}\nolimits_{G}(S) despite the fact that if corresponds to S(G){\mathcal{F}}_{S}(G) to emphasize that its elements are classes of representations of SS.

The theory of {\mathcal{F}}–invariant representations has some similarities with the theory of supercharacters and superclasses [10]. For instance, {\mathcal{F}} determines a partition of SS given by {\mathcal{F}}–conjugacy classes, the set {1}\{1\} is one of the classes and characters of {\mathcal{F}}–invariant representations are constant on each of these GG–conjugacy classes. We can appreciate the difference in the case of D8(Σ4){\mathcal{F}}_{D_{8}}(\Sigma_{4}). The group D8D_{8} is partitioned into four Σ4\Sigma_{4}–conjugacy classes and Example 4.5.4 in [12] shows that any representation of D8D_{8} which is Σ4\Sigma_{4}–invariant is generated by 11, X+ZX+Z, Y+ZY+Z and XYXY, where XX, YY and XYXY are the one-dimensional irreducible representations of D8D_{8} and ZZ is the two-dimensional irreducible representation. If this partition of D8D_{8} were a part of a supercharacter theory, there would exist a supercharacter which would contain 2Z2Z as a subrepresentation. But then the partition of the set of irreducible representations of D8D_{8} would have less than four elements.

On the other hand, it is easier to produce a supercharacter theory over a finite pp-group SS which does not come from a saturated fusion system over SS. The two-part partition of SS where all the nontrivial elements are in the same class always corresponds to a supercharacter theory. But if SS is a pp-group whose nontrivial elements have at least two different orders, the subsets forming this partition can not be {\mathcal{F}}–conjugacy classes.

2. Monoids with unique factorizations

In this section we recall some terminology and results from the theory of reduced commutative monoids in order to study uniqueness of factorization in the monoids of fusion-invariant representations introduced in the previous section. We follow the terminology from [13]. Since our monoids are commutative, we will denote by 0 the identity element. In the rest of the paper, {\mathcal{F}} will denote a saturated fusion system over a pp-group SS.

Definition 2.1.

A commutative monoid is reduced if 0 is the only invertible element.

The commutative monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is reduced because it is a submonoid of Rep(S)\operatorname{Rep}\nolimits(S), which is reduced.

Definition 2.2.

Let MM be a reduced commutative monoid. A non-zero element xx of MM is called an atom if x=y+zx=y+z implies y=0y=0 or z=0z=0.

Definition 2.3.

An atomic monoid is a reduced commutative monoid which is generated by its set of atoms.

The monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is atomic and its atoms are the irreducible {\mathcal{F}}–invariant representations of SS.

Definition 2.4.

An atomic monoid is called factorial if every nonzero element can be expressed as a finite sum of atoms in a unique way up to reordering.

This is also called a unique factorization monoid. It is easy to see that a reduced commutative monoid is factorial if and only if it is free, that is, isomorphic to a monoid of the form

X=xX{\mathbb{N}}X=\bigoplus_{x\in X}{\mathbb{N}}

for some set XX. For a finite group GG, the monoid Rep(G)\operatorname{Rep}\nolimits(G) is factorial because any representation of GG can be decomposed as a direct sum of irreducible representations in a unique way up to reordering. The atoms of RepΣ3(S)\operatorname{Rep}\nolimits_{\Sigma_{3}}(S) are the representations ρ0\rho_{0} and ρ1+ρ2\rho_{1}+\rho_{2} described in Example 1.5 and this monoid is factorial since

mρ0+n(ρ1+ρ2)=mρ0+n(ρ1+ρ2)m\rho_{0}+n(\rho_{1}+\rho_{2})=m^{\prime}\rho_{0}+n^{\prime}(\rho_{1}+\rho_{2})

implies m=mm=m^{\prime} and n=nn=n^{\prime}.

In order to describe a non-factorial example, we introduce the Grothendieck group of Rep()\operatorname{Rep}\nolimits({\mathcal{F}}), which is denoted by R()R({\mathcal{F}}) and called the representation ring of {\mathcal{F}}. We also use the notation RG(S)R_{G}(S) when =S(G){\mathcal{F}}={\mathcal{F}}_{S}(G). In general, R()R({\mathcal{F}}) is a free abelian group whose rank is the number of {\mathcal{F}}–conjugacy classes of elements of SS (see Corollary 2.2 in [4]). However, the monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is not free in general as the following example from [12] shows.

Example 2.5.

Let SS be a 33-Sylow subgroup of Σ9\Sigma_{9}. The monoid RepΣ9(S)\operatorname{Rep}\nolimits_{\Sigma_{9}}(S) is not factorial. The following example was first found in Example 4.5.5 of [12]. The representation ring RΣ9(S)R_{\Sigma_{9}}(S) is the free abelian group generated by the following Σ9\Sigma_{9}–invariant virtual representations

1,a1+a2+X,b0+c0+ZX,b1+b2+c1+c2+2Z,X+Y1,a_{1}+a_{2}+X,b_{0}+c_{0}+Z-X,b_{1}+b_{2}+c_{1}+c_{2}+2Z,X+Y

where aia_{i}, b0b_{0}, cic_{i}, XX, YY and ZZ are representations of SS which do not share any irreducible representations of SS. A more convenient basis is given by

1,a1+a2+X,b0+c0+Z+a1+a2,b1+b2+c1+c2+2Z,X+Y1,a_{1}+a_{2}+X,b_{0}+c_{0}+Z+a_{1}+a_{2},b_{1}+b_{2}+c_{1}+c_{2}+2Z,X+Y

By Remark 2.12, we can conclude that 11, b0+c0+Z+a1+a2b_{0}+c_{0}+Z+a_{1}+a_{2}, b1+b2+c1+c2+2Zb_{1}+b_{2}+c_{1}+c_{2}+2Z and X+YX+Y are irreducible {\mathcal{F}}–invariant representations. On the other hand, using the same criterion with the basis

1,a1+a2+X,b0+c0+Z+Y,b1+b2+c1+c2+2Z,X+Y1,a_{1}+a_{2}+X,b_{0}+c_{0}+Z+Y,b_{1}+b_{2}+c_{1}+c_{2}+2Z,X+Y

we can conclude that a1+a2+Xa_{1}+a_{2}+X and b0+c0+Z+Yb_{0}+c_{0}+Z+Y are irreducible {\mathcal{F}}–invariant representations. Then the expression

(a1+a2+X)+(b0+c0+Z+Y)=(b0+c0+Z+a1+a2)+(X+Y)(a_{1}+a_{2}+X)+(b_{0}+c_{0}+Z+Y)=(b_{0}+c_{0}+Z+a_{1}+a_{2})+(X+Y)

shows that RepΣ9(S)\operatorname{Rep}\nolimits_{\Sigma_{9}}(S) is not factorial.

Another known example of a non-factorial monoid of fusion-invariant representations, shown in Example A.2 of [23], is RepPGL3(𝔽3)(SD16)\operatorname{Rep}\nolimits_{PGL_{3}({\mathbb{F}}_{3})}(SD_{16}).

Remark 2.6.

Let 1{\mathcal{F}}_{1} and 2{\mathcal{F}}_{2} be saturated fusion systems over S1S_{1} and S2S_{2}, respectively. These fusion systems are isomorphic if there is an isomorphism of groups α:S1S2\alpha\colon S_{1}\to S_{2} and an isomorphism of categories α:12\alpha^{\prime}\colon{\mathcal{F}}_{1}\to{\mathcal{F}}_{2} given on objects by α(P)=α(P)\alpha^{\prime}(P)=\alpha(P). It is easy to show that the isomorphism α\alpha induces an isomorphism of monoids Rep(2)Rep(1)\operatorname{Rep}\nolimits({\mathcal{F}}_{2})\to\operatorname{Rep}\nolimits({\mathcal{F}}_{1}) and an isomorphism of rings R(2)R(1)R({\mathcal{F}}_{2})\to R({\mathcal{F}}_{1}).

The following result is well known in the theory of non-unique factorizations, but we include the proof of this particular case for completeness.

Proposition 2.7.

Let MM be an atomic monoid such that its Grothendieck group K(M)K(M) is free abelian of finite rank. Then MM is factorial if and only if the rank of K(M)K(M) equals the cardinality of the set of atoms of MM.

Proof.

Let XX be the set of atoms of MM. If MM is factorial, then it is isomorphic to X{\mathbb{N}}X and therefore K(M)XK(M)\cong{\mathbb{Z}}X.

Now assume that MM is not factorial. Since the set XX of atoms of MM generates K(M)K(M) as an abelian group, there is an epimorphism φ:XK(M)\varphi\colon{\mathbb{Z}}X\to K(M) induced by the universal map from MM to K(M)K(M). Since MM is not factorial, there is a relation nxx=0\sum n_{x}x=0 in K(M)K(M) with at least one nx0n_{x}\neq 0. Hence φ\varphi is not injective and there is an exact sequence

0YXK(M)00\to{\mathbb{Z}}Y\to{\mathbb{Z}}X\to K(M)\to 0

with YY nonempty. Since K(M)K(M) is free abelian of finite rank, if |X||X| is finite we have

|X|=rkK(M)+|Y||X|=\operatorname{rk}\nolimits K(M)+|Y|

and so rkK(M)<|X|\operatorname{rk}\nolimits K(M)<|X|. If XX is infinite, then certainly rkK(M)\operatorname{rk}\nolimits K(M) differs from |X||X|. ∎

In particular, the proof shows that if MM is an atomic monoid such that K(M)K(M) is free abelian of finite rank, the number of atoms in these atomic monoids is greater or equal than the rank of K(M)K(M).

Corollary 2.8.

The number of irreducible {\mathcal{F}}–invariant representation is greater or equal than the rank of R()R({\mathcal{F}}).

Corollary 2.9.

The monoid Rep()Rep({\mathcal{F}}) is factorial if and only if the number of {\mathcal{F}}–conjugacy classes of elements of SS equals the number of irreducible {\mathcal{F}}–invariant representations of SS.

It is illustrative to see an example where these numbers differ.

Example 2.10.

We return to RepΣ9(S)\operatorname{Rep}\nolimits_{\Sigma_{9}}(S) from Example 2.5, which we already know not to be factorial. The rank of RΣ9(S)R_{\Sigma_{9}}(S) is five and it is freely generated as an abelian group by the following Σ9\Sigma_{9}–invariant virtual representations of SS.

1,P=a1+a2+X,Q=b0+c0+ZX,R=b1+b2+c1+c2+2Z,S=X+Y1,P=a_{1}+a_{2}+X,Q=b_{0}+c_{0}+Z-X,R=b_{1}+b_{2}+c_{1}+c_{2}+2Z,S=X+Y

We replace this basis of RΣ9(S)R_{\Sigma_{9}}(S) by

1,P,Q=Q+S=b0+c0+Z+Y,R,S1,P,Q^{\prime}=Q+S=b_{0}+c_{0}+Z+Y,R,S

which is a now a basis of Σ9\Sigma_{9}–invariant representations.

Let ρ=a1+bP+cQ+dR+eS\rho=a1+bP+cQ^{\prime}+dR+eS be an irreducible Σ9\Sigma_{9}–invariant representation, where the coefficients are integers. By the irreducibility, if all the coefficients are nonnegative, then ρ\rho must be one of these five representations, which we already know to be irreducible by Example 2.5.

Next assume that at least one coefficient is negative. This coefficient could not be bb, because then a1a_{1} and a2a_{2} would appear with negative coefficient. For the same reason, it could not be aa, cc or dd. Therefore let us assume that e1e\leq-1. Then we must have e+b0e+b\geq 0 and e+c0e+c\geq 0, since e+be+b is the coefficient of XX in ρ\rho and e+ce+c is the coefficient of YY in ρ\rho. But then

ρ=a1+(e)(PS+Q)+(b+e)P+(c+e)Q+dR\rho=a1+(-e)(P-S+Q^{\prime})+(b+e)P+(c+e)Q^{\prime}+dR

All the coefficients are now nonnegative and PS+Q=a1+a2+b0+c0+ZP-S+Q^{\prime}=a_{1}+a_{2}+b_{0}+c_{0}+Z is a representation. So ρ\rho could also be PS+QP-S+Q^{\prime} which we already knew from Example 2.5 to be irreducible. We have proved that there are exactly six irreducible Σ9\Sigma_{9}–invariant representations.

Given a representation ρ\rho of SS, let us denote by Irr(ρ)\operatorname{Irr}\nolimits(\rho) the set of its irreducible subrepresentations.

Proposition 2.11.

The monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial if and only if R()R({\mathcal{F}}) has a basis {ρ1,,ρk}\{\rho_{1},\ldots,\rho_{k}\} of representations which satisfies

Irr(ρj)ijIrr(ρi)\operatorname{Irr}\nolimits(\rho_{j})-\bigcup_{i\neq j}\operatorname{Irr}\nolimits(\rho_{i})\neq\emptyset

for all jj. Moreover, if this holds, this basis is the set of irreducible {\mathcal{F}}–invariant representations.

Proof.

Given such a basis {ρ1,,ρk}\{\rho_{1},\ldots,\rho_{k}\}, if ρ\rho is an irreducible {\mathcal{F}}–invariant representation, we must have

ρ=n1ρ1++nkρk\rho=n_{1}\rho_{1}+\ldots+n_{k}\rho_{k}

for some njn_{j}\in{\mathbb{Z}}. Since ρj\rho_{j} contains an irreducible subrepresentation which is not shared with any ρi\rho_{i} for iji\neq j, the coefficient njn_{j} must be nonnegative. Since ρ\rho is irreducible, we must have that ρ=ρj\rho=\rho_{j} for some jj and therefore the number of irreducible {\mathcal{F}}–invariant representations is at most kk. By Corollary 2.8, this number is exactly kk and thus Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial by Corollary 2.9. We also obtain the last claim.

On the other hand, if Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial, by Corollary 2.9, the number of irreducible {\mathcal{F}}–invariant representations equal the rank of R()R({\mathcal{F}}), which is finite. Let ρ1,,ρk\rho_{1},\ldots,\rho_{k} be the irreducible {\mathcal{F}}–invariant representations. Given ρj\rho_{j}, if each of its irreducible subrepresentations appeared as a subrepresentation of some ρi\rho_{i} with iji\neq j, then ρj\rho_{j} would be a subrepresentation of

ijniρi\sum_{i\neq j}n_{i}\rho_{i}

for certain nonnegative integers nin_{i}. And this would contradict the fact that Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial. Hence {ρ1,,ρk}\{\rho_{1},\ldots,\rho_{k}\} is a basis of R()R({\mathcal{F}}) which satisfies the desired property. ∎

Remark 2.12.

We can use a similar setup to prove an irreducibility criterion. Assume that R()R({\mathcal{F}}) has a basis {ρ1,,ρk}\{\rho_{1},\ldots,\rho_{k}\} of representations and ρj\rho_{j} satisfies

Irr(ρj)ijIrr(ρi)\operatorname{Irr}\nolimits(\rho_{j})-\bigcup_{i\neq j}\operatorname{Irr}\nolimits(\rho_{i})\neq\emptyset

Let us show that ρj\rho_{j} is an irreducible {\mathcal{F}}–invariant representation.

Let αIrr(ρj)ijIrr(ρi)\alpha\in\operatorname{Irr}\nolimits(\rho_{j})-\bigcup_{i\neq j}\operatorname{Irr}\nolimits(\rho_{i}) and let ρj=γ+ϵ\rho_{j}=\gamma+\epsilon, where γ\gamma and ϵ\epsilon are {\mathcal{F}}–invariant representations. Since ρj\rho_{j} is the only representation in the basis that contains α\alpha as a subrepresentation, we may assume that γ\gamma contains α\alpha with the same multiplicity as in ρj\rho_{j}. When we write γ\gamma as an integral lineal combination of the ρi\rho_{i}, the coefficient of ρj\rho_{j} must be one, hence γ=ρj+γ\gamma=\rho_{j}+\gamma^{\prime}. Then γ+ϵ=0\gamma^{\prime}+\epsilon=0 from where γ=0=ϵ\gamma^{\prime}=0=\epsilon and so ρj=γ\rho_{j}=\gamma.

We say that two virtual representations of SS are disjoint if their decompositions as integral linear combinations of irreducible representations of SS do not share any irreducible representation. More generally, given a finite number of virtual representations of SS, we say that they are disjoint if any two of them are disjoint.

Proposition 2.13.

If R()R({\mathcal{F}}) has a basis of disjoint virtual representations of SS, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial.

Proof.

We first modify this basis by replacing each element of the form ρ-\rho, where ρ\rho is a representation, with ρ\rho. Let α\alpha be a virtual representation in this new basis of R()R({\mathcal{F}}) which is not a representation. If β\beta is an {\mathcal{F}}–invariant representation of SS and we express it in this basis, the coefficient of α\alpha must be zero because β\beta is a representation. Therefore all the {\mathcal{F}}–invariant representations of SS are generated by the set XX of elements in this basis which are representations. In particular, XX is the set of irreducible {\mathcal{F}}–invariant representations of SS by Remark 2.12. This shows that the number of irreducible {\mathcal{F}}–invariant representations is at most the rank of R()R({\mathcal{F}}), hence equal to the rank by Corollary 2.8. Therefore Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial by Corollary 2.9. ∎

Proposition 2.13 is not a characterization of the factoriality of Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) as the following example shows.

Example 2.14.

The representation ring of D8(Σ4){\mathcal{F}}_{D_{8}}(\Sigma_{4}) was determined in Example 4.5.4 of [12]. Let {1,X,Y,Z,XY}\{1,X,Y,Z,XY\} be the set of irreducible representations of D8D_{8}, where ZZ is the two-dimensional irreducible representation. Then RΣ4(D8)R_{\Sigma_{4}}(D_{8}) is the free abelian group generated by 11, X+ZX+Z, Y+ZY+Z and XYXY. By Proposition 2.11, the monoid RepΣ4(D8)\operatorname{Rep}\nolimits_{\Sigma_{4}}(D_{8}) is factorial and these are all the possible irreducible Σ4\Sigma_{4}–invariant representations.

However, it is not possible to find a basis for RΣ4(D8)R_{\Sigma_{4}}(D_{8}) of disjoint virtual representations. Assume there is a basis {v1,v2,v3,v4}\{v_{1},v_{2},v_{3},v_{4}\} of disjoint virtual representations. Let us express the basis we already have in terms of this basis. We can assume ±X\pm X appears in the decomposition of v1v_{1} and then the coefficient of v1v_{1} in Y+ZY+Z, 11 and XYXY would be zero. The analogous conditions must hold, up to reordering, for the pairs (Y,v2)(Y,v_{2}), (1,v3)(1,v_{3}) and (XY,v4)(XY,v_{4}). In particular X+Z=±v1X+Z=\pm v_{1} and Y+Z=±v2Y+Z=\pm v_{2}, which contradicts the disjointness of v1v_{1} and v2v_{2}.

However, Proposition 2.13 is useful in two extreme cases. The following terminology is taken from the Introduction of [17].

Definition 2.15.

A fusion system {\mathcal{F}} is transitive if there are only two {\mathcal{F}}–conjugacy classes of elements of SS.

Lemma 2.16.

If {\mathcal{F}} is transitive, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial.

Proof.

In this proof we work with the equivalent description of R()R({\mathcal{F}}) as the Grothendieck group of {\mathcal{F}}–invariant virtual characters.

The two conjugacy classes must be the class of 11 and the class of all nontrivial elements. Let χR()\chi\in R({\mathcal{F}}) and let m=χ(1)m=\chi(1). Since χ\chi is {\mathcal{F}}–invariant, we must have χ(s)=n\chi(s)=n for all s1s\neq 1. We know that mm is an integer. We will prove that nn is an integer as well. Note that

(χ,1)=1|S|sSχ(s)¯=1|S|(m+n¯(|S|1))=mn¯|S|+n¯(\chi,1)=\frac{1}{|S|}\sum_{s\in S}\overline{\chi(s)}=\frac{1}{|S|}(m+\bar{n}(|S|-1))=\frac{m-\bar{n}}{|S|}+\bar{n}

is an integer, or equivalently

mn|S|+n\frac{m-n}{|S|}+n

On the other hand, since SS is a pp-group, it is nilpotent and so [S,S]S[S,S]\neq S. Therefore its abelianization is not trivial and SS has a nontrivial 11-dimensional representation ρ\rho. We denote its character by ρ\rho as well. Then

(χ,ρ)=1|S|(m+n¯s1ρ(s))=1|S|(m+n¯sSρ(s)n¯ρ(1))=1|S|(mn¯)(\chi,\rho)=\frac{1}{|S|}\left(m+\bar{n}\sum_{s\neq 1}\rho(s)\right)=\frac{1}{|S|}\left(m+\bar{n}\sum_{s\in S}\rho(s)-\bar{n}\rho(1)\right)=\frac{1}{|S|}(m-\bar{n})

is also an integer, hence

mn|S|\frac{m-n}{|S|}

is an integer. Therefore nn is an integer and |S||S| divides mnm-n. Let mn=a|S|m-n=a|S| with aa\in{\mathbb{Z}}. Then

χ=(a+n)1+aregS~\chi=(a+n)1+a\widetilde{\operatorname{reg}\nolimits_{S}}

where regS~\widetilde{\operatorname{reg}\nolimits_{S}} is the reduced regular representation. Since the reduced regular representation does not contain the trivial representation as a subrepresentation, this shows

R()=1regS~R({\mathcal{F}})={\mathbb{Z}}1\oplus{\mathbb{Z}}\widetilde{\operatorname{reg}\nolimits_{S}}

Thus R()R({\mathcal{F}}) has a basis of disjoint characters and it is factorial by Proposition 2.13. ∎

This second extreme case corresponds to the case when every morphism is a restriction of an automorphism of SS.

Lemma 2.17.

If the essential rank of {\mathcal{F}} is zero, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial.

Proof.

If the essential rank of {\mathcal{F}} is zero, then a representation ρ\rho of SS is {\mathcal{F}}–invariant if and only if ρf\rho\circ f is equivalent to ρ\rho for every fAut(S)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S). Note that ρ\rho is irreducible if and only if ρf\rho\circ f is irreducible. Therefore Aut(S)\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S) acts on the set Irr(S)\operatorname{Irr}\nolimits(S) of irreducible representations of SS. Then

R()=R(S)Aut(S)=[Irr(S)]Aut(S)[Irr(S)/Aut(S)]R({\mathcal{F}})=R(S)^{\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S)}={\mathbb{Z}}[\operatorname{Irr}\nolimits(S)]^{\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S)}\cong{\mathbb{Z}}[\operatorname{Irr}\nolimits(S)/{\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S)}]

This shows that R()R({\mathcal{F}}) has a basis of disjoint characters and therefore it is factorial by Proposition 2.13. ∎

If SS is abelian, the essential rank of {\mathcal{F}} is automatically zero and pp-groups of order at most p2p^{2} are abelian, so it is natural to next treat the case when |S|=p3|S|=p^{3}. We will do this in the next section.

Remark 2.18.

If Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial and ρ\rho is an irreducible {\mathcal{F}}–invariant representation, then ρ\rho is a subrepresentation of regSn\operatorname{reg}\nolimits_{S}^{n} for some nn, hence regSn=ρα\operatorname{reg}\nolimits_{S}^{n}=\rho\oplus\alpha. Since regS\operatorname{reg}\nolimits_{S} is {\mathcal{F}}–invariant, so is α\alpha and so we can decompose it as a direct sum of irreducible {\mathcal{F}}–invariant representations

regSn=ρρiki\operatorname{reg}\nolimits_{S}^{n}=\rho\oplus\bigoplus\rho_{i}^{k_{i}}

On the other hand, since regS\operatorname{reg}\nolimits_{S} is {\mathcal{F}}–invariant, it has a decomposition

regS=ωiri\operatorname{reg}\nolimits_{S}=\bigoplus\omega_{i}^{r_{i}}

where the ωi\omega_{i} are irreducible {\mathcal{F}}–invariant representations. By the uniqueness of factorization, ρ=ωj\rho=\omega_{j} for some jj, in particular ρ\rho is a subrepresentation of regS\operatorname{reg}\nolimits_{S}.

The argument in the previous remark shows that any irreducible{\mathcal{F}}–invariant representation is a subrepresentation of the regular representation when Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial. Apart from the examples from [12] and [23], we enumerate in the appendix several examples of saturated fusion systems where Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is not factorial, found using the software GAP. We have checked that many of them still satisfy that any irreducible {\mathcal{F}}–invariant representation is a subrepresentation of the regular representation (we did not include these details in the paper). Hence it is reasonable to pose the following conjecture.

Conjecture 2.19.

Every irreducible {\mathcal{F}}–invariant representation of SS is a subrepresentation of the regular representation.

3. Fusion systems over pp-groups of small order

In this section we determine that the monoids of fusion-invariant representations for saturated fusion systems over pp-groups of order p3p^{3} are all factorial.

We begin with p=2p=2. Let SS be a group of order eight and {\mathcal{F}} a saturated fusion system over SS. If SS is abelian, then the essential rank of {\mathcal{F}} is zero, hence Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial by Lemma 2.17. It is well known that, up to isomorphism, there are three saturated fusion systems over D8D_{8}, the fusion systems of D8D_{8}, S4S_{4} and A6A_{6}, and two over Q8Q_{8}, corresponding to the groups Q8Q_{8} and SL2(𝔽3)SL_{2}({\mathbb{F}}_{3}). For both D8D_{8} and Q8Q_{8}, let us denote by XX, YY and XYXY the one-dimensional irreducible representations and by ZZ the two-dimensional irreducible representation. The monoid of fusion-invariant representations is factorial in all these cases:

  • When the group is D8D_{8} or Q8Q_{8}, because it is the monoid of representations of a group.

  • For Σ4\Sigma_{4}, we proved it in Example 2.14.

  • For A6A_{6}, there are three A6A_{6}–conjugacy classes of elements of D8D_{8}. It is easy to check that the irreducible A6A_{6}–invariant representations are 11, X+XY+ZX+XY+Z and Y+ZY+Z.

  • For SL2(𝔽3)SL_{2}({\mathbb{F}}_{3}), there are three SL2(𝔽3)SL_{2}({\mathbb{F}}_{3})–conjugacy classes of elements of Q8Q_{8}. It is easy to check that the irreducible SL2(𝔽3)SL_{2}({\mathbb{F}}_{3})–invariant representations are 11, X+Y+XYX+Y+XY and ZZ.

For the rest of the section we assume that pp is an odd prime. We first consider the extraspecial group S=p+1+2S=p_{+}^{1+2} with presentation

a,b,ca7=1,b7=1,c7=1,[a,b]=c,[a,c]=1,[b,c]=1\langle a,b,c\mid a^{7}=1,b^{7}=1,c^{7}=1,[a,b]=c,[a,c]=1,[b,c]=1\rangle

Recall that saturated fusion systems over p+1+2p_{+}^{1+2} are determined by the automorphism group of p+1+2p_{+}^{1+2}, the elementary abelian groups of rank 22 which are centric radical and their automorphisms groups (see Corollary 4.5 in [24]). Therefore a representation ρ\rho of SS is {\mathcal{F}}–invariant if and only if the following two conditions are satisfied:

  1. (1)

    fρρf^{*}\rho\cong\rho for all fAut(S)f\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S).

  2. (2)

    hρ|Vρ|Vh^{*}\rho|_{V}\cong\rho|_{V} for all hAut(V)h\in\operatorname{Aut}\nolimits_{{\mathcal{F}}}(V) for each {\mathcal{F}}–centric radical subgroup VV which is elementary abelian of rank 22.

The first condition states that ρ\rho is a fixed point of the action of Aut(S)\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S) on Rep(S)\operatorname{Rep}\nolimits(S). As in the proof of Lemma 2.17, the fixed points form a free monoid, a basis given by adding over the orbits of the irreducible representations of SS under the action of Aut(S)\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S).

By Theorem 5.5.4 in [14], the group SS has p2p^{2} irreducible one-dimensional representations and p1p-1 irreducible representations of dimension pp. The one-dimensional representations come from the quotient

SS/Z(S)/p×/pS\to S/Z(S)\cong{\mathbb{Z}}/p\times{\mathbb{Z}}/p

and are denoted by xiyjx^{i}y^{j}, just like the corresponding representations of /p×/p{\mathbb{Z}}/p\times{\mathbb{Z}}/p. Let ω\omega denote a primitive p2p^{2}-root of unity. The remaining irreducible representations ϕj\phi_{j} for j=1,,p1j=1,\ldots,p-1 are described by their values on aa and bb. Namely,

ϕj(a)=diag(ωj(1+(p1)p),,ωj(1+p),ωj,1)\phi_{j}(a)=\operatorname{diag}\nolimits(\omega^{j(1+(p-1)p)},\ldots,\omega^{j(1+p)},\omega^{j},1)

and ϕj(b)\phi_{j}(b) is the linear transformation that sends e1e_{1} to epe_{p} and ene_{n} to en1e_{n-1} if n2n\geq 2. Here {e1,,ep}\{e_{1},\ldots,e_{p}\} denotes the standard basis of p{\mathbb{C}}^{p}. Note that ϕj(c)=wjI\phi_{j}(c)=w^{j}I.

To determine the orbits of the irreducible representations of SS under the action of Aut(S)\operatorname{Aut}\nolimits_{{\mathcal{F}}}(S), we can use the following lemma.

Lemma 3.1.

Let ff be an automorphism of SS represented by the matrix M=(rrss)M=\left(\begin{array}[]{cc}r&r^{\prime}\\ s&s^{\prime}\end{array}\right) under the isomorphism Out(S)GL2(𝔽p)\operatorname{Out}\nolimits(S)\cong GL_{2}({\mathbb{F}}_{p}). Then f(xiyj)=xir+jsyir+jsf^{*}(x^{i}y^{j})=x^{ir+js}y^{ir^{\prime}+js^{\prime}} and f(ϕj)=ϕjdet(M)f^{*}(\phi_{j})=\phi_{j\det(M)}, where all the indices are taken mod pp.

Hence we will start determining the monoid Rep(S)Out(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{{\mathcal{F}}}(S)} and then impose the conditions given by the elementary abelian subgroups. The elementary abelian subgroups of rank two of SS are the subgroups ViV_{i} generated by cc and abiab^{i} for i=0,,p1i=0,\ldots,p-1, and the subgroup VpV_{p} generated by cc and bb. These subgroups have a natural action of GL2(𝔽p)GL_{2}({\mathbb{F}}_{p}) by identifying each of them with 𝔽p2{\mathbb{F}}_{p}^{2} in such a way that the generating elements mentioned above form the standard basis. Note that the irreducible representations of an elementary abelian group VV of rank two can be identified with elements of Hom𝔽p(V,𝔽p)\operatorname{Hom}\nolimits_{{\mathbb{F}}_{p}}(V,{\mathbb{F}}_{p}) and the action of Aut(V)GL2(𝔽p)\operatorname{Aut}\nolimits(V)\cong GL_{2}({\mathbb{F}}_{p}) under this identification is the dual action, which corresponds to multiplication by the transposed matrix.

We now go through the all the possibilities in Tables 1.1 and 1.2 from [24] to determine the factoriality of the monoid of fusion-invariant representations in each case.

Case 1: G=SWG=S\rtimes W, where pp does not divide the order of WW. It follows from Lemma 2.17.

Case 2: G=(/p)2(SL2(𝔽p)/r)G=({\mathbb{Z}}/p)^{2}\rtimes(SL_{2}({\mathbb{F}}_{p})\rtimes{\mathbb{Z}}/r), where rr divides p1p-1. We have OutG(S)/(p1)×/r\operatorname{Out}\nolimits_{G}(S)\cong{\mathbb{Z}}/(p-1)\times{\mathbb{Z}}/r, and its elements are diagonal matrices. Note that the action of /r{\mathbb{Z}}/r partitions the set of representations of the form yjy^{j} with j0j\neq 0 into d=(p1)/rd=(p-1)/r orbits. Let YiY_{i}, for i=1,,di=1,\ldots,d, be the sum of the representations in each of these disjoint sets. A similar argument applies for representations of the form xiyjx^{i}y^{j} with ij0ij\neq 0, obtaining representations BiB_{i} for i=1,,di=1,\ldots,d. A basis for Rep(S)OutG(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{G}(S)} is given by

{1,X=i=1p1xi,Y1,,Yd,B1,,Bd,Z=j=1p1ϕj}\left\{1,X=\sum_{i=1}^{p-1}x^{i},Y_{1},\ldots,Y_{d},B_{1},\ldots,B_{d},Z=\sum_{j=1}^{p-1}\phi_{j}\right\}

Now let ρ\rho be a GG–invariant representation of SS. Then

ρ=q1+lX+i=1dsiYi+i=1dtiBi+uZ\rho=q1+lX+\sum_{i=1}^{d}s_{i}Y_{i}+\sum_{i=1}^{d}t_{i}B_{i}+uZ

The restriction of ρ\rho to the subgroup V0V_{0} generated by aa and cc is

(q+s1r++sdr)1+l(i=1p1xi)+i=1dtir(i=1p1xi)+u(i=0,j=1p1xiyj)(q+s_{1}r+\ldots+s_{d}r)1+l\left(\sum_{i=1}^{p-1}x^{i}\right)+\sum_{i=1}^{d}t_{i}r\left(\sum_{i=1}^{p-1}x^{i}\right)+u\left(\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)

Since AutG(V0)SL2(𝔽p)/r\operatorname{Aut}\nolimits_{G}(V_{0})\cong SL_{2}({\mathbb{F}}_{p})\rtimes{\mathbb{Z}}/r, the elements xx and xyxy are related through the action of SL2(𝔽p)SL_{2}({\mathbb{F}}_{p}) and therefore u=l+t1r++tdru=l+t_{1}r+\ldots+t_{d}r. Thus

ρ=q1+l(X+Z)+i=1dsiYi+i=1dti(Bi+rZ)\rho=q1+l(X+Z)+\sum_{i=1}^{d}s_{i}Y_{i}+\sum_{i=1}^{d}t_{i}(B_{i}+rZ)

and hence RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial by Proposition 2.11.

Case 3: G=L3(p)G=L_{3}(p), where 33 does not divide p1p-1. In this case OutL3(p)(S)/(p1)×/(p1)\operatorname{Out}\nolimits_{L_{3}(p)}(S)\cong{\mathbb{Z}}/(p-1)\times{\mathbb{Z}}/(p-1), and its elements are the diagonal matrices in GL2(𝔽p)GL_{2}({\mathbb{F}}_{p}). A basis for Rep(S)OutL3(p)(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{L_{3}(p)}(S)} is given by

{1,X,Y=i=1p1yi,M=ij0xiyj,Z}\left\{1,X,Y=\sum_{i=1}^{p-1}y^{i},M=\sum_{ij\neq 0}x^{i}y^{j},Z\right\}

Now let ρ\rho be an L3(p)L_{3}(p)–invariant representation of SS. Then

ρ=q1+rX+sY+tM+uZ\rho=q1+rX+sY+tM+uZ

The restriction of ρ\rho to the subgroup V0V_{0} generated by aa and cc is

q1+r(i=1p1xi)+s(p1)1+t(ij0xi)+u(i=0,j=1p1xiyj)q1+r\left(\sum_{i=1}^{p-1}x^{i}\right)+s(p-1)1+t\left(\sum_{ij\neq 0}x^{i}\right)+u\left(\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)

Since AutL3(p)(V0)GL2(𝔽p)\operatorname{Aut}\nolimits_{L_{3}(p)}(V_{0})\cong GL_{2}({\mathbb{F}}_{p}), its action on V0{1}V_{0}-\{1\} is transitive and so is its action on the set of non-trivial representations. Hence all the coefficients of these non-trivial representations must all be equal, that is, r+(p1)t=ur+(p-1)t=u. By a symmetric argument with the subgroup VpV_{p} generated by bb and cc, we obtain s+(p1)t=us+(p-1)t=u, hence r=sr=s. Therefore

ρ=q1+r(X+Y+Z)+t(M+(p1)Z)\rho=q1+r(X+Y+Z)+t(M+(p-1)Z)

By Proposition 2.11, RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial. In fact, it is the free monoid generated by 11, X+Y+ZX+Y+Z and M+(p1)ZM+(p-1)Z.

Case 4: G=L3(p):2G=L_{3}(p):2, where 33 does not divide p1p-1. Note that an L3(p)L_{3}(p)–invariant representation ρ\rho of SS is (L3(p):2)(L_{3}(p):2)–invariant if and only if fρρf^{*}\rho\cong\rho, where ff is the automorphism of SS that permutes aa and bb. Since ff^{*} fixes XX, YY, MM and ZZ, we obtain that ρ\rho is L3(p)L_{3}(p)–invariant if and only if it is (L3(p):2)(L_{3}(p):2)–invariant. In particular, we see that SS has three L3(p)L_{3}(p)–conjugacy classes.

Case 5: G=2F4(2)G=^{2}F_{4}(2)^{\prime} and J4J_{4} at the prime 33. The fusion systems of these groups contain the fusion system of L3(3):2L_{3}(3):2, hence they either have the same conjugacy classes as L3(3):2L_{3}(3):2 or they have only two conjugacy classes. Therefore the monoid RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial for these two groups by Lemma 2.16 or by the previous case.

Case 6: G=ThG=Th at the prime 55. In this case all the elementary abelian subgroups of rank two are radical and ThTh–conjugate and have GL2(𝔽5)GL_{2}({\mathbb{F}}_{5}) as their group of automorphisms. Given a nontrivial element xx in SS, either x=cjx=c^{j}, in which case xx is conjugate to cc in ThTh, or the subgroup generated by cc and xx is one of the elementary abelian subgroups of rank two of SS and again xx is conjugate to cc in ThTh. Therefore the fusion system of ThTh is transitive and so RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial by Lemma 2.16.

Case 7: G=L3(p)G=L_{3}(p) when 33 divides p1p-1. In this case OutL3(p)(S)/(p1)×/[(p1)/3]\operatorname{Out}\nolimits_{L_{3}(p)}(S)\cong{\mathbb{Z}}/(p-1)\times{\mathbb{Z}}/[(p-1)/3] and it is the subgroup of Out(S)\operatorname{Out}\nolimits(S) generated by the matrices

(ξ00ξ2)(ξ200ξ)\left(\begin{array}[]{cc}\xi&0\\ 0&\xi^{-2}\end{array}\right)\qquad\qquad\left(\begin{array}[]{cc}\xi^{-2}&0\\ 0&\xi\end{array}\right)

where ξ\xi is a generator of 𝔽p×{\mathbb{F}}_{p}^{\times}. A more convenient set of generators is given by

(ξ3001)(ξ200ξ)\left(\begin{array}[]{cc}\xi^{3}&0\\ 0&1\end{array}\right)\qquad\qquad\left(\begin{array}[]{cc}\xi^{-2}&0\\ 0&\xi\end{array}\right)

Let us compute a basis for Rep(S)OutL3(p)(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{L_{3}(p)}(S)}. The orbit of xx contains xξ3x^{\xi^{3}}, hence also xξ3ξ2=xξx^{\xi^{3}\xi^{-2}}=x^{\xi} and all the powers of xx. The same happens for yy. To find the orbit of an element xiyjx^{i}y^{j} with ij0ij\neq 0, note that the action of OutL3(p)(S)\operatorname{Out}\nolimits_{L_{3}(p)}(S) on the set {xiyjij0}\{x^{i}y^{j}\mid ij\neq 0\} is free, hence this set breaks into three orbits of size (p1)2/3(p-1)^{2}/3. If xiyjx^{i}y^{j} is related to xiyjx^{i^{\prime}}y^{j^{\prime}}, then i=iξ3k2ni^{\prime}=i\xi^{3k-2n} and j=jξnj^{\prime}=j\xi^{n}, and so

j1i=j1ξniξ3k2n=j1iξ3k3nj^{\prime-1}i^{\prime}=j^{-1}\xi^{-n}i\xi^{3k-2n}=j^{-1}i\xi^{3k-3n}

This condition breaks the set {xiyjij0}\{x^{i}y^{j}\mid ij\neq 0\} into three subsets, those that satisfy j1i=ξmj^{-1}i=\xi^{m} with mm congruent to 0, 11 and 22 mod 33, respectively. Since each of these three subsets have size (p1)2/3(p-1)^{2}/3, these are the three orbits. Finally, since the determinant of the second matrix is ξ1\xi^{-1}, which is a generator of 𝔽p×{\mathbb{F}}_{p}^{\times}, the orbit of ϕ1\phi_{1} includes all the ϕj\phi_{j}. Hence a basis of Rep(S)OutL3(p)(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{L_{3}(p)}(S)} is given by

{1,X,Y,Z,Mn=j,kxja3k+nyj}\left\{1,X,Y,Z,M_{n}=\sum_{j,k}x^{ja^{3k+n}}y^{j}\right\}

where 0n20\leq n\leq 2, the index jj runs over {1,,p1}\{1,\ldots,p-1\} and the index kk over {1,,(p1)/3}\{1,\ldots,(p-1)/3\}. Now let ρ\rho be an L3(p)L_{3}(p)–invariant representation of SS. Then

ρ=q1+rX+sY+t0M0+t1M1+t2M2+uZ\rho=q1+rX+sY+t_{0}M_{0}+t_{1}M_{1}+t_{2}M_{2}+uZ

The restriction of ρ\rho to the subgroup V0V_{0} generated by aa and cc is

q1+r(i=1p1xi)+s(p1)1+(t0+t1+t2)(p1)3(i=1p1xi)+u(i=0,j=1p1xiyj)q1+r\left(\sum_{i=1}^{p-1}x^{i}\right)+s(p-1)1+\frac{(t_{0}+t_{1}+t_{2})(p-1)}{3}\left(\sum_{i=1}^{p-1}x^{i}\right)+u\left(\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)

Since AutL3(p)(V0)SL2(𝔽p)/[(p1)/3]\operatorname{Aut}\nolimits_{L_{3}(p)}(V_{0})\cong SL_{2}({\mathbb{F}}_{p})\rtimes{\mathbb{Z}}/[(p-1)/3], the elements xx and xyxy are related through the action of AutL3(p)(V0)\operatorname{Aut}\nolimits_{L_{3}(p)}(V_{0}) and therefore the coefficients of xx and xyxy must be equal. That is

u=r+p13(t0+t1+t2)u=r+\frac{p-1}{3}(t_{0}+t_{1}+t_{2})

By a similar argument with the subgroup VpV_{p} generated by bb and cc, we obtain

u=s+p13(t0+t1+t2)u=s+\frac{p-1}{3}(t_{0}+t_{1}+t_{2})

from where r=sr=s. Therefore

ρ=q1+r(X+Y+Z)+t0(M0+p13Z)+t1(M1+p13Z)+t2(M2+p13Z)\rho=q1+r(X+Y+Z)+t_{0}\left(M_{0}+\frac{p-1}{3}Z\right)+t_{1}\left(M_{1}+\frac{p-1}{3}Z\right)+t_{2}\left(M_{2}+\frac{p-1}{3}Z\right)

By Proposition 2.11, the monoid RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial, it is in fact the free monoid generated by 11, X+Y+ZX+Y+Z and Mn+p13ZM_{n}+\frac{p-1}{3}Z for n=0,1,2n=0,1,2.

Case 8: G=L3(p):2G=L_{3}(p):2 when 33 divides p1p-1. The group OutL3(p):2(S)\operatorname{Out}\nolimits_{L_{3}(p):2}(S) is generated by OutL3(p)\operatorname{Out}\nolimits_{L_{3}(p)} and the class of the automorphism which permutes aa and bb. Therefore a basis for Rep(S)OutL3(p):2(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{L_{3}(p):2}(S)}. It is given by

{1,X+Y,M0,M1+M2,Z}\{1,X+Y,M_{0},M_{1}+M_{2},Z\}

Considering the automorphisms of the subgroups V0V_{0} and VpV_{p} as above, we obtain that RepG(S)\operatorname{Rep}\nolimits_{G}(S) is generated by 1,X+Y+Z,M0+p13Z,M1+M2+2(p1)3Z1,X+Y+Z,M_{0}+\frac{p-1}{3}Z,M_{1}+M_{2}+\frac{2(p-1)}{3}Z, hence it is factorial by Proposition 2.11.

Case 9: G=L3(p).3G=L_{3}(p).3 when 33 divides p1p-1. In this case, the group OutL3(p).3(S)\operatorname{Out}\nolimits_{L_{3}(p).3}(S) is the subgroup of diagonal matrices in GL2(𝔽p)GL_{2}({\mathbb{F}}_{p}) and therefore a basis for Rep(S)OutL3(p).3(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{L_{3}(p).3}(S)} is given by {1,X,Y,M,Z}\{1,X,Y,M,Z\}. Considering the automorphisms of the subgroups V0V_{0} and VpV_{p}, we obtain that RepG(S)\operatorname{Rep}\nolimits_{G}(S) is generated by 1,X+Y+Z,M+6Z1,X+Y+Z,M+6Z, hence it is a factorial monoid by Proposition 2.11.

Case 10: G=L3(p).Σ3G=L_{3}(p).\Sigma_{3} when 33 divides p1p-1. Since the fusion system of L3(p).Σ3L_{3}(p).\Sigma_{3} contains the fusion system of L3(p).3L_{3}(p).3, it either has the same conjugacy classes as L3(p).3L_{3}(p).3 or two conjugacy classes. So RepG(S)\operatorname{Rep}\nolimits_{G}(S) is a factorial monoid by the previous case or by Lemma 2.16.

Case 11: G=HeG=He, the Held group, at the prime 77. The group OutHe(S)\operatorname{Out}\nolimits_{He}(S) is generated by the matrices

(0330),(2001),(1002)\left(\begin{array}[]{cc}0&3\\ 3&0\end{array}\right),\left(\begin{array}[]{cc}2&0\\ 0&1\end{array}\right),\left(\begin{array}[]{cc}1&0\\ 0&2\end{array}\right)

Hence Rep(S)OutHe(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{He}(S)} is generated by the elements

{1,Z,A,B,C,D,E}\{1,Z,A,B,C,D,E\}

where A=x+x2+x4+y3+y6+y5A=x+x^{2}+x^{4}+y^{3}+y^{6}+y^{5} and B=x3+x6+x5+y+y2+y4B=x^{3}+x^{6}+x^{5}+y+y^{2}+y^{4}. The rest of representations are given by

C\displaystyle C =xy+x2y+x4y+xy2+x2y2+x4y2+xy4+x2y4+x4y4+x3y3+x6y3+x5y3\displaystyle=xy+x^{2}y+x^{4}y+xy^{2}+x^{2}y^{2}+x^{4}y^{2}+xy^{4}+x^{2}y^{4}+x^{4}y^{4}+x^{3}y^{3}+x^{6}y^{3}+x^{5}y^{3}
+x3y5+x6y5+x5y5+x3y6+x6y6+x5y6\displaystyle+x^{3}y^{5}+x^{6}y^{5}+x^{5}y^{5}+x^{3}y^{6}+x^{6}y^{6}+x^{5}y^{6}
D=xy3+xy6+xy5+x2y3+x2y6+x2y5+x4y3+x4y6+x4y5D=xy^{3}+xy^{6}+xy^{5}+x^{2}y^{3}+x^{2}y^{6}+x^{2}y^{5}+x^{4}y^{3}+x^{4}y^{6}+x^{4}y^{5}
E=x3y+x6y+x5y+x3y2+x6y2+x5y2+x3y4+x6y4+x5y4E=x^{3}y+x^{6}y+x^{5}y+x^{3}y^{2}+x^{6}y^{2}+x^{5}y^{2}+x^{3}y^{4}+x^{6}y^{4}+x^{5}y^{4}

Now let ρ\rho be a HeHe–invariant representation of SS. Then

ρ=q1+rA+sB+tC+uD+vE+wZ\rho=q1+rA+sB+tC+uD+vE+wZ

The restriction of ρ\rho to the subgroup V1V_{1} generated by abab and cc is

(q+3u+3v)1+(r+s+3t+u+v)(i=16xi)+w(i=0,j=1p1xiyj)(q+3u+3v)1+(r+s+3t+u+v)\left(\sum_{i=1}^{6}x^{i}\right)+w\left(\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)

Since Aut(V1)=SL2(𝔽7)\operatorname{Aut}\nolimits_{{\mathcal{F}}}(V_{1})=SL_{2}({\mathbb{F}}_{7}), the elements xx and xyxy are related via the dual action. Therefore w=r+s+3t+u+vw=r+s+3t+u+v and thus

ρ=(q+3t+3u)1+r(A+Z)+s(B+Z)+t(C+3Z)+u(D+Z)+v(E+Z)\rho=(q+3t+3u)1+r(A+Z)+s(B+Z)+t(C+3Z)+u(D+Z)+v(E+Z)

We obtain that RepG(S)\operatorname{Rep}\nolimits_{G}(S) is generated by 1,A+Z,B+Z,C+3Z,D+Z,E+Z1,A+Z,B+Z,C+3Z,D+Z,E+Z, hence it is factorial by Proposition 2.11.

Case 12: G=He:2G=He:2 at the prime 77. A representation ρ\rho of SS is (He:2)(He:2)–invariant if and only if it is HeHe–invariant and fρ=ρf^{*}\rho=\rho, where ff is the order-two automorphism that permutes aa and bb. Note that ff^{*} fixes CC and ZZ, permutes AA and BB, and permutes DD and EE. Hence RepG(S)\operatorname{Rep}\nolimits_{G}(S) is generated by 11, A+B+2ZA+B+2Z, C+3ZC+3Z and D+E+2ZD+E+2Z. By Proposition 2.11, this is a factorial monoid.

Case 13: G=Fi24G=Fi^{\prime}_{24}, the derived Fischer group, at the prime 77. The fusion system of Fi24Fi^{\prime}_{24} contains the fusion system of He:2He:2, but for the derived Fischer group, the subgroups V3V_{3}, V5V_{5} and V6V_{6} are radical with SL2(𝔽7)SL_{2}({\mathbb{F}}_{7}) as group of automorphisms. Let ρ=q1+r(A+B+2Z)+s(C+3Z)+t(D+E+2Z)\rho=q1+r(A+B+2Z)+s(C+3Z)+t(D+E+2Z). The restriction of ρ\rho to the subgroup V6V_{6} generated by ab6ab^{6} and cc is

(q+6s)1+r(2i=16xi+2i=0,j=1p1xiyj)+s(2i=16xi+3i=0,j=1p1xiyj)+t(3i=16xi+2i=0,j=1p1xiyj)(q+6s)1+r\left(2\sum_{i=1}^{6}x^{i}+2\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)+s\left(2\sum_{i=1}^{6}x^{i}+3\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)+t\left(3\sum_{i=1}^{6}x^{i}+2\sum_{i=0,j=1}^{p-1}x^{i}y^{j}\right)

Since xx and xyxy are related through the dual action of SL2(𝔽7)SL_{2}({\mathbb{F}}_{7}), we must have 2r+2s+3t=2r+3s+2t2r+2s+3t=2r+3s+2t, that is, t=st=s and therefore

ρ=q1+r(A+B+2Z)+s(C+D+E+5Z)\rho=q1+r(A+B+2Z)+s(C+D+E+5Z)

Hence RepG(S)\operatorname{Rep}\nolimits_{G}(S) is a factorial monoid by Proposition 2.11 and we see that this fusion system has three conjugacy classes.

Case 14: G=Fi24G=Fi_{24}, the Fisher group, at the prime 77, and =RV1{\mathcal{F}}=RV_{1}. The fusion system of the Fisher group Fi24Fi_{24} and the exotic 77-local fusion system RV1RV_{1} both contain the fusion system of Fi24Fi^{\prime}_{24}, so they either have the same conjugacy classes as Fi24Fi^{\prime}_{24} or two conjugacy classes. So RepG(S)\operatorname{Rep}\nolimits_{G}(S) and Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) are factorial monoids by the previous case or by Lemma 2.16.

Case 15: G=ONG=O^{\prime}N, the O’Nan’s group, at the prime 77. In this case the group Out(S)\operatorname{Out}\nolimits_{{\mathcal{F}}}(S) is generated by the matrices

(3003)(1006)(0110)\left(\begin{array}[]{cc}3&0\\ 0&3\end{array}\right)\quad\left(\begin{array}[]{cc}1&0\\ 0&6\end{array}\right)\quad\left(\begin{array}[]{cc}0&1\\ 1&0\end{array}\right)

and therefore the monoid Rep(S)Out(S)\operatorname{Rep}\nolimits(S)^{\operatorname{Out}\nolimits_{{\mathcal{F}}}(S)} is generated by the representations

{1,X+Y,Z,L,N}\{1,X+Y,Z,L,N\}

where

L=i=16xiyi+i=16xiy7iL=\sum_{i=1}^{6}x^{i}y^{i}+\sum_{i=1}^{6}x^{i}y^{7-i}

and N=MLN=M-L. Let

ρ=m1+n(X+Y)+lL+rN+kZ\rho=m1+n(X+Y)+lL+rN+kZ

The restriction of ρ\rho to the subgroup V0V_{0} generated by aa and cc is given by

m1+n1+n(i=16xi)+2l(i=16xi)+4r(i=16xi)+k(i=0,j=16xiyj)m1+n1+n\left(\sum_{i=1}^{6}x^{i}\right)+2l\left(\sum_{i=1}^{6}x^{i}\right)+4r\left(\sum_{i=1}^{6}x^{i}\right)+k\left(\sum_{i=0,j=1}^{6}x^{i}y^{j}\right)

Since V0V_{0} is centric and radical, then xx and xyxy are related through the action of SL2(𝔽7)SL_{2}({\mathbb{F}}_{7}) and therefore n+2l+4r=kn+2l+4r=k. Similarly, the restriction to the subgroup V1V_{1} generated by abab and cc is given by

(m+6l)1+2n(i=16xi)+l(i=16xi)+4r(i=16xi)+k(i=0,j=16xiyj)(m+6l)1+2n\left(\sum_{i=1}^{6}x^{i}\right)+l\left(\sum_{i=1}^{6}x^{i}\right)+4r\left(\sum_{i=1}^{6}x^{i}\right)+k\left(\sum_{i=0,j=1}^{6}x^{i}y^{j}\right)

and so 2n+l+4r=k2n+l+4r=k. Then n=ln=l and k=3n+4rk=3n+4r. Therefore RepG(S)\operatorname{Rep}\nolimits_{G}(S) is generated by 11, X+Y+M+3ZX+Y+M+3Z and N+4ZN+4Z and so it is factorial by Proposition 2.11. We also see that there are three conjugacy elements of elements of SS.

Case 16: G=ON:2G=O^{\prime}N:2 at the prime 77, and =RV2,RV3{\mathcal{F}}=RV_{2},RV_{3}. Since all these fusion systems contain the fusion system of ONO^{\prime}N, the corresponding monoids are factorial by the previous case or Lemma 2.16.

Case 17: G=MG=M, the Fischer-Griess Monster, at the prime 1313. Following [26], the group OutM(S)3×4S3\operatorname{Out}\nolimits_{M}(S)\cong 3\times 4S_{3} is generated by the matrices

r=(3009)s=(5427)z=(2212)r=\left(\begin{array}[]{cc}3&0\\ 0&9\end{array}\right)\quad s=\left(\begin{array}[]{cc}5&-4\\ -2&7\end{array}\right)\quad z=\left(\begin{array}[]{cc}2&2\\ 1&-2\end{array}\right)

The element rr has order three, while zz has order 2424 and satisfies

z2=(6006)z^{2}=\left(\begin{array}[]{cc}6&0\\ 0&6\end{array}\right)

Moreover, s=zrz1s=zrz^{-1}. Note that 66 is a generator of 𝔽13×{\mathbb{F}}_{13}^{\times}. We first compute the orbit of the irreducible representations of SS under the action of OutG(S)\operatorname{Out}\nolimits_{G}(S), which are

{1,Z,R,T}\{1,Z,R,T\}

where

R\displaystyle R =n=112(z2)nx+n=112(z2)nxy+n=112(z2)nxy3+n=112(z2)nxy9+n=112(z2)nxy7+n=112(z2)nxy8\displaystyle=\sum_{n=1}^{12}(z^{2})^{n}x+\sum_{n=1}^{12}(z^{2})^{n}xy+\sum_{n=1}^{12}(z^{2})^{n}xy^{3}+\sum_{n=1}^{12}(z^{2})^{n}xy^{9}+\sum_{n=1}^{12}(z^{2})^{n}xy^{7}+\sum_{n=1}^{12}(z^{2})^{n}xy^{8}
+n=112(z2)nxy11+n=112(z2)ny\displaystyle+\sum_{n=1}^{12}(z^{2})^{n}xy^{11}+\sum_{n=1}^{12}(z^{2})^{n}y
T=n=112(z2)nxy2+n=112(z2)nxy10+n=112(z2)nxy6+n=112(z2)nxy4+n=112(z2)nxy5+n=112(z2)nxy12T=\sum_{n=1}^{12}(z^{2})^{n}xy^{2}+\sum_{n=1}^{12}(z^{2})^{n}xy^{10}+\sum_{n=1}^{12}(z^{2})^{n}xy^{6}+\sum_{n=1}^{12}(z^{2})^{n}xy^{4}+\sum_{n=1}^{12}(z^{2})^{n}xy^{5}+\sum_{n=1}^{12}(z^{2})^{n}xy^{12}

We obtained these orbits RR and TT as follows. It is clear that the action of z2z^{2} is free on elements of the form xiyjx^{i}y^{j}. On the other hand, the actions of zz and rr allows us to obtain from xx at least the following elements.

x\textstyle{x\ignorespaces\ignorespaces\ignorespaces\ignorespaces}z\scriptstyle{z}x2y2\textstyle{x^{2}y^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}x6y5\textstyle{x^{6}y^{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}z\scriptstyle{z}x4y2\textstyle{x^{4}y^{2}}x5y6\textstyle{x^{5}y^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}z\scriptstyle{z}x3y11\textstyle{x^{3}y^{11}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}x9y8\textstyle{x^{9}y^{8}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}z\scriptstyle{z}y10\textstyle{y^{10}}

It is easy to see that these elements lie in the z2z^{2}-orbits of xyxy, xy3xy^{3}, xy9xy^{9}, xy7xy^{7}, xy8xy^{8}, xy11xy^{11} and yy. Therefore the orbit of xx has at least 9696 elements. On the other hand, 11, rz4=(1003)rz^{4}=\left(\begin{array}[]{cc}1&0\\ 0&3\end{array}\right) and (rz4)2(rz^{4})^{2} fix xx, hence the orbit has at most 9696 elements. This shows that RR is the orbit of xx. To find the orbit of xy2xy^{2}, note that we can obtain the following elements using the actions of zz and rr.

xy2\textstyle{xy^{2}\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}z\scriptstyle{z}r\scriptstyle{r}x7y5\textstyle{x^{7}y^{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}x3y5\textstyle{x^{3}y^{5}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}x8y6\textstyle{x^{8}y^{6}\ignorespaces\ignorespaces\ignorespaces\ignorespaces}r\scriptstyle{r}x9y6\textstyle{x^{9}y^{6}}x11y2\textstyle{x^{11}y^{2}}

Again, it is easy to see that these elements lie in the z2z^{2}-orbits of xy10xy^{10}, xy6xy^{6}, xy4xy^{4}, xy5xy^{5} and xy12xy^{12}. Therefore the orbit of xy2xy^{2} has at least 7272 elements. But there are only 168168 elements of the form xiyjx^{i}y^{j} with (i,j)(0,0)(i,j)\neq(0,0), hence TT is the orbit of xy2xy^{2}.

Now let ρ=a1+nZ+mR+kT\rho=a1+nZ+mR+kT be a GG–invariant representation of SS. The restriction of ρ\rho to the subgroup V1V_{1} generated by abab and cc is

(a+12k)1+ni=0,j=112xiyj+8mi=112xi+5ki=112xi(a+12k)1+n\sum_{i=0,j=1}^{1}2x^{i}y^{j}+8m\sum_{i=1}^{12}x^{i}+5k\sum_{i=1}^{12}x^{i}

Since V1V_{1} is radical and its group of automorphisms contains SL2(𝔽13)SL_{2}({\mathbb{F}}_{13}), the coefficients of xyxy and xx must be equal, that is

n=6m+5kn=6m+5k

Hence the irreducible GG–invariant representations are 11, 8Z+M8Z+M and 5Z+N5Z+N. Therefore RepG(S)\operatorname{Rep}\nolimits_{G}(S) is a factorial monoid by Proposition 2.11, or alternatively, by using the atlas of finite groups [9] to check that there are two conjugacy classes of elements of order 1313 in SS and Corollary 2.9.

Theorem 3.2.

If |S|p3|S|\leq p^{3} and {\mathcal{F}} is a saturated fusion system over SS, then Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is a factorial monoid.

Proof.

If |S|p2|S|\leq p^{2}, then SS is abelian and {\mathcal{F}} has essential rank zero, hence Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial by Lemma 2.17.

Thus we may assume that the order of SS is exactly p3p^{3}. If p=2p=2, then SS is abelian or isomorphic to D8D_{8} or Q8Q_{8} and we checked that Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial in those cases. If p>2p>2, then SS is either abelian or generalized extraspecial. If SS is isomorphic to p+1+2p^{1+2}_{+}, we showed before the proposition that for any saturated fusion system {\mathcal{F}} over SS, the monoid Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial. In SS is not isomorphic to p+1+2p^{1+2}_{+}, then the essential rank of {\mathcal{F}} is zero by Theorem 4.2 in [25], hence Rep()\operatorname{Rep}\nolimits({\mathcal{F}}) is factorial by Lemma 2.17. ∎

Note that this result can not be extended to order p4p^{4}, as the non-factorial examples in [12] and [23] are given over Sylows of order 343^{4} and 242^{4}, respectively.

4. The half-factoriality property

In this section we give an example of a monoid of fusion-invariant representations for which uniqueness of factorization fails in such a way that not even the length of an element is well defined.

Let MM be an atomic monoid. Given a factorization of an element

x=i=0kaix=\sum_{i=0}^{k}a_{i}

as a sum of atoms, we say that kk is the length of the factorization. Following the terminology in [13], but keeping in mind that all our monoids are reduced, we consider half-factorial monoids.

Definition 4.1.

Let MM be an atomic monoid. We say that MM is half-factorial if for each element xMx\in M, every factorization of xx has the same length.

Note that different elements may have factorizations of different lengths. The following proposition is inspired by Proposition 5.4 in [15]. Given an atomic monoid MM, consider the subset Conv(M)\operatorname{Conv}\nolimits(M) of elements of K(M)K(M) which can be expressed in the form

niai\sum n_{i}a_{i}

where the aia_{i} are atoms and ni0\sum n_{i}\neq 0.

Lemma 4.2.

Let MM be an atomic monoid. The following conditions are equivalent.

  1. (1)

    MM is half-factorial.

  2. (2)

    0Conv(M)0\notin\operatorname{Conv}\nolimits(M).

  3. (3)

    Conv(M)K(M)\operatorname{Conv}\nolimits(M)\neq K(M).

Proof.

Assume that MM is not half-factorial. Then there exists xMx\in M that can be expressed as a sum of atoms in two different ways

x=niai=mibix=\sum n_{i}a_{i}=\sum m_{i}b_{i}

with nimi\sum n_{i}\neq\sum m_{i}. Then in K(M)K(M) we have

0=niaimibi0=\sum n_{i}a_{i}-\sum m_{i}b_{i}

and so 0Conv(M)0\in\operatorname{Conv}\nolimits(M).

Assume now that 0Conv(M)0\in\operatorname{Conv}\nolimits(M) and let xK(M)x\in K(M). We must have x=niaix=\sum n_{i}a_{i} where aia_{i} are atoms. If ni0\sum n_{i}\neq 0, then xConv(M)x\in\operatorname{Conv}\nolimits(M). If ni=0\sum n_{i}=0, then since 0Conv(M)0\in\operatorname{Conv}\nolimits(M) we have 0=mibi0=\sum m_{i}b_{i} with mi0\sum m_{i}\neq 0 and therefore

x=x+0=niai+mibix=x+0=\sum n_{i}a_{i}+\sum m_{i}b_{i}

from where xConv(M)x\in\operatorname{Conv}\nolimits(M).

Finally, assume Conv(M)=K(M)\operatorname{Conv}\nolimits(M)=K(M). Since 0Conv(M)0\in\operatorname{Conv}\nolimits(M), we have 0=niai0=\sum n_{i}a_{i} with ni0\sum n_{i}\neq 0. Then

ni>0niai=ni<0(ni)ai\sum_{n_{i}>0}n_{i}a_{i}=\sum_{n_{i}<0}(-n_{i})a_{i}

and note that ni>0ni\sum_{n_{i}>0}n_{i} can not equal ni<0(ni)\sum_{n_{i}<0}(-n_{i}) since ni0\sum n_{i}\neq 0. Hence MM is not half-factorial. ∎

Example 4.3.

We saw in Example 2.5 that RepΣ9(S)\operatorname{Rep}\nolimits_{\Sigma_{9}}(S) is not factorial when SS is a 33-Sylow subgroup of Σ9\Sigma_{9}. But we shall see that it is half-factorial. In Example 2.10 we proved that the atoms of this monoid are given by 11, PP, QQ^{\prime}, RR, SS and PS+QP-S+Q^{\prime}. Let

0=a1+bP+cQ+dR+eS+f(PS+Q)=a1+(b+f)P+(c+f)Q+dR+(ef)S0=a1+bP+cQ^{\prime}+dR+eS+f(P-S+Q^{\prime})=a1+(b+f)P+(c+f)Q^{\prime}+dR+(e-f)S

Since RΣ9(S)R_{\Sigma_{9}}(S) is free abelian with basis {1,P,Q,R,S}\{1,P,Q^{\prime},R,S\}, we obtain a=b+f=c+f=d=ef=0a=b+f=c+f=d=e-f=0 and therefore

a+b+c+d+e+f=a+(b+f)+(c+f)+d+(ef)=0a+b+c+d+e+f=a+(b+f)+(c+f)+d+(e-f)=0

Hence 0Conv(RepΣ9(S))0\notin\operatorname{Conv}\nolimits(\operatorname{Rep}\nolimits_{\Sigma_{9}}(S)) and so RepΣ9(S)\operatorname{Rep}\nolimits_{\Sigma_{9}}(S) is half-factorial by Proposition 4.2.

The following proposition is inspired by this example.

Proposition 4.4.

Let MM be an atomic monoid such that K(M)K(M) is free abelian of finite rank. If there exists a basis of K(M)K(M) such that every atom of MM can be expressed as a convex integral linear combination of the elements of this basis, then MM is half-factorial.

Proof.

Let {b1,,bk}\{b_{1},\ldots,b_{k}\} be a basis of K(M)K(M) fulfilling the condition in the statement. For each atom aa of MM, there exists a convex integral linear combination

a=λj(a)bja=\sum\lambda_{j}(a)b_{j}

Assume we have 0=naa0=\sum n_{a}a, where the sum runs over the set of atoms of MM.

0=anaa=ana(jλj(a)bj)=j(anaλj(a))bj0=\sum_{a}n_{a}a=\sum_{a}n_{a}\left(\sum_{j}\lambda_{j}(a)b_{j}\right)=\sum_{j}\left(\sum_{a}n_{a}\lambda_{j}(a)\right)b_{j}

from where anaλj(a)=0\sum_{a}n_{a}\lambda_{j}(a)=0 for each jj. But then

ana=a(jλj(a))na=j(anaλj(a))=0\sum_{a}n_{a}=\sum_{a}\left(\sum_{j}\lambda_{j}(a)\right)n_{a}=\sum_{j}\left(\sum_{a}n_{a}\lambda_{j}(a)\right)=0

This shows that 0Conv(M)0\notin\operatorname{Conv}\nolimits(M), hence MM is half-factorial. ∎

Let us show that RepΣ6(S)\operatorname{Rep}\nolimits_{\Sigma_{6}}(S) is not half-factorial when SS is a 22-Sylow subgroup of Σ6\Sigma_{6}. We choose SS to be the image of the composition D8×/2Σ4×/2Σ6D_{8}\times{\mathbb{Z}}/2\to\Sigma_{4}\times{\mathbb{Z}}/2\to\Sigma_{6}, where the first map is induced by the action of Σ4\Sigma_{4} on the vertices of a square and the second map is induced by the standard inclusion of Σ4\Sigma_{4} in Σ6\Sigma_{6} and the element (5,6)(5,6). If β\beta is the sign representation of /2{\mathbb{Z}}/2, the irreducible representations of SS are of the form ρ1\rho\otimes 1 or ρβ\rho\otimes\beta, where ρ{1,X,Y,XY,Z}\rho\in\{1,X,Y,XY,Z\} is an irreducible representation of D8D_{8}. We name the latter ρi\rho^{\prime}_{i} according to the order in which they appeared in the previous line. We define for 1i101\leq i\leq 10 the representations

ρi={ρi1 if i5ρi5β if i>5\rho_{i}=\left\{\begin{array}[]{ll}\rho^{\prime}_{i}\otimes 1&\text{ if }i\leq 5\\ \rho^{\prime}_{i-5}\otimes\beta&\text{ if }i>5\end{array}\right.

The group SS has ten conjugacy classes and six Σ6\Sigma_{6}–conjugacy classes. Using the character criterion for a representation to be Σ6\Sigma_{6}–invariant, it is easy to show that

RΣ6(S)={ρ1,ρ2+ρ4+ρ5+ρ7+ρ10,ρ3+ρ5+ρ7+ρ10,ρ4+ρ6ρ7ρ10,ρ4+ρ8+ρ10,ρ9}R_{\Sigma_{6}}(S)={\mathbb{Z}}\{\rho_{1},\rho_{2}+\rho_{4}+\rho_{5}+\rho_{7}+\rho_{10},\rho_{3}+\rho_{5}+\rho_{7}+\rho_{10},-\rho_{4}+\rho_{6}-\rho_{7}-\rho_{10},\rho_{4}+\rho_{8}+\rho_{10},\rho_{9}\}

But a more convenient basis is given by

RΣ6(S)={ρ1,ρ2+ρ4+ρ5+ρ7+ρ10,ρ3+ρ5+ρ7+ρ10,ρ2+ρ5+ρ6,ρ4+ρ8+ρ10,ρ9}R_{\Sigma_{6}}(S)={\mathbb{Z}}\{\rho_{1},\rho_{2}+\rho_{4}+\rho_{5}+\rho_{7}+\rho_{10},\rho_{3}+\rho_{5}+\rho_{7}+\rho_{10},\rho_{2}+\rho_{5}+\rho_{6},\rho_{4}+\rho_{8}+\rho_{10},\rho_{9}\}

Let us call these representations αj\alpha_{j} for j=1,,6j=1,\ldots,6. Note that αj\alpha_{j} with j2j\neq 2 are irreducible Σ6\Sigma_{6}–invariant representations by Remark 2.12. Let α\alpha be an irreducible Σ6\Sigma_{6}–invariant representation of SS. Then

α=niαi\alpha=\sum n_{i}\alpha_{i}

If all the nin_{i} are nonnegative, then α\alpha must be one of the αj\alpha_{j}. On the other hand, the only coefficient that could be negative is n2n_{2} since each of the remaining αi\alpha_{i} contains an irreducible representation of SS which is not shared with any other αk\alpha_{k}. Assume then n2<0n_{2}<0 and ni0n_{i}\geq 0 for all i2i\neq 2. Then

α=n1α1+(n2+n3)α3+(n2+n4)α4+(n2+n5)α5+(n2)(α3+α4+α5α2)+n6α6\alpha=n_{1}\alpha_{1}+(n_{2}+n_{3})\alpha_{3}+(n_{2}+n_{4})\alpha_{4}+(n_{2}+n_{5})\alpha_{5}+(-n_{2})(\alpha_{3}+\alpha_{4}+\alpha_{5}-\alpha_{2})+n_{6}\alpha_{6}

Note that

α3+α4+α5α2=ρ3+ρ5+ρ6+ρ8+ρ10\alpha_{3}+\alpha_{4}+\alpha_{5}-\alpha_{2}=\rho_{3}+\rho_{5}+\rho_{6}+\rho_{8}+\rho_{10}

Let α7=ρ3+ρ5+ρ6+ρ8+ρ10\alpha_{7}=\rho_{3}+\rho_{5}+\rho_{6}+\rho_{8}+\rho_{10}, which is Σ6\Sigma_{6}–invariant by the previous equation. Since α\alpha is irreducible, only one coefficient must be one and the rest must be zero. Apart from the previous possibilities, we also obtain α=α7\alpha=\alpha_{7}. Consider now the basis

{α1,α2,α3,α5,α6,α7}\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{5},\alpha_{6},\alpha_{7}\}

of RΣ6(S)R_{\Sigma_{6}}(S). Note that α2\alpha_{2} and α7\alpha_{7} are the only elements of this basis which contain ρ2\rho_{2} and ρ6\rho_{6}, respectively, hence they are irreducible by Remark 2.12. Therefore RepΣ6(S)\operatorname{Rep}\nolimits_{\Sigma_{6}}(S) is not factorial by Corollary 2.9. Moreover

α3+α4+α5=α2+α7\alpha_{3}+\alpha_{4}+\alpha_{5}=\alpha_{2}+\alpha_{7}

thus RepΣ6(S)\operatorname{Rep}\nolimits_{\Sigma_{6}}(S) is not half-factorial.

Appendix A GAP algorithms for fusion-invariant representations

We include here the algorithms related to fusion-invariant representations that we implemented in GAP for fusion systems of the form S(G){\mathcal{F}}_{S}(G), where SS is a pp-Sylow subgroup of the finite group GG.

  • (1)

    The pattern of GG–fusion of conjugacy classes of SS.

We take the list of SS–conjugacy classes of elements of SS and check which of them fuse to form a single GG–conjugacy class of elements of SS. The output is a list of positive integers with the same size as the list of SS–conjugacy classes of SS such that the integers in two positions coincide if and only if the SS–conjugacy classes in those positions are GG–conjugate. If GG is a symmetric group, we check GG–conjugacy using the cycle structure of elements.

It is easy to obtain from this process the number of GG–conjugacy classes of elements of SS and we add it to the output as an additional element in the list.

  • (2)

    Whether a representation of SS is GG–invariant

Using the pattern of GG–fusion of conjugacy classes of SS, we check for each two classes that were fused whether the characters coincide on these classes. If at some point this is different from zero, we stop and return false. Otherwise it returns true. We have two versions of this algorithm depending on whether the representation is given in terms of the irreducible representations of SS or in terms of the NG(S)N_{G}(S)–invariant representations of SS.

  • (3)

    Whether RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial, assuming Conjecture 2.19 holds.

We compare the number of GG–conjugacy classes of elements of SS, which we can obtain from the pattern of GG–fusion of conjugacy classes of SS, with the number of irreducible GG–invariant representations of SS which are subrepresentations of the regular representations.

In the first version, we run over the list of subrepresentations of the regular representation of SS. We build a list of irreducible GG–invariant representations as follows. If the representation is a subrepresentation of one in the list or a subrepresentation of the complement of a representation in the list in the regular representation, we ignore it. Otherwise, we add it to the list if it is GG–invariant, or ignore it otherwise. If at some point this list contains more elements that GG–conjugacy classes of elements of SS, the function throws false, meaning that RepG(S)\operatorname{Rep}\nolimits_{G}(S) is not factorial. Otherwise, it returns true, which would mean that RepG(S)\operatorname{Rep}\nolimits_{G}(S) is factorial if Conjecture 2.19 holds.

In the second version, we run over the list of those subrepresentations which are NG(S)N_{G}(S)–invariant, which must be determined previously. In the third version, only for symmetric groups, we use the same list as in the second version, but using the pattern of GG–fusion of conjugacy classes that uses the cycle structure of elements.

  • (4)

    The list of irreducible GG–invariant representations of SS which are subrepresentations of the regular representation.

It follows the same algorithm as the previous function, but it does not stop until it checks all the subrepresentations of the regular representation of SS in the first version, and all the subrepresentations which are NG(S)N_{G}(S)–invariant in the second and third versions. It returns the list of irreducible GG–invariant representations of SS in terms of the irreducible representations of SS.

  1. (5)

    A basis for the complexification of the representation ring of GG–invariant representations of SS.

Using the pattern of GG–fusion of conjugacy classes of elements of SS (or the pattern of cycle structure of elements if GG is a symmetric group), we find pairs of elements of SS which determine all GG–conjugacy relations between all the elements of SS. Then we create a matrix that has a row for each of these pairs. The jjth element of the row that corresponds to (s1,s2)(s_{1},s_{2}) contains the element χj(s1)χj(s2)\chi_{j}(s_{1})-\chi_{j}(s_{2}), where χj\chi_{j} is the character of the jjth irreducible representation of SS in the order established by GAP.

Finally we compute the integral nullspace of this matrix. We are regarding an element (a1,,ar)(a_{1},\ldots,a_{r}) in the nullspace as the virtual character a1χ1++arχra_{1}\chi_{1}+\ldots+a_{r}\chi_{r}.

Remark A.1.

Our algorithms to find whether RepG(S)\operatorname{Rep}\nolimits_{G}(S) and the list of irreducible GG–invariant representations of SS may require a lot of computing time.. An alternative method to answer these two questions is to find the complexification of the representation ring as a vector space in GAP using the algorithm described above, then find the representation ring as an abelian group from this object, and finally follow the same process we used in Section 4 to find the irreducible Σ6\Sigma_{6}–invariant representations of a 22-Sylow of Σ6\Sigma_{6}. We illustrate this in the following example.

Example A.2.

Let SS be a 33-Sylow subgroup of A9A_{9}. Using GAP we find that a basis for RA9(S)R_{A_{9}}(S)\otimes{\mathbb{C}} is

X={\displaystyle X=\{ ρ1,ρ2ρ3ρ4+ρ6ρ7+ρ8,ρ2ρ3ρ4+ρ5ρ7+ρ9,ρ2+ρ3+ρ12+ρ15,\displaystyle\rho_{1},-\rho_{2}-\rho_{3}-\rho_{4}+\rho_{6}-\rho_{7}+\rho_{8},-\rho_{2}-\rho_{3}-\rho_{4}+\rho_{5}-\rho_{7}+\rho_{9},\rho_{2}+\rho_{3}+\rho_{12}+\rho_{15},
ρ2ρ3+ρ10+ρ11+ρ14+ρ16,ρ2+ρ3+ρ4+ρ7+ρ13+ρ17}\displaystyle-\rho_{2}-\rho_{3}+\rho_{10}+\rho_{11}+\rho_{14}+\rho_{16},\rho_{2}+\rho_{3}+\rho_{4}+\rho_{7}+\rho_{13}+\rho_{17}\}

where ρi\rho_{i} are the irreducible representations in the order presented by GAP. We see that there are six A9A_{9}–conjugacy classes of elements of SS. It is easy to check that RA9(S)=XR_{A_{9}}(S)={\mathbb{Z}}X. We can also find a basis of representations, namely

{\displaystyle\{ ρ1,ρ6+ρ8+ρ13+ρ17,ρ5+ρ9+ρ13+ρ17,ρ2+ρ3+ρ12+ρ15,\displaystyle\rho_{1},\rho_{6}+\rho_{8}+\rho_{13}+\rho_{17},\rho_{5}+\rho_{9}+\rho_{13}+\rho_{17},\rho_{2}+\rho_{3}+\rho_{12}+\rho_{15},
ρ4+ρ7+ρ10+ρ11+ρ13+ρ14+ρ16+ρ17,ρ2+ρ3+ρ4+ρ7+ρ13+ρ17}\displaystyle\rho_{4}+\rho_{7}+\rho_{10}+\rho_{11}+\rho_{13}+\rho_{14}+\rho_{16}+\rho_{17},\rho_{2}+\rho_{3}+\rho_{4}+\rho_{7}+\rho_{13}+\rho_{17}\}

Let αj\alpha_{j} be the jjth element of this basis. By Remark 2.12, the A9A_{9}–invariant representations αj\alpha_{j} with 1j51\leq j\leq 5 are irreducible. Let α\alpha be an irreducible A9A_{9}–invariant representation of SS expressed as α=niαi\alpha=\sum n_{i}\alpha_{i} in RA9(S)R_{A_{9}}(S). The only nin_{i} that could be negative is n6n_{6} and we can rewrite it as:

α=n1α1+n2α2+n3α3+(n4+n6)α4+(n5+n6)α5+(n6)(α6+α5+α4)\alpha=n_{1}\alpha_{1}+n_{2}\alpha_{2}+n_{3}\alpha_{3}+(n_{4}+n_{6})\alpha_{4}+(n_{5}+n_{6})\alpha_{5}+(-n_{6})(-\alpha_{6}+\alpha_{5}+\alpha_{4})

Let α7=α6+α5+α4\alpha_{7}=-\alpha_{6}+\alpha_{5}+\alpha_{4}. Note that

α7=ρ10+ρ11+ρ12+ρ14+ρ15+ρ16\alpha_{7}=\rho_{10}+\rho_{11}+\rho_{12}+\rho_{14}+\rho_{15}+\rho_{16}

Hence α7\alpha_{7} is a new possibility for α\alpha. Indeed α7\alpha_{7} and α6\alpha_{6} are both irreducible by Remark 2.12 since {α1,α2,α3,α5,α6,α7}\{\alpha_{1},\alpha_{2},\alpha_{3},\alpha_{5},\alpha_{6},\alpha_{7}\} is also a basis of RA9(S)R_{A_{9}}(S). We have found that there are seven irreducible A9A_{9}–invariant representations of SS, hence RepA9(S)\operatorname{Rep}\nolimits_{A_{9}}(S) is not factorial by Corollary 2.9.

Using the algorithms above, we found new examples of non-factorial monoids of the form RepG(S)\operatorname{Rep}\nolimits_{G}(S), which we present in the following table together with the examples from [12] and [23]. We applied them to groups with Sylows of order 242^{4}, 252^{5} and 343^{4} corresponding to the fusion systems in Propositions 4.3 and 4.4 of [1], to some of the fusion systems in Tables 1 and 2 of Chapter 7 of [21], and other classical groups. The column “Irr. GG–invariant reps.” refers to irreducible GG–invariant representations of SS which are subrepresentations of the regular representation. The group (/3×/9)/3({\mathbb{Z}}/3\times{\mathbb{Z}}/9)\rtimes{\mathbb{Z}}/3 in the table is the group from the SmallGroups library of GAP with id [81,9][81,9], which is the semidirect product coming from the action of /3{\mathbb{Z}}/3 determined by the matrix (1001)\left(\begin{array}[]{cc}1&0\\ 0&1\end{array}\right).

Group GG Prime pp pp-Sylow SS GG–conj. classes of SS Irr. GG–invariant reps.
Σ6\Sigma_{6} 22 /2×D8{\mathbb{Z}}/2\times D_{8} 66 77
PSL2(𝔽17)PSL_{2}({\mathbb{F}}_{17}) 22 D16D_{16} 55 77
PSU3(𝔽5)PSU_{3}({\mathbb{F}}_{5}) 22 SD16SD_{16} 55 88
M10M_{10} 22 SD16SD_{16} 66 88
SL3(𝔽3)SL_{3}({\mathbb{F}}_{3}) 22 SD16SD_{16} 55 88
PSL3(𝔽5)PSL_{3}({\mathbb{F}}_{5}) 22 /4/2{\mathbb{Z}}/4\wr{\mathbb{Z}}/2 77 88
PSL2(𝔽31)PSL_{2}({\mathbb{F}}_{31}) 22 D32D_{32} 99 2121
PSU3(𝔽9)PSU_{3}({\mathbb{F}}_{9}) 22 SD32SD_{32} 99 5353
GL3(𝔽3)GL_{3}({\mathbb{F}}_{3}) 22 /2×SD16{\mathbb{Z}}/2\times SD_{16} 1010 1616
Σ8\Sigma_{8} 22 D8/2D_{8}\wr{\mathbb{Z}}/2 1010 ?
Σ9\Sigma_{9} 33 /3/3{\mathbb{Z}}/3\wr{\mathbb{Z}}/3 55 66
A9A_{9} 33 /3/3{\mathbb{Z}}/3\wr{\mathbb{Z}}/3 66 77
PSL3(𝔽19)PSL_{3}({\mathbb{F}}_{19}) 33 (/3×/9)/3({\mathbb{Z}}/3\times{\mathbb{Z}}/9)\rtimes{\mathbb{Z}}/3 77 1616

There are other examples of finite groups where RepG(S)\operatorname{Rep}\nolimits_{G}(S) is not factorial aside from these. For instance, if RepG(S)\operatorname{Rep}\nolimits_{G}(S) is not factorial and HH is any other finite group with pp-Sylow subgroup RR, then RepG×H(S×R)\operatorname{Rep}\nolimits_{G\times H}(S\times R) is not factorial. Another instance comes from finite groups with the same fusion systems (by Remark 2.6), for example the fusion systems of SL3(𝔽3)SL_{3}({\mathbb{F}}_{3}) and M11M_{11} at the prime 22 are isomorphic [20], hence RepM11(S)\operatorname{Rep}\nolimits_{M_{11}}(S) is not factorial when SS is a 22-Sylow subgroup of M11M_{11}.

We also found using these functions that there is uniqueness of factorization for PSp4(𝔽3)PSp_{4}({\mathbb{F}}_{3}) and PSL4(𝔽7)PSL_{4}({\mathbb{F}}_{7}) at the prime 33 and for A8A_{8} at the prime 22.

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