UNIVERSIDADE FEDERAL DE MINAS GERAIS INSTITUTO DE CIENCIASˆ EXATAS DEPARTAMENTO DE MATEMATICA´

Frederico Augusto Menezes Ribeiro

Finite p-groups with three characteristic subgroups and a class of simple anticommutative algebras

BELO HORIZONTE October, 2016 Frederico Augusto Menezes Ribeiro

Finite p-groups with three characteristic subgroups and a class of simple anticommutative algebras

Thesis submitted to Universidade Federal de Minas Gerais for the degree of Doctor of Philosophy in Mathematics

Advisor: Csaba Schneider

Belo Horizonte October, 2016 Abstract

The structure of finite characteristically simple groups is very well known. One might then ask what happens when a group has few characteristic subgroups. UCS groups are one step from characteristically simple groups into the direction of more general groups: UCS groups are groups that have precisely three characteristic sub- groups. These groups however are way more complicated than characteristically simple groups. In this work we take a deeper look at a subclass of UCS groups, and how this subclass is related to the class of irreducible anticommutative algebras (IAC).

We prove that there is a bijection between the set of isomorphism classes of finite- dimensional IAC algebras over the field with p elements and the set of isomorphism classes of finite UCS p-groups of exponent p2. From this bijection we show how to derive some properties of the group by looking at the corresponding algebra, showing for example that a group has proper nontrivial powerfully embeded subgroups if and only if the corresponding algebra has proper nontrivial ideals.

We also investigate the structure of the IAC algebras showing that they are always semisimple (and actually always a direct sum of copies of a simple IAC algebra).

In the last chapter we provide, by using three different methods, examples of infinite families of simple IAC algebras.

i Abstract

A estrutura de grupos simples caracter´ısticamente simples ´ebastante bem conhecida. E´ natural ent˜aoperguntar-se o que ocorre quando um grupo tem poucos subgru- pos caracter´ısticos. Grupos UCS est˜aoa um passo dos grupos caracter´ısticamente simples na dire¸c˜aode grupos mais gerais: Grupos UCS s˜aoos grupos que possuem precisamente trˆessubgrupos caracter´ısticos. Estes grupos por´ems˜aobem mais com- plicados do que grupos caracter´ısticamente simples. Neste trabalho observamos mais precisamente uma subclasse dos grupos UCS, e como esta subclasse est´arelacionada com a classe de ´algebrasirredut´ıveis e anticomutativas (IAC).

N´osprovamos que existe uma bije¸c˜aoentre o conjunto das classes de isomorfismo de ´algebrasIAC de dimens˜aofinita sobre o corpo como p elementos e o conjunto de classes de isomorfismo de p-grupos finitos UCS de expoente p2. A partir desta bije¸c˜aon´osmostramos como obter algumas propriedades do grupo ao olhar para a ´algebracorrespondente, mostrando por exemplo que um grupo tem subgrupos powerfully embeded n˜aotriviais pr´opriosse e somente se a ´algebracorrespondente tem ideais n˜aotriviais pr´oprios.

Tamb´eminvestigamos a estrutura das ´algebraIAC, mostrando que elas s˜aosem- pre semissimples (e na verdade sempre s˜aosoma direta de c´opiasde uma ´algebra IAC simples).

No ´ultimo cap´ıtulon´osapresentamos, usando trˆesm´etodos distintos, exemplos e fam´ıliasinfinitas de ´algebrasIAC simples.

ii Acknowledgements

To my parents for everything.

To my brother for the inspiration.

To my advisor Csaba Schneider, and to Stephen Glasby for the patience.

To my family and friends for the support.

To FAPEMIG and Capes for the financial support.

iii Contents

Abstracti

Abstract ii

Acknowledgements iii

1 Preliminaries1

1.1 Introduction...... 1

1.2 Groups...... 4

1.3 Finite p-groups...... 7

1.4 UCS groups...... 13

1.5 Earlier results about UCS groups...... 14

1.6 Tensor and exterior products...... 18

1.7 Wreath Product...... 19

2 UCS p-groups and IAC algebras 22

2.1 Bijection between UCS groups and IAC algebras...... 22

2.2 Automorphisms of UCS groups...... 30

2.3 ESQ modules...... 32

iv 3 Structure of IAC algebras 34

3.1 Direct powers of UCS p-groups...... 34

3.2 Direct sums of IAC algebras...... 35

3.3 Normal subgroups and ideals...... 38

3.4 nCS groups...... 40

3.5 Abelian nCS Groups...... 42

4 Examples 46

4.1 IAC Algebras of dimension 3...... 46

4.2 A family of IAC algebras with an irreducible cyclic group...... 49

4.3 Another family of IAC algebras...... 52

4.4 The Clebsch-Gordan formula for the exterior product...... 56

v Chapter 1

Preliminaries

1.1 Introduction

Sylow’s theorem states that a finite group always contains p-subgroups of order equal to the largest power of p dividing the order of the group. The structure of these subgroups does have an impact on the structure of the bigger group. Thus understanding the theory of p-groups helps us with the general theory of groups. One problem that arises with this approach is that the number of p-groups is very large. For example, there are 56092 groups of order 28, 10494213 groups of order 29, and 49487365422 groups of order 210 (see [2, Table 1]). It is natural to study subsets of p-groups with some property (see for example [3] which studies p-groups of maximal nilpotency class for their order).

One method used to study not only p-groups, but groups in general is considering their groups of automorphisms, see for example [13] where by considering a subgroup of Aut(G) the authors of the paper determine an upper bound for the exponent of G. One might then ask what happens to groups where the automorphisms do not leave any non-trivial proper subgroups fixed, that is, characteristically simple groups. A finite group is characteristically simple if and only if it is the direct product of copies of a simple group.

A group G is said to be UCS (Unique Characteristic Subgroup) if it has precisely three characteristic subgroups. Since the subgroups 1 and G are always character- istic, a group G is UCS when it has exactly one nontrivial proper characteristic subgroup. Taunt in [26, page 1] says “It was suggested to the author by Prof. Hall that finite groups possessing exactly one proper characteristic subgroup would repay attention”.

1 Taunt in [26] divided the study of UCS groups between solvable and nonsolvable groups, and noticed that finite solvable UCS groups have order divisible by at most two different primes, whereas for finite nonsolvable UCS groups the order is always divisible by at least three primes. In that paper he focused on finite UCS groups whose order is divisible by two primes. He promised a following paper dealing with UCS p-groups, but never delivered. In [8] Glasby, P´alfyand Schneider studied finite UCS p-groups, observing that this study could be done by investigating the Aut(G)-modules Φ(G) and G/Φ(G) where Φ(G) denotes the Frattini subgroup of G (See [8, Theorem 5]). They divided the class of these groups into the groups of exponent p, and the groups of exponent p2, using distinct methods in each case. For groups of exponent p2, the quotient V = G/Φ(G) has the property that the exterior product V ∧V contains a submodule U such that (V ∧ V )/U is isomorphic to V . They called such a module an ESQ (Exterior Self Quotient) module. The study in [8, Section 7] of the UCS p-groups of exponent p2 was done through the investigation of finite groups that admit an irreducible ESQ module of low dimension. In this work we will deal with these ESQ modules, with a natural product that comes from their ESQ structure, as nonassociative algebras whose group of auto- morphisms acts irreducibly on them. We call these algebras IAC. IAC algebras are not necessarily Lie, Malcev, alternative or Jordan algebras. We show the following theorem.

Theorem 1.1.1 There is a bijection G from the set of isomorphism classes of IAC algebras of finite dimension over Fp to the set of isomorphism classes of finite UCS p-groups of exponent p2.

We denote the inverse of the bijection G by L. Given a UCS p-group G of exponent p2, we consider the structure of L(G) to discover some properties of G. We prove the following theorem that shows that the existence of powerfully embeded proper non-trivial subgroups in G is linked to the existence of ideals in L(G).

Theorem 1.1.2 Let G be a finite UCS p-group of exponent p2 and L = L(G). Then there exists a bijection β between the set of powerful subgroups of G that contain Φ(G) and the set of subalgebras of L. Further, the image by β of the set of powerfully embedded subgroups of G containing Φ(G) is the set of ideals of L.

The idea of considering another algebraic structure to learn about a group is not new. We have for example the relation between groups and Lie rings which Khukhro

2 described in his book [12, page xiv]: “The Lie ring method of solving group-theoretic problems consists of three major steps. First, the problem must be translated into a corresponding problem about Lie rings constructed from the groups. Then the Lie ring problem is solved. The results on Lie rings must then be translated back, into required conclusions about groups.” This is what we will do with UCS p-groups of exponent p2 and IAC algebras.

As was remarked by Schafer in [22, page 6] the study of both Lie and associative algebras is done by finding a radical such that the algebra modulo this radical is semisimple. By following the same procedure for IAC algebras, we have proved in this work the following theorem.

n M Theorem 1.1.3 Let L be an IAC algebra of finite dimension. Then L = S, i=1 where S is a simple IAC algebra.

Thus IAC algebras are always semisimple, and the study of IAC algebras can be reduced to the study of simple IAC algebras.

As nonassociative simple algebras are not classified, classifying IAC algebras is an interesting problem even without considering the UCS groups. By the previous theorem, classifying the simple IAC algebras, leads to classifying all IAC algebras. In this thesis we will construct some examples of simple IAC algebras. One of these examples is constructed in a similar manner as in [5]: We use a variation of the Clebsch-Gordan formula in order to find SL(2, p)-modules V that are isomorphic to quotients of their exterior squares. This way we can define an anticommutative product in V , giving V the structure of a nonassociative algebra. In our particular case we do so in a manner such that V is a G-module for some group G that acts irreducibly on V , so the resulting algebra will have G as a subgroup of its automorphism group, making the algebra IAC. On the other hand in [5] they see V as a module for the Lie algebra sl(2), and conclude that sl(2) acts by derivations on the resulting algebra.

This work is divided into four chapters. In the first, we introduce the nota- tion used throughout this text while presenting results that will be useful in later chapters.

The second chapter covers the bijection of Theorem 1.1.1 between UCS p-groups 2 of exponent p and IAC algebras of finite dimension over Fp. We also make a brief introduction to ESQ modules and show in Section 2.3 how they can actually be perceived as IAC algebras.

3 We investigate the structure of IAC algebras in Chapter 3. We show that the direct sum of copies of a simple IAC algebra is itself IAC (Proposition 3.2.1). We also show that any nonsimple IAC algebra can be written as such a direct sum (Theorem 3.2.4). The fact that direct sums of copies of an IAC algebra is itself IAC is not surprising as Taunt showed in [26, Theorem 2.1] that direct powers of UCS groups under certain conditions are UCS. We then show that this theorem of Taunt is valid not only for UCS groups, but can be extended to groups that have a finite number of characteristic subgroups and they form a chain (Theorem 3.4.3). Finally, to close that chapter we explore what happens to normal subgroups of UCS p-groups of exponent p2, and ideals of IAC algebras (Section 3.3).

In Chapter 4 we explore examples of IAC algebras. In the first section we show that IAC algebras of dimension 3 are actually simple Lie algebras (Theorem 4.1.1). We then close this text by showing some examples of how IAC algebras can be constructed, in the following two sections by a direct manner, and in Section 4.4 by constructing a family of irreducible ESQ modules using a result similar to the Clebsch-Gordan formula (Theorem 4.4.2).

1.2 Groups

In this section we will show some basic definitions and preliminary results about groups.

We use the notation xy = y−1xy for conjugates and [x, y] = x−1y−1xy for commu- tators. We warn the reader that as some authors use the notation [x, y] = xyx−1y−1, the same results may appear differently in different references.

Lemma 1.2.1 The following identities are valid for any elements x, y, z of a group.

1. xy = x[x, y];

2. [y, x] = [x, y]−1;

3. [x, zy] = [x, y][x, z]y and [xz, y] = [x, y]z[z, y];

4. [x, y−1] = [y, x]y−1 and [x−1, y] = [y, x]x−1 ;

5. [[x, y−1], z]y[[y, z−1], x]z[[z, x−1], y]x = 1 and [[x, y], zx][[z, x], yz][[y, z], xy] = 1.

4 6. Let [x, ny] for n ∈ N denote the commutator [... [[x, y], y] . . . , y] where y ap- pears n times.

n n n n [x, y ] = [x, y] [x, 2y](2)[x, 3y](3) ... [x, ny] mod N

where N is the normal closure in G of the subgroup generated by the set of all commutators in {y, [x, y]} of weight at least 2 in [x, y].

Proof For the last item, see [16, Corollary 1.1.7]. The first equation of 5. can be found in [20, 5.1.5], see [16, Proposition 1.1.6] for the rest.

Item 5 of the previous lemma is called the Hall-Witt identity, and it is related to the Jacobi identity in Lie rings.

n Lemma 1.2.2 Let G be a group such that G0 ⊆ Z(G). Then (gh)n = gnhn[h, g](2).

Proof For a proof, see [20, 5.3.5].

Definition 1.2.3 Let G be a group. A subgroup H of G is called characteristic if Hϕ = H for all ϕ ∈ Aut(G). We say that G is characteristically simple if its only characteristic subgroups are {1} and G.

Theorem 1.2.4 1. A direct product of finitely many isomorphic simple groups is characteristically simple.

2. A characteristically simple group which has at least one minimal normal sub- group is a direct product of isomorphic simple groups.

Proof See [20, 3.3.15].

Corollary 1.2.5 If G is a finite group, then G is characteristically simple if and only if G is isomorphic to a direct product of isomorphic simple groups.

Note that since the only finite abelian simple groups are the cyclic groups with prime order, a finite abelian characteristically simple group is elementary abelian.

5 n Y ∼ Lemma 1.2.6 If G is a characteristically simple group and G = Ti, with Ti = T i=1 m Y for all i and T simple, then a normal subgroup N of G is isomorphic to T with i=1 n Y m ≤ n. Further, if G is nonabelian, then N = Si where Si is equal to either Ti i=1 or 1.

n ∼ Y Proof If G is abelian, then G = Cp. As every element of G has order p, and G i=1 m Y is abelian, then every subgroup of G is isomorphic to Cp with some m ≤ n. For i=1 the assertion concerning non-abelian groups, see [20, 3.3.16].

Corollary 1.2.7 A normal subgroup of a finite characteristically simple group is a characteristically simple group.

Proof This follows imediatelly from the previous Lemma and Corollary 1.2.5.

Definition 1.2.8 If G is a group and p is a prime, we can define the following subgroups of G:

• Gp = hxp | x ∈ Gi.

• G0 = h[x, y] | x, y ∈ Gi.

• Φ(G), also known as the Frattini subgroup of G, is the intersection of all maximal subgroups of G if G has maximal subgroups, and equal to G otherwise.

It it is easy to see from the definitions that these three subgroups are always characteristic in G.

6 Definition 1.2.9 An action of a group G on a set Ω is a homomorphism ϕ : G → Sym Ω. When Ω is an algebraic structure such as a group, ring, vector space or algebra, then by an action of G on Ω, we mean a homomorphism ϕ : G → Aut(Ω). We say that a group G acts on Ω if a homomorphism ϕ as above is given. If g ∈ G and ω ∈ Ω, the element ω(gϕ) is said to be the image of ω by g, and will be denoted by ωg.

1.3 Finite p-groups

Throughout this text, p will always denote a prime number. In this work we will need a few results about finite p-groups, so we will begin by introducing them.

Definition 1.3.1 Let p be a positive prime integer. A group G is said to be a p-group if every element of G has order equal to a power of p.

The trivial group is a p-group for every prime p.

The following theorem is a direct consequence of Cauchy’s Lemma.

Theorem 1.3.2 Every finite p-group has order pk for a non negative integer k.

Theorem 1.3.3 If G is a finite p-group, then Φ(G) = GpG0.

Proof See [20, Theorem 5.3.2].

In the case G is a p-group, the quotient by any normal subgroup N ≤ G contains only elements of order p if and only if Gp ≤ N. Further N contains G0 if and only if G/N is abelian. Therefore for p-groups the quotient G/Φ(G) is the largest quotient of G that is elementary abelian.

Any elementary abelian group can be seen as an Fp vector space, where the multiplication by a scalar n is just the repeated product of the element n times.

Theorem 1.3.4 (Burnside’s basis theorem) Let G be a finite p-group, and let Φ(G) be its Frattini subgroup.

7 • A subset S of G is a generating set for G if and only if the image of S in G/Φ(G) generates G/Φ(G) as an Fp vector space. • A subset S of G is a minimal generating set for G if and only if the image of S in G/Φ(G) is a vector space basis for G/Φ(G).

Proof See the proof of [20, Theorem 5.3.2]

Definition 1.3.5 A finite p-group G is called extra-special if G0 = Z(G) and |G0| = p.

Extra-special groups appear frequently in representation theory.

The construction of UCS groups is often done by taking a quotient of a relatively free group; see [8, page 87]. Hence we introduce some base terminology in the theory of varieties. See [19] for more.

Let X = {x1, x2,..., } a countably infinite set and let F∞ denote the free group freely generated by X; see [19, 1.1] or [20, 2.1]. If w is an element of F∞ then w is said to be a word. Suppose that w = w(x1, . . . , xd) is a word in the generators x1, . . . , xd and a1, . . . , ad are elements of a group G. Then we can substitute a1, . . . , ad in w and obtain an element w(a1, . . . , ad) ∈ G. If w(a1, . . . , ad) = 1 for all a1, . . . , ad ∈ G then G is said to satisfy the law w.

Definition 1.3.6 A variety of groups is the class of all groups that satisfy each one of a given set of laws.

Let V be a variety, let Xn = {x1, x2, . . . , xn}, and let Fn denote the free group freely generated by Xn. If we denote by W the set of laws defining V, and by W (Fn) the group generated by all substitutions of elements of Fn in the words of W , then the group Fn/W (Fn) is a group in V. The group Fn/W (Fn) is called the relatively free group in the variety V with rank n.

Theorem 1.3.7 A finitely generated group G belongs to a variety V with a set of laws W if and only if it is isomorphic to a factor of Fn/W (Fn) for some non-negative integer n.

8 Proof See [19, Theorem 14.23].

Consider the groups that have exponent dividing p2, nilpotency class at most 2, and have the property that all p-th powers are central. The class of these groups is the variety defined by the set of laws {xp2 , [[x, y], z], [xp, y]}. Consider then the group Hp,r which we define as the r-generator free group in this variety. In Proposition 1.5.6 we will prove that every finite UCS p-group is a quotient of some Hp,r.

First of all we will prove that Hp,r is a finite p-group. The group Hp,r may be seen as the quotient of the free group Fr with r generators by the subgroup generated p2 p by all the elements of the form x , [[x, y], z] and [x , y] where x, y, z ∈ Fr (see the p2 discussion before Theorem 1.3.7). If we look at the exponent of Hp,r, since h = 1 2 0 for all h ∈ Hp,r, clearly exp(Hp,r) | p . Since Hp,r/Hp,r is abelian, finitely generated (because it is generated by the images of the generators of Hp,r), and has finite 0 0 exponent, we can then conclude Hp,r/Hp,r is finite. Further, since Hp,r is central it is abelian and finitely generated (in fact generated by [x, y] where x and y are 0 elements of a finite generating set of Hp,r) with finite exponent. Thus Hp,r/Hp,r and 0 2 Hp,r are finite which makes Hp,r finite. Since exp(Hp,r) | p , Hp,r is a finite p-group.

Even better, we can calculate the order of Hp,r, though it is a long process, and relies on theory about rewriting, which we will not present here. The proof of Theorem 1.3.8 will use various concepts and results from [23, chapters 2 and 9].

Theorem 1.3.8 Let

Γ = hx1, . . . , xr, y1, . . . , yr, z1,2 . . . z1,r, z2,3, . . . zr−1,r | p p xi = yi; yi = 1; for all 1 ≤ i ≤ r p zi,j = 1; xjxi = xixjzi,j; for all 1 ≤ i < j ≤ r

yixj = xjyi; for all 1 ≤ i, j ≤ r

zi,jxk = xkzi,j; zi,jyk = ykzi,j; for all 1 ≤ k ≤ n and 1 ≤ i < j ≤ r

zi,jzk,l = zk,lzi,j for pairs (i, j) and (k, l) where i ≤ l or i = l and j < ki .

2r+ r Then |Γ| = p (2).

Proof The given presentation for Γ is a polycyclic presentation as defined in [23, Sec- tion 9.4]. By [23, Chapter 9, Proposition 4.1] Γ is a polycyclic group with polycyclic generating sequence given by the generators in the presentation above in the order −1 −1 −1 they appear there. By adding the rules xixi = 1, yiyi = 1 and zijzij = 1 to the presentation above, we have a monoid polycyclic presentation for the group Γ. This

9 monoid presentation gives us a rewriting system with respect to the wreath product order for the polycyclic sequence. We claim that this rewriting system is confluent (defined in [23, Section 2.2]). Once we prove this claim, it follows by [23, Chapter 2, Proposition 2.6] that every word has a unique normal form, i. e., after applying the rewriting process (described in [23, page 52]) we obtain a new word for the same α1 αr β1 βr γ1,2 γr−1,r element of the form x1 . . . xr y1 . . . yr z1,2 . . . zr−1,r with 0 ≤ αi, βi, γi,j ≤ p − 1 for all 1 ≤ i, j ≤ r. As distinct words in normal form represent distinct elements in the group Γ, |Γ| will equal the number of different normal words we can form with 2r+ r its generators, which is p (2). To finish this proof it suffices to show the claim, that is, to show that the rewriting system we obtain is confluent. By [23, Proposition 8.3 of Chapter 9], in order to show that the rewriting system is confluent, we need only to show that collecting the following pairs of words results in the same word in the generators of Γ.

1.( akaj)ai, ak(ajai), i < j < k; p p−1 2.( aj )ai, aj (ajai), i < j;

p p−1 3. aj(ai ), (ajai)ai , i < j; p p 4. ai(ai ), (ai )ai, where ai is the i-th generator in the presentation of Γ, that is, ai = xi for i = 1, . . . , r, ai+r = yi for i = r + 1,... 2r and for i > 2r, ai = zj,k for some j, k = 1, . . . , n. It remains then to show that the result of the collection of these words is inde- pendent on where we began in each of the four overlaps above. We will show this by taking each word and using the relations in the presentation of Γ, using → to denote the use of one relation and →∗ to denote the use of several relations. Also, to highlight where we apply the next relation we underline the elements. Notice also that the generators yi and zi,j of Γ are always in the center whereas xi is not.

We will start now with overlap 1. First suppose that ak and aj are in the center of Γ. In this case, we get

akajai → ajakai → ajaiak → aiajak, and akajai → akaiaj → aiakaj → aiajak, and so the end results are equal. Suppose now that only ak is in the center. We have then aj = xj and ai = xi. Then

akxjxi → xjakxi → xjxiak → xixjzi,jak;

10 akxjxi → akxixjzi,j → xiakxjzi,j → xixjakzi,j.

If ak appears before zi,j in the presentation of Γ, we will need another relation in the first rewriting, whereas if zi,j appears before ak, we will need another relation in the second. If ak = zi,j, we need nothing more, and in all cases we get the same final result on both rewritings. Suppose now that ak = xk, aj = xj and ai = xi.

xkxjxi → xjxkzj,kxi → xjxkxizj,k → xjxixkzi,kzj,k →

→ xixjzi,jxkzi,kzj,k → xixjxkzi,jzi,kzj,k;

xkxjxi → xkxixjzi,j → xixkzi,kxjzi,j → xixkxjzi,kzi,j → xixjxkzj,kzi,kzi,j →

→ xixjxkzi,kzj,kzi,j → xixjxkzi,kzi,jzj,k → xixjxkzi,jzi,kzj,k, and once again we get the same result on both rewritings.

Let us now treat overlap 2. If aj ∈ Z(Γ),

p aj ai → ai;

p−1 ∗ p aj ajai → aiaj → ai.

If aj = xj and ai = xi, then

p xj xi → yjxi → xiyj;

p−1 ∗ p ∗ p p p xj xjxi → xi(xjzi,j) → xixj zi,j → xiyjzi,j → xiyj, which covers overlap 2.

Now, for overlap 3, if ai ∈ Z(Γ), then

p ajai → aj;

p−1 ∗ p ajaiai → ai aj = aj.

If ai = xi and aj ∈ Z(Γ), then p ajxi → ajyi; p−1 ∗ p ajxixi → xi aj → yiaj, and applying one last rewriting rule on one of the rewritings, or none in case aj = yi, makes both results equal. If now ai = xi and aj = xj, then

p xjxi → xjyi;

11 p−1 p−1 ∗ p−1 ∗ p p p xjxixi → xixjzi,jxi → xixjxi zi,j → xi xjzi,j → xi xj → yixj → xjyi.

Finally, on overlap 4, if ai ∈ Z(Γ)

p aiai → ai;

p ai ai → ai, and if ai = xi p xixi → xiyi p xi xi → yixi → xiyi. Thus the theorem is proved.

∼ Proposition 1.3.9 Hp,r = Γ, where Γ is the group defined in Proposition 1.3.8.

Proof Notice first that the group Γ is generated by the elements xi. We claim that we can define a surjective homomorphism from Γ to Hp,r by mapping xi 7→ hi when {h1, . . . , hr} is a minimal generating set for Hp,r. Indeed, the laws defining Hp,r p imply the relations defining Γ with the exception of the ones of the form zi,j = 1, p but these follow from applying item 6 of Lemma 1.2.1 to [hj, hi ] and the fact that all commutators in Hp,r are central. Hence we can define a surjective homomorphism by xi 7→ hi.

Since Hp,r is a free group in the variety of groups that satisfy the laws {xp2 , [[x, y], z], [xp, y]} (See Section 1.3), by Theorem 1.3.7, it suffices to show that Γ also satisfies these laws, that is, Γ has exponent dividing p2, nilpotency class at most 2 and p-th powers are central in Γ.

If a, a1, . . . an ∈ {xi, yi, zi,j} and w = a1a2 . . . an, by item 3 of Lemma 1.2.1 and the fact that [a, an] is always central in Γ, using induction on n we show that [a, w] = [a, an][a, an−1] ... [a, a1]. Thus [a, w] can be written as a product of commutators in the generators of Γ, and is a central element. We can now use induction once again on n and show that [w, g] = [a1, g][a2, g] ... [an, g] for any g ∈ Γ. Thus every commutator in Γ is central, that is Γ satisfies the law [[x, y], z]. This has also shown that any commutator [g1, g2] with g1, g2 ∈ Γ can be written as [g1, g2] = [a1, a2][a3, a4] ... [an−1, an] for ak ∈ {xi, yi, zi,j}. A group satisfies the law [xp, y] if and only if it satisfies [x, yp]. From item 6 of p Lemma 1.2.1 and from the fact that Γ satisfies the law [[x, y], z], we have [g1, g2] = p [g1, g2] . As discussed in the previous paragraph, [g1, g2] = [a1, a2][a3, a4] ... [an−1, an]

12 where each ak is equal to one xi, yi or zij. Since the commutators on the right hand p side are all central and have order dividing p,[g1, g2] = 1. Thus Γ satisfies the law [xp, y]. Finally, to show that Γ satisfies the law xp2 , let g ∈ Γ and suppose g = α1 αr β1 βr γ1,2 γr−1,r x1 . . . xr y1 . . . yr z12 . . . zr−1,r (every element can be written as a word in this form by the proof of Theorem 1.3.8). Then gp2 can be partially collected in the following manner: We will use only the relations xjxi = zi,jxixj, yixj = xjyi and zi,jxk = xkzi,j. First we collect all x1 to the left of the word, starting with the left- most appearance of x1, and then proceeding to the right. Then for each i = 2, . . . , r we repeat the process, each time collecting xi to the right of the letters xi−1 and to the left of the rest of the word. As no xi appears or vanishes in this process 2 2 2 α1p α2p αrp for any i, the final result is a word of the form x1 x2 . . . xr w where w is a p2 p2 word in yi and zi,j. From xi = 1 follows g = w. Since w is a word in ele- ments in the center of Γ, all of which have order p, what we have to do now is for each generator yi and zi,j count how many times it appears in w and show that this number is a multiple of p. Since, during the collection process no new 2 instances of yi appear, we have exactly βip times the letter yi in w. For zi,j we 2 have all the γi,jp times they appeared before collecting, and each time the relation xjxi = zi,jxixj was used a new instance of zi,j appeared. These were used a total 2 2 2 of αiαj + 2αiαj + ··· + (p − 1)αiαj = αiαj(p − 1)p /2 times, which always is a multiple of p.

1.4 UCS groups

Definition 1.4.1 A group G having exactly one characteristic subgroup different from G and {1} is called a UCS group.

UCS groups were first studied by Taunt in [26]. He found sufficient conditions for the direct power of a UCS group to be a UCS group and discussed solvable UCS groups. Glasby, P´alfyand Schneider, in [8], studied UCS p-groups, separating these in the cases where the group has exponent p or p2. In the second case these groups are associated with a special class of modules, called ESQ modules. These modules for instance can be naturally associated with a class of algebras, which we will discuss in Chapter2. In the next section we will present the main results of Taunt and also those of P´alfy, Glasby and Schneider. For now we will make an introduction to UCS groups. Some examples of UCS groups are:

13 Example 1.4.2 The cyclic group with 4 elements, C4, is a UCS group for, if hgi = 2 C4, then the only proper non-trivial subgroup of C4 is hg i.

Example 1.4.3 The symmetric group of a set with n elements, Sn with n ≥ 3 and n 6= 4, is also UCS. Indeed, An, the alternating group, is its unique proper nontrivial normal subgroup, thus it is clearly characteristic.

Example 1.4.4 The quaternion group, Q8, has as its unique proper nontrivial char- acteristic subgroup its center, so Q8 is UCS.

Example 1.4.5 If p is a prime, then Cp2 × · · · × Cp2 is UCS as the set of elements of order at most p is a subgroup, and it is the only proper nontrivial characteristic subgroup. Also, every finite abelian UCS p-group is of this form, for, if G is abelian, then, for every integer k > 0, the set of elements whose order divides k is a charac- teristic subgroup of G. Thus the order of every element of G has at most 3 divisors. We are then left with three possibilities:

• G = Cp × · · · × Cp, which is characteristically simple, hence not UCS;

• G = Cp × · · · × Cp × Cp2 × ...Cp2 , which has the set of all elements of order | {z } | {z } n times m times dividing p as a characteristic subgroup, and also Gp is characteristic, so this is also not UCS;

• Finally we have Cp2 × · · · × Cp2 , which is UCS.

1.5 Earlier results about UCS groups

In this section we will present the main results about UCS groups found in the literature. The following simple lemma and its proof appear implicitly in the second paragraph of [26, page 25].

Lemma 1.5.1 Let G be a UCS group with proper nontrivial characteristic subgroup K. Then K and G/K are characteristically simple.

Proof Since any characteristic subgroup of K must be characteristic in G, then K must be characteristically simple. Additionaly, if G ≥ L ≥ K is such that L/K is a characteristic subgroup of G/K, then L must be characteristic in G.

14 By Burnside p-q theorem ([20, 8.5.3]), non-solvable UCS finite groups, just like all non-solvable finite groups, must have order divisible by at least 3 different primes. On the other hand, if G is a solvable finite UCS group with nontrivial proper charac- teristic subgroup N, then N and G/N both have to be solvable and characteristically simple by Lemma 1.5.1. Hence N and G/N are elementary abelian, and so |G| is divisible by at most 2 primes. We have then for solvable UCS groups the subclass of UCS p-groups, and the subclass of UCS groups where the order of the proper non- trivial characteristic subgroup is prime to its index. The p-groups can once again be subdivided. Combining Theorems 1.3.3 and 1.3.4, we obtain that Gp 6= G. Hence the characteristic subgroups Gpi are proper subgroups of G. Thus we have that UCS p-groups have either exponent p or p2.

Taunt focused his article on the solvable UCS groups of order prqs, that is, the ones that are not p-groups. He showed equivalent conditions for a group with order prqs with p, q primes and r, s positive integers to be UCS.

Theorem 1.5.2 [26, Theorem 6.1] Let G be a UCS group of order prqs, having proper nontrivial characteristic subgroup N of order pr. Then the direct product of a number of groups each isomorphic to G is itself a UCS group, except possibly if s = 1 and q | p − 1.

Taunt also showed some examples of such UCS groups.

Consider the group G1 given by

p p q y −λ0 y −λi−1 G1 = x1, . . . , xe, y | x1 = ··· = xe = y = 1; x1 = xe ; xi = xi−1xe .

Theorem 1.5.3 [26, Theorem 6.3] Suppose that G is a UCS group of order prqs, having characteristic subgroup of order pr. Then, if the order of p (mod q) is e,

1. r = ce for some integer c and s ≤ c;

c 2. G is isomorphic to a subgroup of the direct product G1, where G1 is as defined above;

c 3. if e > 1, G1 is itself a UCS group.

Theorem 1.5.4 [26, Theorem 7.2] Let p and q be distinct primes and let the order of p (mod q) be e. Then the number of distinct UCS groups of order prq with characteristic subgroups of order pr is equal to the number of divisors of (r, q − 1) which are multiples of e.

15 The case of UCS p-groups was further studied in [8], by Glasby, P´alfyand Schnei- der. Their main results are summarized in the following theorem:

Theorem 1.5.5 Let G be a nonabelian finite UCS p-group where |G/Φ(G)| = pr.

1. If r = 2, then G belongs to a unique isomorphism class.

2. If r = 3, then G belongs to a unique isomorphism class if p = 2, and to one of two distinct isomorphism classes if p > 2.

3. If r = 4 and either p = 2 or G has exponent p, then G belongs to one of eight distinct isomorphism classes.

4. Suppose that r = 4 and G has exponent p2. Then p 6= 5. Further, if p ≡ ±1 (mod 5) then G belongs to a unique isomorphism class; while if p ≡ ±2 (mod 5) then G belongs to one of two distinct isomorphism classes.

5. Let p be an odd prime, and let k be a positive integer. Then there exist non- abelian UCS p-groups of exponent p2

(a) of order p6k for all p and k; (b) of order p10 if and only if p5 ≡ 1 (mod 11); (c) of order p14k for all p and k.

In the article [8] presentations are given for the groups in parts 1, 2, 3 and 4 of Theorem 1.5.5.

Throughout this work we will need various results from [8] which were used to prove the previous theorem. We will use the rest of this section to present them.

By Lemma 1.3.3 when G is a finite p-group, its Frattini subgroup is always equal to GpG0 and it is a characteristic subgroup. For Φ(G) to be equal to {1} we need G0 = Gp = {1}, that is, G must be abelian and every non-identity element of G must have order p, or in other words, G must be elementary abelian, hence not UCS. So in the case of UCS p-groups, the proper nontrivial characteristic subgroup is always the Frattini subgroup. Also, as Φ(G) = GpG0, Φ(G) is the smallest normal subgroup such that G/Φ(G) is elementary abelian.

Throughout this work we will denote G = G/Φ(G).

Notice that, since for UCS p-groups the center is nontrivial and it is character- istic, we have Φ(G) ≤ Z(G). Consequently Φ(G) is abelian. In the case of UCS

16 p-groups, Φ(G)p is a proper characteristic subgroup of Φ(G), and consequently is characteristic in G. We have then that Φ(G)p = 1, and thus in the case of UCS p-groups Φ(G) is elementary abelian.

Since Φ(G) is characteristic, the action of Aut(G) on G induces actions of Aut(G) on both Φ(G) and G. We can consider then these two groups as Aut(G)-modules.

Recall that Hp,r is the r-generator free group in the variety of groups that have exponent dividing p2, nilpotency class 2, and have the property that all p-th powers are central. See Section 1.3.

Proposition 1.5.6 Every finite UCS p-group G is a quotient of Hp,r for some r.

Proof As Gpi is characteristic in G for every integer i > 0, then for G to be UCS it must have exponent dividing p2. Similarly, as the terms of the lower central series are characteristic, then G must have nilpotency class at most 2. Finally, Gp < G is characteristic, and Z(G) ≤ G is nontrivial characteristic for UCS p-groups, then Gp ≤ Z(G).

The following theorem is a consequence of [8, Theorem 4]. It gives us a charac- terization of the UCS p-groups which will be useful in the next chapters.

Theorem 1.5.7 Let p be a prime, let r be an integer, set H = Hp,r, and let G = H/N where N ≤ Φ(H). Then G is UCS if and only if the actions of Aut(G) on G and Φ(G) are irreducible.

Proposition 1.5.8 If p is an odd prime and G is a nonabelian finite UCS p-group of exponent p2 generated minimally by r elements, then |G| = p2r and |Φ(G)| = |G| = pr.

Proof It follows from [8, Theorem 4] that |G| = p2r. By Theorem 1.3.4 (Burnside’s Basis Theorem), |G| = pr, thus |Φ(G)| = |G|/|G| = pr.

Assume now that G is a nonabelian UCS p-group of exponent p2. Then the groups G0, Gp and Φ(G) = G0Gp are all non trivial, proper, and are also character- istic. This proves the following result:

Lemma 1.5.9 If G is a finite nonabelian UCS group of exponent p2, then G0 = Φ(G) = Gp 6= 1.

17 1.6 Tensor and exterior products

In [8] UCS groups of exponent p2 were characterized using exterior square represen- tations of certain irreducile linear groups. In this section we introduce the necessary terminology concerning tensor products, exterior squares and symmetric squares that is needed to discuss the related results. See [21, Chapter 14] for more on this topic.

Definition 1.6.1 Let U and V be F-vector spaces. Let FU×V denote the vector space of formal linear combinations in U × V . Let S be the subspace of FU×V genereated by all vectors of the form

1. α(u, v) + β(u0, v) − (αu + βu0, v);

2. α(u, v) + β(u, v0) − (u, αv + βv0);

0 0 where α, β ∈ F, u, u ∈ U and v, v ∈ V . The quotient space FU×V /S is called the tensor product of U and V and denoted U ⊗ V . The coset (u, v) + S is denoted by u ⊗ v.

If U and V are vector spaces, then U ⊗ V is generated as a vector space by the elements of the form u ⊗ v where u ∈ U and v ∈ V . Further, any element of U ⊗ V Pn can be written as a linear combination i=1 ui ⊗ vi where ui ∈ U and vi ∈ V .

Definition 1.6.2 Let V be a vector space and consider its tensor square V ⊗ V . Let

φ : V ⊗ V → V ⊗ V X X (ui ⊗ vi) 7→ (vi ⊗ ui).

We define SV = {a ∈ V ⊗V | aφ = a} and V ∧V = {a ∈ V ⊗V | aφ = −a}. We say that an element of SV is symmetric, and an element of V ∧V is antisymmetric. We also call V ∧ V the exterior square of V .

We will be interested in V ⊗ V in the cases where V has finite dimension and char F 6= 2. By [21, Theorem 14.2], if U and V have finite dimension, then dim(U ⊗ V ) = dim(U) dim(V ). By [21, Theorem 14.18], if dim(V ) = n, then dim(SV ) = n+1
n
2 2 and dim(V ∧V ) = 2 , hence dim(SV )+dim(V ∧V ) = dim(V ⊗V ) = n . From

18 the definition of SV and V ∧ V it follows that if char F 6= 2, then SV ∩ (V ∧ V ) = 0. Thus if char F 6= 2 and V has finite dimension, then V ⊗ V = SV ⊕ (V ∧ V ). If the group G acts on the vector space V , then we can define an action induced on V ⊗ V setting, for all v, w ∈ V and g ∈ G,(v ⊗ w)g = (vg) ⊗ (wg). Notice that ((v ⊗ w)g)φ = (vg ⊗ wg)φ = wg ⊗ vg = (w ⊗ v)g = ((v ⊗ w)φ)g for φ as in Definition 1.6.2. Thus both SV and V ∧ V are invariant under g. We will denote the action induced by g on V ∧ V by g ∧ g.

1.7 Wreath Product

Wreath product is an important tool in group and semigroup theory, providing means to construct groups or semigroups with certain properties. In this work we will use it to show that, in some cases, the direct product of copies of a UCS group is UCS, and also that the sum of copies of an IAC algebra is also IAC.

Let G be a group and denote by Gn the direct product of n copies of G. Let H n be a subgroup of the symmetric group Sn. There is a natural action of H on G given by

n (g1, . . . , gn)h = (g1h−1 , . . . , gnh−1 ) for all (g1, . . . , gn) ∈ G and h ∈ H. (1.1)

Definition 1.7.1 Let G be a group and H be a subgroup of the symmetric group Sn for some n. The wreath product of G by H denoted G o H, is the semidirect product Gn o H where H acts on the group Gn as in Equation 1.1. The subgroup B = {(g, 1) | g ∈ Gn} ∼= Gn of G o H is called the base group of the wreath product, whereas the subgroup

T = {(1, h) | h ∈ H} ∼= H is called the top group.

n If G acts on Ω and H ≤ Sn, then W = G o H acts on Ω as follows. If n (ω1, . . . , ωn) ∈ Ω and (g1, . . . , gn, h) ∈ W , then

(ω1, . . . , ωn)(g1, . . . , gn, h) = (ω1h−1 g1h−1 , . . . , ωnh−1 gnh−1 ). This action is called the product action. For a proof that it is indeed an action, see [6, page 50].

19 We will consider G actions on sets Ω in the case when Ω has the structure of a group, an algebra or a vector space and each g ∈ G induces an automorphism of Ω. If Ω is a group, then Ωn denotes the direct product of n copies of Ω, while if Ω is a vector space or an algebra, Ω⊕n denotes the direct sum of n copies of Ω. In this case an element of G o H induces an automorphism of Ωn or Ω⊕n. If the action of G on Ω is faithful, that is, only the identity of G induces the trivial map on Ω, then the action of G o H is also faithful (for a proof of this fact, see once again [6, page 50]).

If V is a vector space then GL(V ) denotes the group of invertible linear trans- formations of V . If V = Fn for some field F and n ≥ 1 then we write GL(n, F) for GL(V ). A subgroup K ≤ GL(V ) is said to be irreducible if 0 and V are the only subspaces of V that are invariant under K.

Theorem 1.7.2 Suppose that G ≤ GL(V ) is a non-trivial linear group and K ≤ Sl is a transitive permutation group. Then G o K is irreducible on V ⊕l if and only if G is irreducible on V .

Proof If G is irreducible on V , then [25, Lemma 4 of Section 15] shows that G o K is irreducible on V ⊕l.

If G is not irreducible on V , then let 0 < U < V be a subspace fixed by every element of G. Notice that U ⊕l is fixed by elements of G o K of the forms (1, k) and (g, 1) where k ∈ K and g ∈ Gn and every element of G o K can be written as (g, k) = (g, 1)(1, k), then U ⊕l is fixed by G o K. Further 0 < U ⊕l < V ⊕l and so G o K considered as a subgroup of GL(V ⊕l) is not irreducible.

Taunt in [26] proved the next theorem about the direct power of a UCS group. The proof of this theorem can be presented differently from how it was presented in [26] using the language of wreath products as follows.

Theorem 1.7.3 [26, Theorem 2.1] Let G be a UCS group with proper nontrivial characteristic subgroup N, satisfying the following conditions:

1. no element of N other than 1 is fixed under every automorphism of G;

2. no element of G/N other than N is fixed under every automorphism of G/N induced by an automorphism of G.

Then for each n ≥ 1, the direct product Gn is a UCS group.

20 Proof Given a UCS group G as in the theorem, consider the group n G = {(x1, . . . , xn)|xi ∈ G}. We will denote the i-th coordinate subgroup, that is, the subgroup of the elements {(1,..., 1, xi , 1,..., 1) | xi ∈ G}, by Gi. |{z} i-th entry Suppose then that K ≤ Gn is a nontrivial characteristic subgroup of Gn. We will show that K must be equal to either N n or to Gn.

n As we remarked after Definition 1.7.1, Aut(G)oSn is a subgroup of Aut(G ). The base group B acts as Aut(G) in each coordinate, whereas the top group T permutes the coordinates.

We will divide the proof in two cases. The arguments in both cases are very similar.

Case 1: Suppose first that K 6≤ N n, that is, K has an element of the form k = (k1, . . . , kn), ki ∈/ N for some i. Suppose without loss of generality that k1 ∈/ N. Therefore k1N 6= N, and so there exists θ ∈ Aut(G) such that (k1N)θ 6= k1N, that −1 −1 is, (k1θ)k1 ∈/ N. Let θb denote the element (θ, 1,..., 1) ∈ B. Then l = (kθb)k = −1 ((k1θ)k1 , 1,..., 1). That is, l ∈ K contains precisely one entry different from 1 and this entry lies outside N. This means that K∩G1 6= N. Since the intersection K∩G1 needs to be characteristic in G1 (because every automorphism ϕ of G1 can be lifted n to an element (ϕ, 1,..., 1) ∈ B ≤ Aut(G )), and K ∩ G1 is nontrivial and different from N, must be equal to G1, hence G1 ≤ K. As the top group T ≤ Aut(G) o Sn permutes the coordinate groups, we conclude that Gj ≤ K for all j ∈ {1, . . . , n}. Thus K = Gn.

n Case 2: Suppose now that K ≤ N . In this case there is k ∈ K, k = (k1, . . . , kn) with ki 6= 1. Suppose without loss of generality that k1 6= 1. Then there exists −1 θ ∈ Aut(G) such that k1θ 6= k1, that is, (k1θ)k1 6= 1. Let θb denote the element −1 −1 (θ, 1,..., 1) ∈ B. Then l = (kθb)k = ((k1θ)k1 , 1 ..., 1) ∈ K. Since N and K ∩ G1 are characteristic in the UCS group G1, K ∩G1 = N, and thus N ≤ K. Using again that the top group T ≤ Aut(G) o Sn permutes the coordinate groups we find that N n ≤ K. Hence K = N n.

We proved that if Gn has a proper nontrivial characteristic subgroup, it has to be N n. To complete the proof, we need to show that Gn is not characteristically simple, but as G is normal in Gn and G is not characteristically simple, by Lemma 1.2.7 Gn is not characteristically simple, which completes the proof.

21 Chapter 2

UCS p-groups and IAC algebras

In this chapter we will show how the class of the finite UCS p-groups of exponent p2 can be bijectively associated with a certain class of nonassociative algebras.

Recall that if G is a group we denote by G the quotient G/Φ(G).

2.1 Bijection between UCS groups and IAC alge- bras

The next lemma shows that in a UCS p-group G of exponent p2 the Frattini sub- group Φ(G) and the quotient G = G/Φ(G) are isomorphic Aut(G)-modules via an isomorphism Ψ.

Lemma 2.1.1 Let G be a finite p-group, p > 2, such that Φ(G) ≤ Z(G) and Φ(G)p = 1. The map

Ψ: G → Φ(G) gΦ(G) 7→ gp is such that

1. Ψ is a well defined group homomorphism;

2. if G0 ⊆ Gp, then Ψ is a surjective homomorphism of Aut(G)-modules;

22 3. if G is a UCS p-group of exponent p2, then Ψ is an isomorphism of Aut(G)- modules.

Proof

1. Ψ is well defined since, from Φ(G) ≤ Z(G) and from Φ(G)p = 1, we have that p p p p p if g1 = g2z where g1, g2 ∈ G and z ∈ Φ(G), then g1 = (g2z) = g2z = g2. To show that it is a homomorphism, let g and h be the classes of g and h in G respectively. Using Lemma 1.2.2 it follows that

p p p p p p (gh)Ψ = (gh) = g h [h, g](2) = g h = (gΨ)(hΨ),

where the third equality occurs because [h, g] ∈ G0 ⊆ Φ(G), and thus it has order p or 1. Hence Ψ is a group homomorphism.

p p p p 2. In this case Φ(G) = G . If h ∈ Φ(G) we have h = g1g2 . . . gk, with some g1, . . . , gk ∈ G, and thus h = Ψ(g1g2 . . . gk), therefore Ψ is surjective. If ϕ is an automorphism of G, then ϕ induces automorphisms on both G and Φ(G). Then, for every g ∈ G, ((gΦ(G))ϕ)Ψ = (gϕ)p = gpϕ = (gΦ(G))Ψϕ. Thus Ψ is a homomorphism of Aut(G)-modules.

3. For Ψ to be an isomorphism (of groups) we only need now that |Φ(G)| = |G|, which we have by Proposition 1.5.8.

When working with UCS p-groups of exponent p2, we will frequently need to use the isomorphism Ψ defined in the theorem above. If g ∈ G we will denote gΨ by gp. We will also denote for g ∈ Φ(G) the image gΨ−1 by g1/p. Hence g1/p ∈ G. Further we have for each g ∈ G and for each h ∈ Φ(G) that (gp)1/p = g and (h1/p)p = h. Hence if G is a UCS p-group of exponent p2, then h 7→ h1/p is the inverse of g 7→ gp. By part 3 of Lemma 2.1.1, g 7→ gp is an isomorphism of Aut(G)-modules, thus h 7→ h1/p is also an isomorphism of Aut(G)-modules.

We are now ready to associate to each UCS p-group of exponent p2 a nonasso- ciative algebra.

Definition 2.1.2 Given a finite UCS p-group G of exponent p2, p > 2, we define the algebra L(G) associated with the group G as follows:

23 • (L(G), +) = (G, ·) as Fp vector spaces, seeing the sum in G as the usual multiplication, and the product by elements of Fp as repeated addition. • The product in L(G) is given by v, w = [v, w]1/p, where v and w are repre- sentatives in G of the classes v andJ w Kin G.

Since the sum in L(G) is defined as the product in G, the zero element of L(G) is the trivial coset 1. We will show in Proposition 2.1.3 that the product defined above is well defined. In the case when G is an abelian UCS p-group of exponent p2, we can readily see that the algebra defined above will have u, v = 0 for all u, v ∈ L(G). We will now prove some properties of L(G), which laterJ weK will use to define a class of algebras:

Proposition 2.1.3 Let G be a finite UCS p-group with exponent p2 and let (L(G), +, , ) be the algebra associated with G, as defined above. J K

1. , is well defined. J K 2. (L(G), +, , ) is a non-associative algebra over Fp. J K 3. v, v = 0 for all v ∈ L(G). J K 4. L(G) is an irreducible Aut(L(G))-module. 5. L(G), L(G) = 0 or L(G), L(G) = L(G). J K J K Proof

0 0 1. Let v and v be two representatives of v, then vz1 = v for some z1 ∈ Φ(G). In 0 0 the same way, let w and w be representatives of w, then wz2 = w for some z2 ∈ Φ(G). Since z1, z2 ∈ Φ(G) ⊆ Z(G), then [v, w] = [vz1, wz2]. Hence, by Lemma 1.2.1

1/p 1/p 0 0 1/p v, w = [v, w] = [vz1, wz2] = [v , w ] , J K and thus , is well defined. J K ∼ 2. Since (L(G), +) = (G, ·), it is a vector space over Fp. We have already shown that , is well defined. All that is left to show then is that , is distribu- tive overJ K the sum. Using part 3 of Lemma 1.2.1, u, v + w J=K [u, vw]1/p = ([u, w][u, v]w)1/p. As the commutators are elementsJ of Φ(G) ⊆K Z(G), we have [u, v]w = [u, v], and thus [u, vw]1/p = ([u, v][u, w])1/p = [u, v]1/p[u, w]1/p = u, v + u, w . J K J K 24 3. v, v = [v, v]1/p = 11/p = 1. The result follows since 1 ∈ G is the zero of L(G). J K 4. As, by Theorem 1.5.7 Aut(G) is irreducible on G, it suffices to show that every automorphism induced on G by an automorphism of G is also an auto- morphism (of algebras) of L(G). If α ∈ Aut(G), and α is the automorphism induced on G, note that for u ∈ G, u α = uα. Let us then calculate for u, v ∈ G, uα, vα . As uα is a representative of uα and vα is a representative of vα, J K uα, vα = [uα, vα]1/p = ([u, v]α)1/p. J K Since the map x 7→ x1/p is an isomorphism of Aut(G)-modules (see the dis- cussion before Definition 2.1.2), we have ([u, v]α)1/p = [u, v]1/pα = u, v α as we wanted. J K 5. If G is abelian, then trivially L(G), L(G) = 0. Suppose then G nonabelian. Since u, v = [u, v]1/p and theJ map x 7→ x1K/p is a group isomorphism, then, to show thatJ KL(G), L(G) = L(G), it suffices to show that h[u, v]|u, v ∈ L(G)i = Φ(G). ThisJ is true as GK0 = Φ(G).

Since these algebras are irreducible and anticommutative, we will call an alge- bra that satisfies the properties of the previous proposition IAC (Irreducible Anti- Commutative).

For better clarity in the next definition, we remind the reader that a nonassocia- tive algebra is an algebra that may or may not be associative (so associative algebras are a particular case of nonassociative algebras).

Definition 2.1.4 An IAC algebra over a field F is a nonassociative F-algebra (L, +, , ) that satisfies J K 1. , is anticommutative, that is, v, v = 0 for all v ∈ L. J K J K 2. Aut(L) is irreducible over L.

Notice that, since L, L is a subspace stable under all automorphisms of L, we have that condition 2J impliesK L, L = L or L, L = 0. J K J K Definition 2.1.5 If L is an IAC algebra such that u, v = 0 for all u, v ∈ L, we will call L an abelian IAC algebra. J K

25 We have, by Proposition 2.1.3, that given an UCS p-group G of exponent p2, we can construct an IAC algebra. The reverse process is also possible.

Theorem 2.1.6 Given an IAC algebra L of finite dimension over Fp, there exists a finite UCS p-group G of exponent p2 such that L(G) ∼= L. Also, G will be abelian if and only if L is abelian.

We will denote the group G constructed in this theorem by G(L). In Theorem 2.1.7 we will also show that for any UCS p-group of exponent p2, G(L(G)) = G.

Proof Suppose first that L is abelian and has dimension r. Then by taking G = 2 ∼ Cp2 × Cp2 × · · · × Cp2 , G is UCS, has exponent p and L(G) = L. Notice that in | {z } r times this case G is abelian. In the rest of the proof we assume that L is a non-abelian algebra.

Let v1, . . . , vr be a basis for L and let H = Hp,r be the r-generator free group in the variety V of groups that have exponent dividing p2, nilpotency class at most 2 and have the property that all p-th powers are central; see Proposition 1.3.9 for a (i,j) description of H. Assume that ck are the structure constants of L; that is,

r X (i,j) vi, vj = ck vk (2.1) J K k=1 for all 1 ≤ i < j ≤ r. Suppose that h1, . . . , hr are generators of H. Considering the (i,j) field elements ck ∈ Fp as integers in {0, . . . , p − 1}, define N ≤ H as the subgroup

* r + Y p (i,j) −1 ck N = [hi, hj] (hk) | 1 ≤ i < j ≤ r . (2.2) k=1

Since H ∈ V, Φ(H) ≤ Z(H). Further N ≤ Φ(H) ≤ Z(H), and so N E H. Further, considering Φ(H) as an Fp-vector space, the given generators of N are linearly r independent, and hence |N| = p(2).

Set G = H/N. Then G is a finite p-group, and so the Frattini subgroup of G is

GpG0 = (H/N)p(H/N)0 = (HpH0N)/N = Φ(H)/N.

On the other hand, since Φ(G) = GpG0 and the relations in N imply that G0 ≤ Gp, we find that Φ(G) = Gp. Since Φ(G) ≤ Z(G) and Φ(G)p = 1, the map g 7→ gp is a homomorphism between the Aut(G)-modules G and Φ(G). Also, since |G| =

26 |Φ(G)| = pr, the map g 7→ gp is an isomorphism. Although G is not known to be UCS at this point, this isomorphism allows us to construct the anti-commutative algebra L(G), as in Definition 2.1.2. Set gi = hiN for 1 ≤ i ≤ r, and define the linear map ξ : L → G, by letting viξ = gi for all i. We claim that ξ is an isomorphism of algebras. Indeed,

r !1/p r Y p (i,j) Y (i,j) 1/p ck ck viξ, vjξ = gi, gj = [gi, gj] = (gk) = (gk) = vi, vj ξ. J K J K k=1 k=1 J K

We claim that G is a UCS p-group of exponent p2. Since G is clearly a p-group of exponent p2 and G is a quotient of H by a normal subgroup contained in Φ(H), by Theorem 1.5.7, it suffices to show that Aut(G) acts irreducibly on G and on Φ(G). Let α ∈ Aut(L) and assume, for all i ∈ {1, . . . , r}, that r X (i) viα = ck vk k=1 (i) ∗ where ck ∈ Fp. Define the homomorphism α : H → H by setting r (i) ∗ Y ck hiα = hk . k=1

Since α ∈ Aut(L), the elements v1α, . . . , vrα form a basis for L. Hence the elements ∗ hiα Φ(H) of the quotient H/Φ(H) considered as an Fp-vector space form a minimal generating set for H/Φ(H). Thus by Burnside’s Basis Theorem (Theorem 1.3.4), α∗ is surjective and hence, since H is finite, α∗ ∈ Aut(H). We claim that N is invariant under α∗. Define ϕ : H0 → Hp as the linear map r p (i,j) 0 Q ck that maps the generators of H as [hi, hj]ϕ = k=1(hk) for all i < j. Note that the subgroup N can be viewed as N = {h−1(hϕ) | h ∈ H0}. It suffices to ∗ ∗ p p show that ϕα = α ϕ. Now observe that γ : H → L defined by hi γ = vi is a 0 linear isomorphism. Similarly γ ∧ γ : H → (L ∧ L) defined by [hi, hj] 7→ vi ∧ vj is a linear isomorphism to the exterior square L ∧ L of L. Suppose that M denotes the multiplication map from L ∧ L to L; that is, M :(L ∧ L) → L, defined by (u ∧ v)M = u, v . The different maps are illustrated in diagram (2.3): J K ϕ

γ∧γ γ−1 H0 L ∧ L M L Hp

α∗ α∧α α α∗ (2.3) H0 L ∧ L L Hp. γ∧γ M γ−1

ϕ

27 We claim that each square piece of diagram (2.3) is commutative. Indeed, for hi, hj ∈ H,

r r X (i) X (j) [hi, hj](γ ∧ γ)(α ∧ α) = (vi ∧ vj)(α ∧ α) = ck vk ∧ ck vk k=1 k=1

" r r # r r (i) (j) ∗ Y ck Y ck X (i) X (j) [hi, hj]α (γ ∧ γ) = hk , hk (γ ∧ γ) = ck vk ∧ ck vk. k=1 k=1 k=1 k=1

For vi, vj ∈ L,

r r tX (i) X (j) | (vi ∧ vj)Mα = vi, vj α = ck vk, ck vk J K k=1 k=1

r r ! r r X (i) X (j) tX (i) X (j) | (vi ∧ vj)(α ∧ α)M = ck vk ∧ ck vk M = ck vk, ck vk k=1 k=1 k=1 k=1 and also r p Y p (i) −1 ∗ ∗ ck viγ α = hi α = (hk) k=1 r ! r X (i) Y p (i) −1 −1 ck viαγ = ck vk γ = (hk) , k=1 k=1

Hence the claim about the square pieces is verified. Also, notice that for hi, hj generators of Hp,r,

−1 −1 −1 [hi, hj](γ ∧ γ)Mγ = (vi ∧ vj)Mγ = vi, vj γ = rJ K ! r X (i,j) Y p (i,j) −1 ck = ck vk γ = (hk) = [hi, hj]ϕ, k=1 k=1 thus ϕ = (γ ∧ γ)Mγ−1. Hence,

ϕα∗ = (γ ∧ γ)Mγ−1α∗ = α∗(γ ∧ γ)(α ∧ α)−1Mγ−1α∗ = = α∗(γ ∧ γ)(α ∧ α)−1(α ∧ α)Mα−1γ−1α∗ = = α∗(γ ∧ γ)Mα−1αγ−1(α∗)−1α∗ = α∗(γ ∧ γ)Mγ−1 = α∗ϕ.

This implies that α∗ induces an automorphismα ˜ of G. Further, the automorphism α induced byα ˜ on G coincides with ξ−1αξ. Hence ξ−1Aut(L)ξ is contained in the group induced by Aut(G) on G, and so Aut(G) acts irreducibly on G. Since G and

28 Φ(G) are isomorphic as Aut(G)-modules, Aut(G) is irreducible on Φ(G) also. By Theorem 1.5.7, G is UCS.

1/p Notice that, since (L, +, , ) is isomorphic to (G, +, , G), where , G = [ , ] , J K J K J∼K and (L(G), +, , ) is also isomorphic to (G, +, , G), we have that L = L(G). J K J K To show that G is not abelian in this case, notice that this would be equivalent to N being equal to H0, which cannot occur as L is not abelian.

Let IACp denote the set of isomorphism classes of finite-dimensional IAC algebras over the field with p elements Fp, and let UCSp2 denote the set of isomorphism classes of finite UCS p-groups of exponent p2. The previous theorem shows that for an IAC ∼ algebra L of finite dimension over Fp the isomorphism L(G(L)) = L holds. Hence by the following theorem, the map L : UCSp2 → IACp defined in Definition 2.1.2 is the inverse to the map G : IACp → UCSp2 defined in Theorem 2.1.6.

Theorem 2.1.7 If G is a finite UCS p-group of exponent p2, then G(L(G)) ∼= G.

Proof Let G be a finite UCS p-group of exponent p2 and set K = G(L(G)). We need to show that K ∼= G. The group K was constructed in Theorem 2.1.6 as Hp,r/Nk where Nk is the normal subgroup of Φ(Hp,r) defined in equation 2.2. Since ∼ G = Hp,r/N for some N ≤ Φ(Hp,r), then it suffices to show that N = NK ,

Given a set {g1, . . . , gn} of generators for G, the algebra L(G) is constructed by 1/p L(G) = (G, ·, , ), with gi, gj = [gi, gj] for every i, j = 1, . . . , n. If gi, gj = r J K J K J K X (i,j) ck gk in L(G), then the group K = G(L(G)) is constructed as Hp,r/NK where k=1 * r + Y p (i,j) −1 ck NK = [hi, hj] (hk) . k=1

r X (i,j) On the other hand, for gi, gj = ck gk to happen in L(G), we must have J K k=1

r (i,j) 1/p Y ck [gi, gj] = gk . k=1

r Y p (i,j) ck Applying the p-th powering on both sides, we get [gi, gj] = (gk) , and as k=1 r p (i,j) −1 Q ck G = Hp,r/N, we conclude that [hi, hj] k=1 (hk) ∈ N.

29 Now we have N ≤ NK . Since |N| = |Hp,r|/|G|, by Theorem 1.3.8 and Propo- (r) sitions 1.3.9 and 1.5.8, N has order p 2 . As NK is, by the proof of the previous r
theorem, a vector subspace of Φ(Hp,r) generated by the 2 linearly independent r p (i,j) −1 Q ck vectors of the form [hi, hj] k=1 (hk) , we have |N| = |NK |, hence N = NK .

Now we can prove the main theorem stated in the introduction.

Proof of Theorem 1.1.1 By theorems 2.1.6 and 2.1.7 the maps G : IACp → UCSp2 and L : UCSp2 → IACp are inverses of one another. Thus they both are bijections.

2.2 Automorphisms of UCS groups

In this section we investigate the relation between the automorphisms of UCS groups and the automorphisms of the associated IAC algebra.

Proposition 2.2.1 Let G be a finite UCS p-group of exponent p2, and L(G) its associated IAC algebra. Every automorphism of the group G induces an automor- phism of L(G), and every automorphism of L(G) can be lifted to an automorphism G.

Proof Given an automorphism α of the group G, we can take the map induced on G, which we will denote by α. The argument in the proof of Proposition 2.1.3 (4) shows that α ∈ Aut(L(G).

The converse follows readily from the proof of Theorem 2.1.6, for, given α an ∗ automorphism of L, we can construct the automorphism α of Hp,r which induces an automorphism of G.

Note that the processes are not necessarily inverses of one another. Even though starting from an automorphism of L(G), then constructing one of G and then con- structing one of L(G) gives us the very same initial automorphism, if we start from an automorphism of G, construct one of L(G) and then construct the automor- phism of G, we have no guarantee that the final result will be the same initial automorphism. By Proposition 2.2.1, Aut(L(G)) is a quotient of Aut(G) and in the rest of this section we will further characterize the relationship between the two automorphism groups.

30 For the remainder of this section, G will denote a UCS p-group of exponent p2, L = L(G) will be its associated IAC algebra, and K will denote the subgroup of Aut(G) consisting of all automorphisms that act trivially on G.

Theorem 2.2.2 Aut(L) ∼= Aut(G)/K.

Proof Notice that K is a normal subgroup of Aut(G), and that if ϕ, ψ ∈ Aut(G), are elements such that ϕK 6= ψK in Aut(G)/K, then ϕψ−1 ∈/ K, which implies that ϕ and ψ act differently on G. As, by Proposition 2.2.1 the automorphisms of L are always restrictions to G of an automorphism of G, we have that Aut(L) ∼= Aut(G)/K.

Let J be the kernel of the map ρ from Aut(G) to Aut(Φ(G)) = GL(Φ(G)) that takes ϕρ = ϕ|Φ(G) .

Lemma 2.2.3 K ≤ J.

Proof Let ϕ ∈ K. Since gϕ = g for all g ∈ G, we have that gϕ = gfg, where p p p p p p p fg ∈ Φ(G). Since Φ(G) ≤ Z(G), then (g )ϕ = (gfg) = g fg = g . Thus g ϕ = g p for all g ∈ G. As G = Φ(G), we have ϕ|Φ(G) = 1.

In the proof of Lemma 2.2.3 we defined for g ∈ G and ϕ ∈ K an element fg ∈ Φ(G). Note that fg depends only on the class g, and not on the representative.

Indeed, if g = hf with h ∈ G and f ∈ Φ(G), as ϕ|Φ(G) = 1 we have fϕ = f, from where we get

hffg = gfg = gϕ = (hf)ϕ = (hϕ)(fϕ) = hfhf = hffh, from where we conclude that fg = fh. Also, note that fg depends on the automor- phism ϕ, but we supress it in the notation.

For every ϕ ∈ K, define the homomorphism ζϕ ∈ HomFp (G, Φ(G)) by

ζϕ : G → Φ(G)

g 7→ fg.

Note that we have xϕ = x(xζϕ).

∼ Proposition 2.2.4 K = HomFp (G, Φ(G)) as groups.

31 Proof Since in both G and Φ(G), Fp acts by repeated multiplication, we only need to show that the map

Λ: K → Hom(G, Φ(G))

ϕ 7→ ζϕ is a group isomorphism.

To show that Λ is a homomorphism, we take ϕ, ψ ∈ K. If gϕ = gfg and 0 0 gψ = gfg, then gζϕ = fg and gζψ = fg. Hence

0 0 g(gζϕψ) = gϕψ = (gfg)ψ = (gψ)(fgψ) = gfgfg = gfgfg = g(gζϕ)(gζψ).

Hence gζϕψ = (gζϕ)(gζψ), which implies ζϕψ = ζϕζψ thus Λ is a homomorphism.

For the injectivity, we take ϕ ∈ ker(Λ), that is ϕ ∈ K such that gζϕ = 1 for all g ∈ G. Then gϕ = g(gζϕ) = g for all g ∈ G, which implies that ϕ is the trivial automorphism. Hence Λ is injective.

Finally, to show Λ is surjective, we take, for each ζ ∈ HomFp (G, Φ(G)), ϕ : G → G g 7→ g(gζ).

This is a homomorphism, as, for every g, h ∈ G,

(gh)ϕ = gh(ghζ) = gh(gζ)(hζ) = g(gζ)h(hζ) = (gϕ)(hϕ).

Since ϕ is an endomorphism of a finite group, to show it is an automorphism we only need to show it is injective. Let then g ∈ G such that gϕ = 1. Then g(gζ) = 1, which implies g = (gζ)−1 ∈ Φ(G). Thus, g = 1, from where we get gζ = 1 and consequently g = 1. Thus ϕ is an isomorphism. As ϕ ∈ K and ϕΛ = ζ, Λ is surjective.

2.3 ESQ modules

We have estabilished in the previous section a correspondence for an odd prime 2 p between IAC algebras over the field Fp and UCS p-groups of exponent p . The idea for this correspondence comes from [8], where the authors proved that the existence of UCS p-groups of exponent p2 is equivalent to the existence of certain

32 ESQ modules. In this section we will introduce ESQ modules, and clarify their relation to IAC algebras.

Recall that, as noted at the end of Section 1.6, if a group G acts on a vector space V , then G acts on the exterior square V ∧ V by (v ∧ w)g = vg ∧ wg.

Definition 2.3.1 Let F be a field, V be an F-vector space and G a subgroup of GL(V ). The G-module V is said to be an ESQ module (exterior self quotient) if there is a G-submodule U of V ∧ V such that (V ∧ V )/U and V are isomorphic as G-modules. If this is the case then the group G is said to be an ESQ subgroup of GL(V ). An ESQ subgroup G ≤ GL(V ) is said to be maximal ESQ if whenever H ≤ GL(V ) is ESQ and G ≤ H, then G = H.

Let L be a finite dimensional IAC algebra over a field F. Set A = Aut(L). Let ϕ : L ∧ L → L denote the map induced by multiplication, that is, (v ∧ u)ϕ = v, u . Then ϕ can be viewed as a surjective homomorphism of A-modules. Further, (JL ∧KL)/(ker ϕ) ∼= L, and so L is an ESQ A-module and A ≤ GL(L) is an ESQ subgroup.

Conversely, if V is an ESQ G-module for some G ≤ GL(V ), then there is some ϕ ∈ HomG(V ∧V,V ) such that ϕ is surjective. The homomorphism ϕ can be viewed as an anticommutative multiplication on V and hence (V, , ϕ) is an anticommu- tative algebra where u, v ϕ = ϕ(u ∧ v). Since G is supposedJ K to be irreducible, (V, , ϕ) is IAC. J K J K

Theorem 2.3.2

1. Suppose that L is a finite dimensional IAC algebra over F and A = Aut(L). Then A is an ESQ subgroup of GL(V ).

2. Suppose G is an ESQ subgroup of GL(V ) for a finite dimensional vector space V over a field F. Then the algebra (V, , ϕ) in the last paragraph before the theorem is IAC and G ≤ Aut(V, , ϕ). J K J K

Proof Follows from the discussion before the theorem.

33 Chapter 3

Structure of IAC algebras

In this chapter we will explore the structure of IAC algebras, showing that IAC algebras are always semisimple and they are a direct sum of copies of a simple IAC algebra. We will also study the relation between the automorphisms of a UCS p-group of exponent p2 and the automorphisms of its associated IAC algebra L(G).

3.1 Direct powers of UCS p-groups

In this section we will show that direct powers of UCS finite p-groups are also UCS, while in Section 3.2 we will show how this fact, in the particular case of UCS p-groups of exponent p2, relates to IAC algebras.

Recall that extraspecial groups were defined in Definition 1.3.5.

Theorem 3.1.1 Let G be a UCS p-group, p ≥ 3. Then for any positive integer a, the direct product Ga of a copies of G is UCS.

∼ Proof If G is abelian, then, as noted in Example 1.4.5, G = Cp2 × · · · × Cp2 , and | {z } k copies a ∼ consequently G = Cp2 × · · · × Cp2 , which is UCS. | {z } ak copies Suppose that G is nonabelian. We claim that no element of G/Φ(G) and of Φ(G) is invariant under all automorphisms of G. By Theorem 1.7.3, this implies that Ga is UCS.

34 First we show this claim for G/Φ(G). Since G is UCS, G acts irreducibly on

G/Φ(G). Hence the claim is valid if dimFp (G/Φ(G)) ≥ 2. If dimFp (G/Φ(G)) = 1 then G/Φ(G) is cyclic and so G is also cyclic (see Theorem 1.3.4) which is impossible since G is nonabelian. Hence the claim is valid for G/Φ(G).

Let us show this claim for Φ(G). As in the previous paragraph, if |Φ(G)| ≥ p2 then the claim follows from the irreduciblity of Aut(G) on Φ(G). Suppose that |Φ(G)| = p. Since G is nonabelian UCS, Φ(G) = G0 = Z(G) and so G is extra- special. Thus by [20, Exercise 7] G is one of two isomorphism types: one having exponent p2, the other having exponent p. If G has exponent p2, then by [27, Corollary 1], Aut(G) has fixed points in G/Φ(G), which is impossible since G is UCS. Thus G must have exponent p. By [20, 5.3.8] G is the central product of extraspecial groups of order p3 and exponent p. Thus, up to isomorphism there is a unique such group and hence G is of the form

δij p p p G = x1, . . . , xn, y1, . . . , yn, z | [xi, yj] = z ; xi = yi = z = 1;

[xi, xj] = [yi, yj] = [xi, z] = [yi, z] = 1i .

−1 In this case the map defined by xi 7→ yi, yi 7→ xi, z 7→ z is an automorphism which does not fix any element of Φ(G) = hzi.

3.2 Direct sums of IAC algebras

Taunt showed in Theorem 1.7.3 that under certain conditions the direct powers of UCS groups are UCS. In Theorem 3.1.1 we showed that for p ≥ 3 a UCS p-group always satisfies those conditions. This raises the question: Is the direct sum of copies of an IAC algebra also IAC? The answer is yes as shown in Proposition 3.2.1. We will also show in this section that IAC algebras are semisimple, and are always isomorphic to the direct sum of copies of a simple IAC algebra.

Proposition 3.2.1 If L is a simple IAC algebra and K = L⊕n, that is, K is the sum of a finite number of copies of L, then K is IAC.

Proof Since L is anticommutative, so must be K. Assume first that L = F2 (the ⊕n one-dimensional abelian algebra over F2). Then K = F2 and Aut(K) = GL(K) which is irreducible on K. Hence K is IAC in this case. Suppose now that L 6= F2. Then Aut(L) is a non-trivial irreducible subgroup of GL(L). Hence by Theorem

35 1.7.2, Aut(L) o Sn is an irreducible subgroup of GL(K). By the discussion after Defintion 1.7.1, Aut(L) o Sn ≤ Aut(K). Hence K is IAC, as claimed.

The question then arises: is L⊕n the IAC algebra associated with the group G(L)n = G(L) × · · · × G(L)? The answer is yes, as shown by the next theorem.

Theorem 3.2.2 Let G be a finite UCS p-group of exponent p2 for some prime p. Then L(Gn) = L(G)⊕n.

Proof Since Gn is finite, Φ(Gn) = Φ(G)n, the direct product of a copy of Φ(G) in each copy of G in Gn. Thus Gn/Φ(Gn) ∼= (G/Φ(G))n.

To show that L(Gn) ∼= L(G)⊕n, recall the definition of the product in (G/Φ(G))n, 1/p that is, g, h = [g, h] , as stated in Definition 2.1.2. If G is generated by g1, . . . , gr, n then G J is generatedK by gij = (1,..., 1, gj , 1,..., 1) with 1 ≤ i ≤ n and |{z} i-th entry 1 ≤ j ≤ r. Since [gij, gkl] = 1 if i 6= k, then the algebra multiplication of elements in different copies of G/Φ(G) always equals 0. We have then that if Gi is a coordinate n copy of G in G , then Gi with the multiplication , , which is isomorphic to L(G), n can be seen as a subalgebra of L(G ), and if Gi andJ K Gj are diferent copies of G in n G , then L(Gi), L(Gj) = 0. Since every element of L(G) can be written as a sum n J K n X M ⊕n gi with gi ∈ Gi, we have L(G) = L(Gi) = L(G) . i=1 i=1

Definition 3.2.3 If L is an IAC algebra, we define Z(L) = {v ∈ L | v, u = 0 for all u ∈ L}, and call it the center of L. J K

Notice that for any IAC algebra L, Z(L) is invariant under Aut(L), which implies that either Z(L) = L (and L is abelian), or Z(L) = 0.

With the next theorem we see that the sum of isomorphic IAC algebras is actually the only possible case of nonsimple IAC algebras.

Theorem 3.2.4 Let L be an IAC algebra of finite dimension. Then L = S⊕n, where S is a simple IAC algebra.

36 Proof If L is an abelian F-algebra, and has dimension n, then L = F⊕n, and each copy of F is a simple IAC algebra of dimension 1.

Suppose that L is nonabelian. Let I0 be a minimal ideal of L, and consider the set I = {I0α | α ∈ Aut(L)}. Let X J = I. (3.1) I∈I Since J contains all the possible images of I, the image of J by any automorphism of L must be J. Thus, since L is irreducible, we have L = J. X We show that the sum (3.1) is direct. Let I1 ∈ I and define J1 = I. We

I∈I\{I1} must show that I1 ∩ J1 = 0. If K ∈ I\{I1}, then K,I1 = 0. Hence I1,J1 = 0. Since I1 ∩ J1,L = I1 ∩ J1,I1 + J1 = I1 ∩ J1,I1J + IK1 ∩ J1,J1 =J 0, I1 ∩KJ1 ⊆ Z(L) =J 0. Thus KI1 ∩JJ1 = 0 and consequentlyK J the sumK (J3.1) is direct.K Thus L is the direct sum of the elements of I. Since L is finite-dimensional, |I| is finite and so I = {I0,I1,...,In} and L = I0 ⊕ I1 ⊕ · · · ⊕ In. Furthermore every automorphism of L permutes the subspaces I0,...,In. To finish this proof we only need to show that Aut(I0) is irreducible on I0. Suppose that I0 has a proper nonzero subspace S0 stable under all automorphisms of I0. Take ϕ ∈ Aut(L). Then I0ϕ = Ik for some k ∈ {0, . . . , n}. We claim that every automorphism of L that sends I0 to Ik sends also S0 to S0ϕ. Indeed, if ψ is such that I0ψ = Ik, but S0ψ 6= S0ϕ, −1 −1 then S0ϕψ 6= S0, thus ϕψ ∈ Aut(L) induces an automorphism of I0 such that −1 S0ϕψ 6= S0, a contradiction. This proves that the image of S0 in Ik by any automorphism of L that takes I0 to Ik is unique. For every k ∈ {0, . . . , n}, define Sk n X to be the image of S0 by the automorphisms of L that take I0 to Ik. Then Sk is k=0 a proper subspace of L stable under every automorphism of L, a contradiction.

Corollary 3.2.5 Let L be an IAC algebra of finite dimension and I0 be a minimal ideal of L. Set I = {I0α | α ∈ Aut(L)} to be the set of all images of I0 under automorphisms of L. Then

M 1. L = I; I∈I 2. The set I is the set of all minimal ideals of L.

37 Proof Item 1 follows from the proof of the previous theorem. To prove 2, if J/∈ I is a minimal ideal of L, then J, I ⊆ J ∩I, which, as both are minimal and J 6= I must M be equal to 0. As L = JI, thenK J, L = 0, that is J ≤ Z(L) = 0, a contradiction. I∈I J K

3.3 Normal subgroups and ideals

Since we have proved that there is a bijection between UCS p-groups of exponent 2 p and IAC algebras over Fp it is only natural to ask whether there is in each case some kind of correspondence between the normal subgroups, and the ideals of the algebra. This section is dedicated to answering this question.

We will start by defining powerful groups and powerfully embedded subgroups. These were first defined by Lubotzky and Mann in [18].

Definition 3.3.1 A finite p-group G such that p is odd and G0 ≤ Gp, or p = 2 and G0 ≤ G4 is said to be powerful.

A subgroup N of a p-group G is said to be powerfully embedded in G if p is odd and [N,G] ≤ N p or p = 2 and [N,G] ≤ N 4.

Notice that a powerfully embedded subgroup is always a normal subgroup.

Theorem 3.3.2 Let G be a UCS p-group of exponent p2 and L = L(G). Then there exists a bijection β between the set of powerful subgroups of G that contain Φ(G) and the set of subalgebras of L. Further, the image by β of the set of powerfully embedded subgroups of G containing Φ(G) is the set of ideals of L.

Proof Recall that subspaces of L are precisely subgroups of G. For a subgroup H ≤ G such that Φ(G) ≤ H, we denote H/Φ(G) by H. Note that H is a subgroup of G and can be viewed as a subspace of L. Set Hβ = H = H/Φ(G). Suppose that H is powerful. We claim that Hβ is a subalgebra of L. First note that Hβ is an 1/p Fp-subspace of L. Let x, y ∈ Hβ such that x, y ∈ G. Then x, y = [x, y] . Since p p J K H is powerful [x, y] can be written as [x, y] = h1 . . . hs with h1, . . . , hs ∈ H. Since p p h1, . . . , hs ∈ Z(G),

1/p p p 1/p [x, y] = (h1 . . . hs) = h1 ... hs ∈ H.

38 Thus H is closed for the product , and so H is a subalgebra of L. We will not distinguish in this proof between HJ,K the subgroup of G and H, the subalgebra of L.

For each H powerful subgroup of G containing Φ(G), set Hβ = H = H/Φ(G). The map β is injective by the isomorphism theorem.

We claim the image of β contains all subalgebras of L. Indeed, if H0 is a subalge- bra of L, then by setting H0 the preimage of H0 by the projection of G into G then H0 contains Φ(G) and H0β = H0. Hence we just need to show that H0 is powerful. As H0 is a subalgebra of L, then for every h1, h2 ∈ H0 and for any representatives 1/p h1 and h2 of h1 and h2, h1, h2 = [h1, h2] = h3 is an element of H0. By applying J K p p-th powering on both sides we get [h1, h2] = h3 for h3 a representative of h3 (thus h3 ∈ H0). Therefore H0 is powerful and hence β is surjective. Suppose now that H is powerfully embedded in G. As G is UCS of exponent p2, then it is either abelian, or Z(G) = Φ(G) = GpG0. Either way, Gp is abelian, as is Hp, hence every element of Hp is of the form hp for some h ∈ H. For every h ∈ H and g ∈ G, if h and g are respectively representatives of h and g in G then p p 1/p p 1/p [h, g] = h0 ∈ H , thus h, g = [h, g] = (h0) = h0. Hence H, G ⊆ H, that is H is an ideal of L. J K J K

If, on the other hand H is an ideal of L, then for every h ∈ H and g ∈ G, 1/p h, g = h0 ∈ H. Hence, for any representatives h of h and g of g,[h, g] = h0, J K p and by applying p-th powering to both sides we have [h, g] = h0, which, once again, as h and g are arbitary, then H is powerfully embedded in G.

We already know from Theorem 3.2.4 that every nontrivial proper ideal of an IAC algebra is a direct summand of the algebra. Something equivalent should then be true for UCS groups. We will prove that whenever a UCS group has a proper powerfully embedded subgroup, we can take one particular powerfully embedded subgroup and show that it is a direct factor of the group.

Corollary 3.3.3 Let G be a finite UCS p-group with odd exponent p2. There exists subgroup H of G satisfying the following conditions:

1. H is powerfully embedded in G and Φ(G) < H;

2. H is minimal among the subgroups of G satisfying 1.

k 2 Moreover G = H0 for some UCS subgroup H0 of exponent p , k ≥ 1, such that H0Φ(G) = H.

39 Proof Note that G is powerfully embedded in G and Φ(G) < G. Thus G satisfies (1). Since G is finite G contains a subgroup H minimal by inclusion that satisfies (1). Therefore the existence of H is established. Set L = L(G) and let I0 = H/Φ(G). By Proposition 2.1.3 L is IAC. Since H is powerfully embedded in G, Theorem 3.2.2 implies that I0 E L. Since H is minimal subject to (1), I0 is a minimal ideal of L. ⊕n Thus by Corollary 3.2.5 L = I0 for some n ≥ 1. Let H0 = G(I0). Then by Theorem 2 n n ⊕n 2.1.6 H0 is UCS of exponenet p and so is H0 . However L(H0 ) = L(H0) = ⊕n ⊕n n ∼ L(G(I0)) = I0 = L. Therefore L(H0 ) = L(G) and hence, by Theorem 3.2.2 n ∼ ∼ H0 = G. Finally L0 = HΦ(G) = H0/Φ(H0) implies that H0Φ(G) = H.

3.4 nCS groups

In Theorem 1.7.3, which was proved by Taunt in [26], we have conditions for a UCS group G that imply that its direct power Gn is also UCS. In this section we will prove that a similar result is valid for groups in which all the characteristic subgroups form a chain.

Definition 3.4.1 Let G be a group. We say that G is nCS if G has exactly n + 1 characteristic subgroups N0,...,Nn and {1} = N0 < N1 < ··· < Nn = G.

The only 0CS group is the trivial group. 1CS groups are the characteristically simple groups. 2CS groups are the UCS groups defined by Taunt.

Lemma 3.4.2 Let G be an nCS group, and N be characteristic in G. Then N is mCS for some m ≤ n.

Proof It follows trivially from noticing that a characteristic subgroup of N is a characteristic subgroup of G, and thus part of the chain of subgroups that we have for G.

The following theorem is a generalization of Theorem 1.7.3, and the proof uses a very similar argument.

Theorem 3.4.3 Let G be a finite nCS group with proper characteristic subgroups {1} = N0 < N1 ··· < Nn = G, such that for every i ∈ {0, . . . , n − 1} no element

40 of Ni+1/Ni other than Ni is fixed by every automorphism of Ni+1/Ni induced by an automorphism of G. Then the direct product of a finite number of groups, each isomorphic to G, is an mCS group for some 2 ≤ m ≤ n. Further, if for every i ∈ {0, . . . , n − 2} the characteristically simple groups Ni+2/Ni+1 and Ni+1/Ni are isomorphic to powers of nonisomorphic simple groups, then G is nCS.

a Proof Given the nCS group G, consider the group G = {(x1, . . . , xa) | xi ∈ G}. Suppose then that K ≤ Ga is a nontrivial characteristic subgroup of Ga. We will a a show that K must be equal to one of the Ni . Let i be minimal such that K ≤ Ni a a a and fix this i. Such an i exists, as K ≤ Nn = G . We need to show that K ≥ Ni .

a We recall that, as mentioned in Section 1.7, Aut(G)oSa is a subgroup of Aut(G ), each element of the base group B act as an element of Aut(G) on each coordinate a subgroup of G and the top subgroup S = {(1, s) | s ∈ Sa} acts by permuting the coordinates.

By the minimality of i, K contains an element k = (k1, . . . , kn) such that kj ∈ Ni\Ni−1 for some j. Suppose without loss of generality that j = 1. Let φ ∈ Aut(G) be such that (k1Ni−1)φ 6= k1Ni−1. Such φ ∈ Aut(G) exists by hypothesis as k1Ni−1 6= Ni−1. Set (φ1, 1) ∈ B where φ1 = (φ, 1,..., 1). Consider the element, −1 −1 ∗ ∗ l = (k(φ1, 1))k = ((k1φ)k1 , 1,..., 1). Then l ∈ K ∩ Ni and l∈ / K ∩ Ni−1, where ∗ ∗ Ni is the copy of Ni contained in G , the first copy of G in the direct product. Since for every automorphism ψ of G we can take an element of the base group of ∗ Aut(G) o Sa that acts as ψ on the first entry and as the identity in all others, K ∩ G ∗ ∗ ∗ ∗ ∗ and Ni both are characteristic in G , K ∩ Ni = (K ∩ G ) ∩ Ni also is characteristic ∗ ∗ ∗ ∗ in G . Since K ∩ Ni is not contained in Ni−1, K ∩ Ni cannot be contained in any ∗ ∗ ∗ characteristic subgroup of G smaller than Ni , thus it must be Ni . As the images ∗ of Ni by elements of the top group S are copies of Ni in each copy of G in the direct a product, we have that K ≥ Ni . This shows that the only possible characteristic a a subgroups of G are of the form Ni . We have then proved that Ga is mCs with m ≤ n. To show that m ≥ 2 we need only to show that Ga is not characteristically simple, but as G∗ is normal in Gn and G is not characteristically simple, by Lemma 1.2.7, Ga is not characteristically simple.

si Suppose now that the characteristically simple groups Ni/Ni−1 = Ti are such that Ti  Ti+1. Since a a ∼ asi Ni /Ni−1 = Ti a a we have that each Ni /Ni−1 is characteristically simple and isomorphic to the product of copies of Ti.

41 a a a a a Suppose that Ni 6= G is characteristic in G and Ni+1 is not. If Ni+k is the a a a minimal characteristic subgroup of G that strictly contains Ni , then Ni+1/Ni and a a a a Ni+2/Ni both are normal subgroups of the characteristically simple group Ni+k/Ni , and by Lemma 1.2.6 are direct products of isomorphic simple groups, a contradiction.

3.5 Abelian nCS Groups

In this section we will classify the finite abelian nCS groups.

To avoid confusion, we will denote in this section the direct product of a group G by itself n times by G×n, while Gn will denote the subgroup of G given by Gn = hgn | g ∈ Gi.

×d Lemma 3.5.1 Consider the group G = Cpn = Cpn × · · · × Cpn . If g1, g2 ∈ G are | {z } d times such that |g1| = |g2|, then there is an automorphism ϕ of G such that g1ϕ = g2.

α1 αd Proof Let Cpn = hzi. The elements of G can be written as the d-tuples (z , . . . , z ) n where 0 ≤ αi < p for all i. To prove this lemma we just need to construct for each g ∈ G of order pn−k with 0 ≤ k ≤ n, an automorphism ϕ ∈ Aut(G) such that g = (zpk , 1,..., 1)ϕ. Since |g| = pn−k, then g ∈ Gpk . Hence there are non-negative k k p q1 p qd n−k integers q1, . . . , qd such that g = (z , . . . , z ). As |g| = p , there exists i0 such that p - qi0 . For 1 ≤ i ≤ d set ei = (1,..., 1, z , 1,..., 1). Set also |{z} i-th entry q1 qd g = (z , . . . , z ). By Theorem 1.3.4, the sets X = {e1, . . . , ei0−1, g, ei0+1, . . . , ed} and Y = {e1, . . . , ed} are both minimal generating sets for G. Since the elements of X and the elements of Y satisfy the same relations, then there is ϕ ∈ Aut(G) that takes

• e1ϕ = g;

• eiϕ = ei−1 for i = 2, . . . , i0 ;

• eiϕ = ei for i = i0 + 1, . . . , d.

k k k k k k p p p p p q1 p qd Now (z , 1,..., 1)ϕ = e1 ϕ = (e1ϕ) = (g) = (z , . . . , z ) = g, which completes the proof.

42 The following theorem is a direct consequence of [11, Theorem 3.3] but the proof here uses a different technique.

Theorem 3.5.2 Let p be a prime. A finite abelian p-group G is nCS if and only ×d ×k ×(d−k) if G has, for some integer d > 0, the form, G = Cpn or G = Cpm × Cpm−1 where 1 ≤ k < d and n = 2m − 1.

×d ×k ×(d−k) Proof We will prove first that the groups G = Cpn and G = Cpm × Cpm−1 with 1 ≤ k < d and n = 2m − 1 are nCS, and then prove these are the only nCS abelian groups.

×d Let first G = Cpn and N be a characteristic subgroup of G. Let g ∈ N an element of maximal order and assume |g| = pk. By Lemma 3.5.1 N contains all elements of order pk, thus Gpn−k ≤ N. On the other hand, Gpn−k contains all elements g ∈ G such that |g| ≤ pk. Thus N ≤ Gpn−k and so N = Gpn−k . Hence the characteristic subgroups of G are the groups of the form Gpn−k with 0 ≤ k ≤ n. These n + 1 groups form a chain, and so G is nCS.

×k ×(d−k) Let now G = Cpm × Cpm−1 where 1 ≤ k < d. For each pair of automorphisms ×k ×(d−k) φ1 ∈ Aut(Cpm ) and φ2 ∈ Aut(Cpm−1 ) we have an automorphism of G,(φ1, φ2), ×k ×(d−k) that acts as φ1 on Cpm and φ2 on Cpm−1 . If N is a characteristic subgroup of G, ×k ×(d−k) let g ∈ N and write g = g1g2 with g1 ∈ Cpm and g2 ∈ Cpm−1 . If Cpm = hz1i and Cpm−1 = hz2i, set for i = 1, . . . , k, the elements ei = (1,..., 1, z1 , 1,..., 1) |{z} i-th entry and for i = k + 1, . . . , d, the elements fi = (1,..., 1, z2 , 1,..., 1). Then |{z} i-th entry m m−1 p /|g1| p /|g2| ×k |e1 | = |g1| and |fd | = |g2|. By Lemma 3.5.1, there exist φ1 ∈ Aut(Cpm ) m m−1 ×(d−k) p /|g1| p /|g2| and φ2 ∈ Aut(Cpm−1 ) such that g1φ1 = e1 and g2φ2 = fd . Hence m m−1 m m−1 p /|g1| p /|g2| p /|g1| p /|g2| setting φ = (φ1, φ2), gφ = e1 fd . Therefore e1 fd ∈ N. Sup- pose now |g1| > |g2| and consider the sets S1 = {e1, . . . , ek, fk+1, . . . , fd} and S2 = − |g1| p|g2| {e1fd , e2, . . . , ek, fk+1, . . . , fd} (notice that −|g1|/p|g2| is an integer). As the sets S1 and S2 both generate G and as both satisfy the same relations, then there is an au- tomorphism ψ of G mapping the elements from S1 to the elements of S2 in the order m m−1 m p /|g1| p /|g2| p /|g1| they are presented above. Notice that (e1 fd )ψ = e1 . Since N is char- m m−1 m p /|g1| p /|g2| p /|g1| acteristic and e1 fd ∈ N, then e1 ∈ N, and, once again by Lemma −1 3.5.1 we get that (g1, 1) ∈ N. This then implies that (g1, g2)(g1, 1) = (1, g2) ∈ N. m ×k p /|g1| ∼ ×k We then have N ≥ (C m ) × 1 = C × 1, and similarly, N also contains p |g1| ×(d−k) m−1 ×(d−k) p /|g2| ∼ 1 × (C m−1 ) = 1 × C . If |g | ≥ |g |, we can apply a similar argument p |g2| 2 1

43 −p|g2|/|g1| by taking S2 = {e1, . . . , ek, fk+1, . . . , fd−1, e1 fd}, and conclude once again m ×(d−k) m−1 ×(d−k) ×k p /|g1| ∼ ×k p /|g2| ∼ that N ≥ (C m ) × 1 = C × 1, and N ≥ 1 × (C m ) = 1 × C . p |g1| p −1 |g2| a Thus, by taking a maximal such that (g1, g2) ∈ N and |g1| = p , and b also maximal b ×k pm−a ×(d−k) pm−1−b such that (h1, h2) ∈ N and |h2| = p , we have N ≤ (Cpm ) × (Cpm−1 ) and ×k pm−a ×(d−k) pm−1−b ∼ ×k ×(d−k) conclude that N = (Cpm ) × (Cpm−1 ) = Cpa × Cpb .

We claim that a − b = 0 or a − b = 1. To show this, take S1 = {e1, . . . , ek, fk+1, . . . , fd} and S2 = {e1fd, e2, . . . , ek, fk+1, . . . , fd}. Once again, both these sets generate G, and as they satisfy the same relations, there is an automorphism φ that takes the elements of S1 to the elements of S2 in the order they are shown pm−a pm−a −pm−a pm−a above. Since e1 ∈ N, then (e1 )φe1 = fd ∈ N, so m − a ≥ m − 1 − b from where we get a − b ≤ 1. Similarly, for S1 = {e1, . . . , ek, fk+1, . . . , fd} p and S2 = {e1, e2, . . . , ek, fk+1, . . . , fde1}, the resulting automorphism φ is such that pm−1−b pm−1−b pm−b fd φfd = e1 from where we get m − b ≥ m − a, thus a ≥ b. The candidates for characteristic subgroups left are the groups of the form N ∼= ×k ×(d−k) ∼ ×k ×(d−k) Cpa × Cpa−1 and N = Cpa × Cpa , but these are all characteristic as they are pm−a pm−a given respectively by G and Ωm−a(G) = {g ∈ G | g = 1} for 0 ≤ i ≤ m − 1. They also form a chain as

pm−1 pm−2 p0 1 = Ω0 < G < Ω1 < G < ··· < Ωm−1 < G = G.

Thus in this case, as G has 2m characteristic subgroups and n = 2m − 1, G is nCS as claimed.

It remains to prove that if G is finite, abelian and nCS, it is of one of the forms ×d ×k ×(d−k) G = Cpn or G = Cpm × Cpm−1 . As G is abelian, by the fundamental theorem of abelian groups, we can write

∼ ×m1 ×mc G = Cpµ1 × · · · × Cpµc with µ1 > ··· > µc > 0 and mi ≥ 1 for all i. Suppose c ≥ 3. Then pµ3 ∼ G = Cpµ1−µ3 × Cpµ2−µ3 is a characteristic subgroup of G, as also is p ∼ ×m1 ×mc Ω1(G) = {g ∈ G | g = 1} = Cp × · · · × Cp . For G to be nCS we need pµ3 pµ3 that G ≤ Ω1(G) or Ω1(G) ≤ G . The second cannot occur due to Ω1(G) having pµ3 a minimal generating set with m1 + ··· + mc generators while G can be generated pµ3 by m1 + m2 < m1 + ··· + mc generators. For G ≤ Ω1(G), we need µ1 − µ3 ≤ 1 and µ2 − µ3 ≤ 1, but µ1 − µ3 ≥ 2. From this contradiction we conclude that c ≤ 2.

×m1 ×m1 If c = 1, then G = Cpµ1 , and as G is nCS and we have proved that Cpµ1 m1 is µ1CS we have µ1 = n, thus G = Cpn as desired. Suppose then c = 2. Then µ p 2 ∼ ×m1 G = Cpµ1−µ2 and once again this must contain or be a subgroup of Ω1(G). Since

44 pµ2 G has a minimal generating set with m1 generators, then the only possibility pµ2 is G ≤ Ω1(G), but this implies µ1 − µ2 ≤ 1. Hence µ2 = µ1 − 1 and G = ×m1 ×m2 µ Cp 1 × Cpµ1−1 . As we have shown, this group is (2µ1 − 1)CS, and as G is nCS, we ∼ ×k ×(d−k) have 2µ1 − 1 = n, that is µ1 = m and G = Cpm × Cpm−1 as we wanted.

45 Chapter 4

Examples

First in Section 4.1, we classify 3-dimensional nonabelian IAC algebras. The main result of this section is that these algebras are simple Lie algebras.

Then in sections 4.2 and 4.3 we construct two infinite families of simple IAC algebras.

In the final Section 4.4 we adapt the Clebsch-Gordan formula for exterior square of a representation of the group SL(2, F) to obtain a family of simple IAC algebras whose automorphism groups contain SL(2, F) acting irreducibly.

Throughout this chapter, if a is invertible in Z/bZ we will denote the multiplica- tive order of a (mod b) by ordb(a).

4.1 IAC Algebras of dimension 3

Let L be an IAC algebra. If dim L = 1, then dim(L∧L) = 0 and hence in this case L is abelian. If dim L=2 and L = hu, vi, then L ∧ L = hu ∧ vi and so dim(L ∧ L) = 1. Thus L, L = L is not possible and we obtain that L is abelian. Thus there are no nonabelianJ K IAC algebras of dimensions 1 or 2. This section is dedicated to the nonabelian IAC algebras of smallest possible dimension.

The proof of the following theorem uses ideas from the proof of [8, Lemma 7].

Theorem 4.1.1 Let F be a field with char F 6= 2 and L be a nonabelian IAC F- algebra of dimension 3. Then L is a simple Lie algebra.

46 Proof Let e = {e1, e2, e3} be a basis for L. Since L, L = L, we have f = {f1, f2, f3} where f1 = e2, e3 , f2 = e3, e1 and f3 = e1,J e2 (laterK it will be clear why this order) is alsoJ a basisK for LJ. For KL to be a LieJ algebra,K it must additionally satisfy the Jacobi identity, that is,

e1, e2, e3 + e2, e3, e1 + e3, e1, e2 = e1, f1 + e2, f2 + e3, f3 = 0. J J KK J J KK J J KK J K J K J K 3 X If we write fi = αijej, we have that the Jacobi identity is equivalent to j=1

α12 e1, e2 + α13 e1, e3 + α21 e2, e1 + α23 e2, e3 + α31 e3, e1 + α32 e3, e2 = 0, J K J K J K J K J K J K which, in turn, is equivalent to

α12f3 − α13f2 − α21f3 + α23f1 + α31f2 − α32f1 = 0, that is α12 = α21, α13 = α31, α23 = α32. Then L is a Lie algebra if and only if the matrix A = [αij] is symmetric. We will then show that A is symmetric.

Let g ∈ Aut(L), and let [g]e and [g]f denote the matrices that represent g in the bases e and f.

If e1g = g11e1 + g12e2 + g13e3;

e2g = g21e1 + g22e2 + g23e3;

e3g = g31e1 + g32e2 + g33e3; i.e.   g11 g12 g13 [g]e =  g21 g22 g23  . g31 g32 g33 We can see that

f1g = e2, e3 g = e2g, e3g J K J K = (g22g33 − g23g32) e2, e3 + (g23g31 − g21g33) e3, e1 + (g21g32 − g22g31) e1, e2 J K J K J K = (g22g33 − g23g32)f1 + (g23g31 − g21g33)f2 + (g21g32 − g22g31)f3; and similarly f2g = e3, e1 g = (g32g13 − g33g12)f1 + (g33g11 − g31g13)f2 + (g31g12 − g32g11)f3; J K f3g = e1, e2 g = (g12g23 − g13g22)f1 + (g13g21 − g11g23)f2 + (g11g22 − g12g21)f3. J K

47 Thus   g22g33 − g23g32 g23g31 − g21g33 g21g32 − g22g31 [g]f =  g32g13 − g33g12 g33g11 − g31g13 g31g12 − g32g11  . g12g23 − g13g22 g13g21 − g11g23 g11g22 − g12g21 T Therefore the matrix ([g]f ) is equal to the classical adjoint of [g]e (see [10, page 353] for the definition of the classical adjoint matrix). By [10, Chapter VI, Proposition 3.7], T [g]f [g]e = det[g]eI, where I is the identity 3 × 3 matrix, which implies that

−T [g]f = det[g]e[g]e −T where [g]e is the inverse transpose matrix of [g]e. For v ∈ L, let [v]e and [v]f denote vector representations of v in the bases e and f respectively. The matrix A defined in the first paragraph of the proof acts as a basis transformation matrix −1 from the basis f to the basis e and [v]f A = [v]e. Therefore [g]e = A [g]f A and so −1 −T −1 −1 T [g]e = A det([g]e)([g]e) A, which is equivalent to det[g]eA = [g]eA [g]e . Thus Aut(L) = {g ∈ GL(L) | det(g)A−1 = gA−1gT }.

Consider now the group Γ = {g ∈ GL(L) | g(A−1 − A−T )gT = det(g)(A−1 − A−T )}. Clearly Aut(L) is a subgroup of Γ. Notice now that Γ fixes the subspace W = {v ∈ L | v(A−1 − A−T ) = 0}, because if w ∈ W and g ∈ Γ, then wg(A−1 − A−T ) = wg(A−1 − A−T )gT g−T = det(g)w(A−1 − A−T )g−T = 0. Since Aut(L) acts irreducibly on L, we have W = {0} or W = L. The first case implies that A−1 − A−T is invertible, while the second that A−1 − A−T = 0, which is what we need to finish the proof. We then need to show that A−1 − A−T is not invertible. Indeed, det(A−1 − A−T ) = det((A−1 − A−T )T ) = det(A−T − A−1) = (−1)3 det(A−1 − A−T ) and as the characteristc of the field F is not 2 we have det(A−1 − A−T ) = 0.

As is well-known, over an algebraically closed field or over a finite field there is a unique isomorphism type of simple Lie algebras of dimension 3 (See [24, Theorem 3.2]). Over R, there are two such isomorphism types, while over Q there are infinite isomorphism types. Hence we obtain an alternative proof for the following theorem, which has been proved in [8, Theorem 8, (c)].

Theorem 4.1.2 For an odd prime p, there is a unique isomorphism type of non- abelian UCS p-groups of exponent p2 and order p6.

48 4.2 A family of IAC algebras with an irreducible cyclic group

In This section we will construct an infinite family of simple IAC algebras.

Lemma 4.2.1 Let q and p be primes such that ordq(p) = q−1. Then the polynomial q−1 q x + ··· + x + 1 = (x − 1)/(x − 1) is irreducible in Fp.

Proof As xq−1 + ··· + x + 1 = (xq − 1)/(x − 1), then in the algebraic algebraic q−1 closure F of Fp, x + ··· + x + 1 has as roots the q-th roots of unity different from 1. Since q is a prime, there are q −1 primitive q-th roots of 1. Thus xq−1 +···+x+1 is the q-th cyclotomic polynomial over Fp (defined in [17, Definition 2.44]). By [17, Theorem 2.47] g(x) is irreducible.

Lemma 4.2.2 Let p and q be prime numbers for which ordq(p) = q − 1. Let ζ be p pq−2 2 q−1 a primitive q-th root of unity in Fpq−1 . Then {ζ, ζ . . . , ζ } and {ζ, ζ , . . . , ζ } are bases for Fpq−1 over Fp.

Proof Let f(x) = xq−1 + xq−2 + ··· + x + 1. Then ζ is a root of f(x). If φ denotes the Frobenius automorphism, as f(ζ) = 0, then 0 = f(ζ)φ = f(ζp), thus every element in the orbit of ζ under φ is also a root of f(x). As ordq(p) = q − 1, we have that pq−1 ≡ 1 mod q and pi 6≡ 1 mod q if 1 ≤ i ≤ q − 2. Consider the elements ζ, ζp, . . . , ζpq−2 . By the argument above, these are roots of f(x). Further, we claim that these elements are pairwise distinct. Assume that ζpi = ζpj with 0 ≤ i < j ≤ q − 2. Then 1 = ζpj −pi = ζpi(pj−i−1). Since ζ is an element of order q, q | pi(pj−i − 1). Since q and p are distinc primes q | pj−i − 1. Hence pj−i ≡ 1 mod q. Since 1 ≤ j − i < q − 1, this is a contradiction. Thus ζ, ζp, . . . , ζq−2 are pairwise distinct as claimed. As all of these are q-th roots of unity, and all are different from 1, we have {ζ, ζp, . . . , ζpq−2 } = {ζ, ζ2, . . . , ζq−1}. Suppose that {ζ, ζ2, . . . , ζq−1} is linearly dependent, then there is a linear combination such that 2 q−1 α1ζ + α2ζ + ··· + αq−1ζ = 0. Then dividing both sides by ζ we obtain 0 = q−2 q−2 α1 + α2ζ + ··· + αq−1ζ , from where we get g(x) = α1 + α2x + ··· + αq−1x is a polynomial of degree q − 2 with ζ as a root. On the other hand, ζ is a root of the polynomial f(x) = xq−1+xq−2+···+x+1 which is irreducible by Lemma 4.2.1. Hence (x−ζ) is a divisor of f(x) and g(x). Since f(x) is irreducible and degf(x) > degg(x), gcd(f(x), g(x)) = 1, which is a contradiction. Thus {ζ, ζ2, . . . , ζq−1} is a basis of p pq−2 Fpq−2 as claimed. By the equality above, {ζ, ζ , . . . , ζ } is also a basis of Fpq−2 .

49 Theorem 4.2.3 Let q 6= 2 and p be prime numbers for which ordq(p) = q − 1, and let k, l be integers satisfying 0 ≤ k, l ≤ q − 1 and pk + pl ≡ 1 mod q. Then there is an IAC algebra L of dimension q − 1 over Fp given by:

1. (L, +) = Fpq−1 as Fp vector spaces;

2. The product is given by the Fp-bilinear alternating map

, : a, b = apk bpl − apl bpk . J K J K Additionally, there is a cyclic subgroup of Aut(L) that acts irreducibly on L. Also, L is abelian if and only if k = l.

Note that integers k 6= l satisfying the conditions above always exist if q > 3, k for, as p is primitive in Fq, we can take k and l smaller than q − 1 such that p ≡ 2 mod q and pl ≡ −1 mod q.

We will show later, in Proposition 4.2.4, that there exist infinitely many pairs of primes p and q satisfying the conditions of the theorem and hence this construction leads to an infinite family of nonabelian IAC algebras!

Proof Set L = Fpq−1 , and define a bilinear product , on L as in assertion 2 of the q−1 J K theorem. By hypothesis, p ≡ 1 mod q hence the field Fpq−1 contains a primitive q-th root of unity. Fix one specific such root ζ and define an Fp-linear map β by β : L → L aβ = aζ.

Note that aβ, bβ = aζ, bζ = (aζ)pk (bζ)pl − (aζ)pl (bζ)pk = (apk bpl − apl bpk )ζpk+pl and since,J by hypothesis,K J pk K+ pl ≡ 1 mod q, then aβ, bβ = a, b β, thus β is an endomorphism of the algebra L. J K J K

2 q−1 By Lemma 4.2.2, {ζ, ζ , . . . , ζ } is an Fp-basis of L. Since multiplication by × 2 q−1 an element a ∈ Fpq−1 is Fp-linear and invertible, {aζ, aζ , . . . , aζ } is also a basis for L for all a ∈ L\{0}. Let then S be a nonzero subspace of L invariant under hβi. If s ∈ S, s 6= 0, then sβ = sζ, . . . , sβq−1 = sζq−1 ∈ S, thus S contains the basis {sζ, . . . , sζq−1} for L , hence S = L. Thus hβi ≤ Aut(L) acts irreducibly on L.

We now prove that the resulting algebra is abelian if and only if k = l. If k = l, then a, b = 0 for all a, b ∈ L and the algebra is abelian. On the other J K × hand, if k 6= l, then take a generator z for Fpq−1 . We will show that the product

50 z, z2 = zpk+2pl − z2pk+pl is nonzero. Indeed, it will be zero only if zpk+2pl = z2pk+pl , whichJ K is equivalent to the following: zpk−pl = 1 (pq−1 − 1)|pk − pl (pq−1 − 1)|pk(1 − pl−k) (pq−1 − 1)|pl−k − 1. Since we can assume l > k and we had l, k ≤ q − 1, we have that the last cannot occur. So we conclude that z, z2 6= 0, hence L is nonabelian in this case. J K Even though we have the restriction that p must be a prime and a primitive root in Fq, we still have that these algebras exist for every q due to the following result:

Proposition 4.2.4 For every prime q there are infinitely many primes p such that ordq(p) = q − 1.

Proof Let a ∈ {1, . . . , q − 1} be an integer such that a considered as an element × of Fq is a generator for Fq . Since a ≤ q − 1 and q is prime, we have gcd(a, q) = 1. Thus, by Dirichlet’s Theorem on primes in arithmetic progression (see [1, Theorem 7.9]) there are infinitely many primes p of the form a + kq with k ∈ N. If p is such a prime, then p = a + kq ≡ a mod q. Since ordq(a) = q − 1, we have that ordq(p) = q − 1.

Proposition 4.2.5 The nonabelian algebras constructed in Theorem 4.2.3 are sim- ple.

Proof Suppose the algebra L constructed as in Theorem 4.2.3 is nonabelian and let I be a minimal ideal of L. By Corollary 3.2.5, L is the direct sum of the images of I by automorphisms of L. Let β be the automorphism of L defined in the proof of Theorem 4.2.3. Note that β has prime order q. By the proof of Theorem 4.2.3, if q−1 X a ∈ I, a 6= 0, then {aβ, . . . , aβq−1} is a basis for L, so L = Iβj. Hence L is the j=1 sum of the minimal ideals in the hβi-orbit O = {I,Iβ,...,Iβq−1}. Since |β| = q is a prime, O contains either 1 or q elements. If |O| = q, then by Corollary 3.2.5, the direct sum of the elements of O is contained in L. Since dim(L) = q − 1, this is a contradiction. If |O| = 1 then Iβ = I for all i ∈ {0, . . . , q − 1} and so I = L. Hence I = L and L is simple.

51 4.3 Another family of IAC algebras

In this section we will construct another infinite family of IAC algebras.

We start the construction by first setting some parameters. Let b ≥ 2 be an 2 integer, n > 1 a divisor of b + b − 1 and suppose r = ordn(b) (that is, r is the minimal positive integer such that br ≡ 1 mod n) and r > 2. Let p be a prime such that p - n. Then gcd(p, n) = 1, and hence p is invertible modulo n. Hence there exists some a such that pa ≡ 1 mod n, that is n|(pa − 1). Let q = pa.

Set ζ to be a primitive n-th root of 1 in Fq and also set the following matrices

 0 1 ... 0   ζ 0 ... 0   . . .. .   0 ζb ... 0   . . . .    A =   and B =  . . .. .  .  0 0 ... 1   . . . .  1 0 ... 0 0 0 . . . ζbr−1

It is immediate to see that both Ar and Bn are equal to the identity r × r matrix. Note that  0 ζ 0 ... 0   0 0 ζb ... 0    BA =  ......  ,  . . . . .  ζbr−1 0 0 ... 0  ζb−1 0 ... 0   0 ζ 0 ... 0  b −1  0 ζ . . . 0   0 0 ζ ... 0  b     AB = A  . . .. .  =  ......  .  . . . .   . . . . .  0 0 . . . ζbr−2 ζb−1 0 0 ... 0 Since br ≡ 1 mod n, ζbr = 1, and so ζb−1 = ζbr−1 . Thus BA = ABb−1 , that is, BA = Bb−1 .

r Let {e0, . . . , er−1} be the canonical basis of Fq. Notice that each ei is an eigen- vector for B associated with the eigenvalue ζbi . We claim that the eigenvalues ζ, ζb, . . . , ζbr−1 are pairwise distinct. Assume that ζbi = ζbj with some 0 ≤ i < j ≤ r − 1. Then 1 = ζbj −bi = ζbi(bj−1−1). Since the order of ζ is n, this implies that n | bi(bj−i − 1). As n | b2 + b − 1, gcd(n, b) = 1, and so n | bj−i − 1, that is, bj−i ≡ 1 mod n. As ordn(b) = r and 1 ≤ j − i < r − 1, this is a contradiction. Therefore the elements ζ, ζb, . . . , ζbr−1 are pairwise distinct, as claimed.

If S is a subspace fixed by B and s = α0e0 + ··· + αr−1er−1 ∈ S such that αi 6= 0, we claim that ei ∈ S. We prove this by induction on the number of

52 non-zero coefficients of s. If this number is 1, there is nothing to show. Assume that the claim holds if s has k ≥ 1 non-zero coefficients. Suppose now that s has k + 1 non-zero coefficients including αi. Suppose that j 6= i and αj 6= 0. Then 0 bj bj br−1 bj s := sB − ζ s = α0(ζ − ζ )e0 + ··· + αr−1(ζ − ζ )er−1. As was shown above, bi bj 0 0 ζ 6= ζ . Assuming that s = β0e0 + ··· + βr−1er−1 we have that s ∈ S is such that βk = 0 if and only if αk = 0 for all k 6= j, but βj = 0. Hence the induction hypothesis 0 applies to s and so ei ∈ S. Hence we obtain that S is of the form S = hei | i ∈ Ii for I a subset of {0, . . . , r − 1}.

For the space hei | i ∈ Ii to be fixed also by A, as eiA = ei+1 for i < r − 1 and er−1A = e0, we have that the set I must be either empty, or equal to {0, . . . , r − 1}, r that is, a fixed subspace by the two linear maps A and B is either equal to Fq or 0, r so G = hA, Bi acts irreducibly on Fq.

r Let V = Fq as a G-module. Then, as noted in Section 1.6 the exterior square V ∧ V is also a G-module with the action (v1 ∧ v2)g = (v1g) ∧ (v2g). Moreover, as vector spaces, V ∧ V = U1 ⊕ U2 where

U1 = hei ∧ ej | j − i ≡ 1 mod ri = he0 ∧ e1, e1 ∧ e2, . . . , er−2 ∧ er−1, er−1 ∧ e0i ,

U2 = hei ∧ ej | j − i 6≡ 1 mod ri . r
Clearly, dim(U1) = r and dim(U2) = 2 − r = r(r − 3)/2.

Proposition 4.3.1 Under the conditions above

1. V ∧ V = U1 ⊕ U2 is G-invariant, that is, both U1 and U2 are invariant under G;

2. The map φ : V → U1 defined by eiφ = ei+1 ∧ei+2 if i < r −2, er−2φ = er−1 ∧e0 and er−1φ = e0 ∧ e1 is a G-module isomorphism. ∼ 3. V = V ∧ V/U2 as G-modules.

Proof

1. Since eiA = ei+1, and j − i = (j + 1) − (i + 1), U1 and U2 are invariant under A. Since ζ ∈ Fq, any subspace of the form hei ∧ eji is invariant under B, and as both U1 and U2 are generated by elements of the form ei ∧ ej for some i and j, we have that both are invariant under B. Hence U1 and U2 are invariant under G.

53 2. To show that φ is a G-module homomorphism, since G = hA, Bi, we only need to show it commutes with both A and B for the elements in the basis of V . Considering the indexes below modulo r,

(eiA)φ = ei+1φ = ei+2 ∧ ei+3;

(eiφ)A = (ei+1 ∧ ei+2)A = ei+2 ∧ ei+3; bi bi (eiB)φ = (ζ ei)φ = ζ (ei+1 ∧ ei+2);

bi+1+bi+2 (eiφ)B = (ei+1 ∧ ei+2)B = ζ (ei+1 ∧ ei+2) = bi(b+b2) bi = ζ (ei+1 ∧ ei+2) = ζ (ei+1 ∧ ei+2);

Where the last equality follows from the hypothesis that n = |ζ| is a divisor of b2 + b − 1. Finally, to show it is an isomorphism, it suffices to see that both modules have the same dimension r and the image of the basis {e0, . . . , er−1} of V is a basis of U1. ∼ ∼ 3. Since V ∧ V = U1 ⊕ U2 and U1 = V by item 2, we have that (V ∧ V )/U2 = ∼ U1 = V .

The G-module V is an irreducible ESQ module, and thus we can associate with it an IAC algebra and a UCS p-group. Since e0, e1 = e2 6= 0, these algebras are not abelian (and consequently, the groups are alsoJ notK abelian).

2 Corollary 4.3.2 Let b ≥ 2 be an integer, n > 1 a divisor of b +b−1, let r = ordn(b) be such that r > 2 and let q be a power of a prime such that n|q − 1. Then there is a simple IAC algebra L over Fq with dimension r given by:

1. L as an Fq-vector space has a basis {e0, . . . , er−1}; 2. The anticommutative product for the basis above is given by:

ei, ej = ei−1 if j − i ≡ 1 mod r J K ei, ej = 0 if j − i 6≡ ±1 mod r; J K where the indexes above are modulo r.

54 Proof Let V be the ESQ module from the previous proposition and set ψ as the projection from V ∧V to V ∧V/U2 where U2 is as defined in the previous proposition. Then the algebra L = (V, ϕ) as defined in Section 2.3 is IAC, has basis {e0, . . . , er−1} and product as in the statement of this corollary, thus all we need to show is that L is simple. Let I be an ideal of L, suppose v = v0e0 + ··· + vr−1er−1 6= 0 is an element of I, and suppose without loss of generality that v1 6= 0. Then

v, e1 , e0 = v0er−1 − v2e0, e0 = v0er−2. JJ K K J K This means er−2 ∈ I, and as ei−2 = ei−1, ei for i = 2, . . . , r − 1 and er−1 = e0, e1 , then all elements of the basis {e0, . . .J , er−1} areK elements of I from where we concludeJ K that I = L.

To prove Proposition 4.3.5 we will need the law of quadratic reciprocity, which was conjectured by Legendre and Euler, and later proved by Gauss in [7].

Definition 4.3.3 Let p be a prime number. An integer a is a quadratic residue modulo p if it is congruent to a perfect square modulo p and is a quadratic non- residue otherwise. The Legendre symbol is a function of a and p defined by

 1 if a is a quadratic residue modulo p and a 6≡ 0 mod p, a  = -1 if a is a quadratic non-residue module p, p  0 a ≡ 0 mod p.

Theorem 4.3.4 (Quadratic reciprocity) If p and q are distinct odd primes, then

p q  = (−1)(p−1)(q−1)/4. q p

Proof See [1, Theorem 9.8].

Proposition 4.3.5 If p is a prime that divides b2 + b − 1 for some integer b, then p is congruent to 0, 1 or −1 mod 5.

Proof Let p be a prime that divides b2 + b − 1 for some integer b. Clearly p 6= 2. 2 2 1 2 5 Since b + b − 1 ≡ 0 mod p, then b + b − 1 = (b + 2 ) − 4 is congruent to 0 mod p. 1 2 Also, since 0, (b + 2 ) and 4 are all squares in Fp, then 5 needs also to be a square. 5 p 5−1 4 (p−1) p−1 By quadratic reciprocity, ( p )( 5 ) = −1 = (−1) = 1, and since 5 is a square

55 5 p in Fp, then ( p )=1. Therefore ( 5 ) = 1, that is, p must be congruent to a square in F5, but the only squares in F5 are 0, 1 and 4 = −1.

Notice that this construction for IAC algebras and the one in the previous section cover different cases, for example, the construction in the previous section only leads to algebras of dimension q − 1 where q is a prime, whereas the one in this section does not have this restriction, for example, if b = 3, n = 11, r = 5, and q = 35 = 243 we have an algebra of dimension 5. On the other hand, the previous proposition shows that this construction is not a generalization of that made in the previous section, for that one could be done for any prime q 6= 2, but here the equivalent parameter, n, is always congruent to 1, −1 or 0 modulo 5.

4.4 The Clebsch-Gordan formula for the exterior product

The Clebsch-Gordan formula describes tensor products of irreducible representa- tions of classical groups. It has an important role in quantum mechanics. Here we will prove a similar result for the exterior product of irreducible representations of SL(2, F), and use it to find a class of ESQ modules.

Let F be a field. For every integer m ≥ 0, we can define Vm as the F-vector space of the polinomials in F[x, y] that are homogeneous of degree m. The monomials i m−i x y for i = 0, . . . , m form a basis for Vm, from where we obtain that dim(Vm) = m + 1.

We will show that for some m, Vm with an SL(2, F) action is an ESQ-module.

First notice that we can give F[x, y] an F-linear SL(2, F)-action defined on the  a b  monomials xαyβ by xαyβ = (ax + by)α(cx + dy)β. Notice that if f, g ∈ c d F[x, y], and s ∈ SL(2, F), then (fg)s = (fs)(gs). This action also is such that Vms = Vm for all m. In this section we will always consider Vm as an SL(2, F)- module.

Lemma 4.4.1 Let char F = 0 or char F = p with p > m. Then Vm with the SL(2, F)-action defined above is irreducible.

Proof See [4, Lemma 5.3.1].

56 Here we present the Clebsch-Gordan Formula for SL(2, F). For its proof, see [15, Theorem 2.6.3]. We follow the argument in [15] modifying it whenever necessary for the case of characteristic p.

Theorem 4.4.2 (Clebsch-Gordan Formula) Let F be a field with char F = 0. For any integers, m ≥ n ≥ 0, the tensor product Vm ⊗ Vn is completely reducible and decomposes as ∼ Vm ⊗ Vn = Vm+n ⊕ Vm+n−2 ⊕ · · · ⊕ Vm−n as SL(2, F)-modules.

Lemma 4.4.3 The group SL(2, F) is generated by

{ua, la | a ∈ F} where  1 a   1 0  u = , l = . a 0 1 a a 1

Proof See [20, 3.2.10].

For the rest of this section the following map will play an important role.

δm : Vm−1 ⊗ Vm−1 → Vm ⊗ Vm

vi ⊗ vj 7→ (vi ⊗ vj)(x ⊗ y − y ⊗ x).

Definition 4.4.4 Let w ∈ Vm ⊗ Vm. Then the left degree of w in x is the biggest degree in x of the monomials that appear on the left hand sides of the tensor sign.

For example, the left degree in x of x3y2 ⊗ y5 + xy4 ⊗ x4y is 3.

Lemma 4.4.5 δm is an injective homomorphism of SL(2, F)-modules.

Proof To show that δm is injective, let v be a nonzero element of Vm−1 ⊗ Vm−1. We can see that the left degree in x of vδm is one bigger that the left degree in x of v, and thus vδm 6= 0.

i m−i−1 For i ∈ {0, . . . , m−1} set ei = x y . Then e0, . . . , em−1 form a basis for Vm−1. To show that δm is a homomorphism of SL(2, F)-modules, since SL(2, F) is generated

57 by {ua, la | a ∈ F} (Lemma 4.4.3) we just need to verify, for every element of the basis ei ⊗ ej of Vm−1 ⊗ Vm−1, and every generator s of SL(2, F) that (ei ⊗ ej)δms = (ei ⊗ ej)sδm.

i m−i−1 j m−j−1 Since ei = x y , ej = x y , we obtain

 1 0  (e ⊗ e ) = xi(ax + y)m−i−1 ⊗ xj(ax + y)m−j−1, i j a 1 thus  1 0  (e ⊗ e ) δ = xi+1(ax + y)m−i−1 ⊗ xjy(ax + y)m−j−1− i j a 1 m − xiy(ax + y)m−i−1 ⊗ xj+1(ax + y)m−j−1.

Similarly,

i+1 m−i−1 j m−j i m−i j+1 m−j−1 (ei ⊗ ej)δm = x y ⊗ x y − x y ⊗ x y , and so,

 1 0  (e ⊗ e )δ = xi+1(ax + y)m−i−1 ⊗ xj(ax + y)m−j− i j m a 1 − xi(ax + y)m−i ⊗ xj+1(ax + y)m−j−1 = = xi+1(ax + y)m−i−1 ⊗ axj+1(ax + y)m−j−1 + xi+1(ax + y)m−i−1 ⊗ xjy(ax + y)m−j−1− − axi+1(ax + y)m−i−1 ⊗ xj+1(ax + y)m−j−1 − xiy(ax + y)m−i−1 ⊗ xj+1(ax + y)m−j−1 = = xi+1(ax + y)m−i−1 ⊗ xjy(ax + y)m−j−1 − xiy(ax + y)m−i−1 ⊗ xj+1(ax + y)m−j−1.

We have then showed that  1 0   1 0  (e ⊗ e )δ = (e ⊗ e ) δ . i j m a 1 i j a 1 m The proof that  1 a   1 a  (e ⊗ e )δ = (e ⊗ e ) δ i j m 0 1 i j 0 1 m is similar.

The reader should recall at this point the definitions of symmetric tensors and antisymmetric tensors, as well as the subspaces V ∧ V and SV , which were defined in Section 1.6.

58 Lemma 4.4.6 If m > 0, then (Vm−1 ∧ Vm−1)δm ≤ SVm and SVm−1 δm ≤ Vm ∧ Vm.

Proof To prove this lemma we only need to calculate the image of elements of SVm−1 and Vm−1 ∧ Vm−1. Let φm−1 : Vm−1 ⊗ Vm−1 → Vm−1 ⊗ Vm−1 and φm : Vm ⊗ Vm → Vm ⊗ Vm be the maps defined by f ⊗ g 7→ g ⊗ f as in Definition 1.6.2. Suppose that P v ∈ SV (or v ∈ Vm−1 ∧ Vm−1) and v = fi ⊗ gi with fi, gi ∈ Vm−1. Then, since m−1 P P v ∈ SVm−1 (or v ∈ Vm−1 ⊗ Vm−1), we have v = fi ⊗ gi = ± gi ⊗ fi where the actual sign of the last expression depends on whether v ∈ SVm−1 (in which case it is “+”) or v ∈ Vm−1 ∧ Vm−1 (when it is “-”). Let us apply the map δm and obtain

X  X  vδm = v(x ⊗ y − y ⊗ x) = fi ⊗ gi x ⊗ y − fi ⊗ gi y ⊗ x X  X  X X = ± gi ⊗ fi x ⊗ y ∓ gi ⊗ fi y ⊗ x = ± gix ⊗ fiy ∓ giy ⊗ fix.

Let us now compute X X  X X vδmφ = fix ⊗ giy − fiy ⊗ gix φ = giy ⊗ fix − gix ⊗ fiy.

Hence if v ∈ SVm−1 , then vδmφ = −vδm and hence vδm ∈ Vm ∧ Vm. When v ∈

Vm−1 ∧ Vm−1 we obtain vδmφ = vδm and vδm ∈ SVm .

Lemma 4.4.7 Let F be a field with char F = p where p is a prime, let m be an integer such that p > 2m and Vm be as defined before. Then Vm ⊗ Vm is completely reducible and decomposes as the sum of simple modules

m ∼ M Vm ⊗ Vm = V2m−2i. i=0

Proof We will prove this by induction. When m = 0, then the result follows trivially ∼ ∼ since V0 ⊗ V0 = F = V0. Suppose then the result is valid for m − 1. Define the map

ψ : Vm ⊗ Vm → V2m f ⊗ g → fg.

Since the action of SL(2, F) preserves the product of polynomials, this map is a homomorphism of SL(2, F)-modules. We will define W as the submodule of Vm ⊗Vm m m m m 2m generated by x ⊗ x . Since (x ⊗ x )ψ = x 6= 0 and V2m is irreducible (by Lemma 4.4.1), then by Schur’s Lemma ([15, Lemma 2.2.6]), the restriction of ψ to

59  a b  W is surjective. Thus dim(W ) ≥ dim(V ) = 2m + 1. If g = ∈ SL(2, ), 2m c d F then (xm ⊗ xm)g = (ax + by)m ⊗ (ax + by)m. Observe that the expression (ax + by)m ⊗ (ax + by)m is a linear combination of i m−i j m−j expressions of the form αijx y ⊗ x y for αij ∈ F. We say that the x-degree of such an expression is i + j. The x-degree of these expressions varies from 0 to 2m. Hence we may write 2m m m X (ax + by) ⊗ (ax + by) = fi i=0 i 2m−i where fi is the sum of all the terms with x-degree equal to i. However fi = a b gi for all i where gi is a polynomial in x and y which only depends on m and i. Since W is the SL(2, F)-submodule generated by xm ⊗ xm, W is generated as an F-vector space by the gi for i = 0,..., 2m. Hence dim(W ) ≤ 2m + 1, from where we conclude that dim(W ) = 2m+1. Thus, the restriction of ψ to W is an isomorphism. Observe also that for f ⊗ g ∈ Vm−1 ⊗ Vm−1, ((f ⊗ g)δm)ψ = (xy − yx)fg = 0, thus (Vm−1 ⊗ Vm−1)δm ≤ ker ψ. As Im(ψ) = V2m has dimension 2m + 1, then 2 2 dim(ker ψ) = (m + 1) − (2m + 1) = m , which is equal to the dimension of Im(δm). Thus Im(δm) = ker ψ. Since ψ restricted to W is an isomorphism, ker ψ ∩ W = {0}. As dim(Vm−1 ⊗ Vm−1) + dim(W ) = dim(Vm ⊗ Vm), ∼ Vm ⊗ Vm = (Vm−1 ⊗ Vm−1)δm ⊕ W = (Vm−1 ⊗ Vm−1) ⊕ V2m. (4.1)

By the induction hypothesis we get the decomposition in the statement of the the- orem. Since p > 2m, by Lemma 4.4.1 all modules V2m−2i are irreducible.

Theorem 4.4.8 Suppose m ≥ 0 and char F = 0 or char F = p 6= 2 a prime such that p > 2m. Define for m even

Im = {0, 4, 8,..., 2m} and Jm = {2, 6, 10,..., 2m − 2} and if m is odd,

Im = {2, 6, 10,..., 2m} and Jm = {0, 4, 8,..., 2m − 2}

Then ∼ M ∼ M Vm ∧ Vm = Vj and SVm = Vj. (4.2)

j∈Jm j∈Im

60 Proof We prove this theorem by induction on m. Suppose first m = 0. Then m+1
I0 = {0} and J0 = ∅. Thus (4.2) is true since dim(Vm ∧ Vm) = 2 = 0. Suppose then m > 0 and that the result is true for m − 1. Note that

Jm−1 ∪ {2m} = Im, and Im−1 = Jm. (4.3)

The following equalities and isomorphisms follow respectively from (4.1), the fact that δm is injective (Lemma 4.4.5), the induction hypothesis and from (4.3):

∼ Vm ⊗ Vm = W ⊕ (Vm−1 ∧ Vm−1)δm ⊕ (SVm−1 )δm = V2m ⊕ (Vm−1 ∧ Vm−1) ⊕ SVm−1 ∼ M M M M = V2m ⊕ ( Vj) ⊕ ( Vj) = ( Vj) ⊕ ( Vj),

j∈Jm−1 j∈Im−1 j∈Im j∈Jm where W is as defined in the proof of Lemma 4.4.7. Notice that in the last expression the left summand is the image of W ⊕ ((Vm−1 ∧ Vm−1)δm) by the isomorphisms, whereas the right summand is the image of SVm−1 δm. By Lemma 4.4.6,(Vm−1 ∧

Vm−1)δm ≤ SVm and SVm−1 δm ≤ Vm ∧ Vm. Since W is generated, as an SL(2, F)- m m module, by x ⊗ x ∈ SVm , then W ≤ SVm . Thus the theorem holds.

Corollary 4.4.9 Let F be a field such that char F 6= 2, suppose m is an integer such that m ≡ 2 mod 4, and if char F = p 6= 0 suppose also 2m < p. Let Vm be the F-vector space of the polynomials in F[x, y] that are homogeneous of degree m. Then Vm with the SL(2, F)-action defined before is an irreducible ESQ module.

∼ Proof By Theorem 4.4.8, Vm ∧Vm = V2m−2 ⊕V2m−6 ⊕· · ·⊕V2. As m ≡ 2 mod 4 then Vm appears in the right hand side. If we take U as the kernel of the projection from ∼ Vm ∧ Vm into the submodule isomorphic to Vm, we have Vm ∧ Vm/U = Vm. Thus we conclude that Vm is ESQ. By Theorem 4.4.1 Vm is an irreducible SL(2, Fp)-module.

Theorem 4.4.10 Let F be a field such that char F 6= 2, let m be an integer such that m ≡ 2 mod 4, and if char F = p 6= 0 then suppose 2m < p. Let then Vm be the F-vector space of the polynomials F[x, y] that are homogeneous of degree m. Define the IAC algebra (Vm, , ϕ) as in Section 2.3. Then the algebra (Vm, , ϕ) is a simple IAC algebra. J K J K

Proof Let L = (Vm, , ϕ) as defined for an ESQ module in Section 2.3. By Corollary 4.4.9, Vm is an irreducibleJ K ESQ module, and so L is IAC.

61 We claim that L is simple. By Corollary 3.2.5 L = I1 ⊕ · · · ⊕ Ik where X = {I1,...,Ik} is the set of minimal ideals of L and the same corollary claims that X is an Aut(L)-orbit. Since SL(2, F) is an irreducible subgroup of Aut(L), SL(2, F) permutes the elements of X transitively. Further Z(SL(2, F)) induces scalar trans- formations on L = Vm, and so Z(SL(2, F)) lies in the kernel of the SL(2, F)- action on X. Thus we obtain a transitive PSL(2, F)-action on X. Note that dim(L) = k dim(I1) and I1 is a nonabelian IAC algebra, and hence dim(I1) ≥ 3. Thus k ≤ dim(L)/3 = (m + 1)/3. It is well known that PSL(2, F) is a simple group whenever |F| ≥ 4 (see [9, Theorem 1.13]). In these cases, the PSL(2, F)-action on X must be faithful, or trivial, and when this action is faithful PSL(2, F) can be viewed as a subgroup of Sk. Thus when F (and PSL(2, F)) is infinite, the action must be trivial and transitive which implies k = 1. Let us now assume that F is finite and set |F| = q. Let p = char F. The minimal degree d of a nontrivial transi- tive permutation representation of PSL(2, q) is q + 1 except for q = 3, 5, 7, 9, 11, in these cases d = 3, 5, 7, 6, 11 respectively; see [14, Table 5.2.A, page 175] for the cases ∼ when q > 3, and notice that PSL(2, 3) = A4 when q = 3. In all the cases we have 2 2 2 3 p ≤ 3 q ≤ d. Our hypothesis 2m < p implies 3 m < p. If k > 1, then

2p m + 1 3 p + 1 ≤ d ≤ k ≤ ≤ 2 . 3 3 3

1 As 2 p < 1 is false for all primes p, we conclude that k = 1 and L is simple.

Finally, we claim that the family of IAC algebras from the previous theorem is not contained and does not contain the families presented in sections 4.2 and 4.3. To show it is not contained, first notice that it contains algebras over fields of characteristic 0 which does not happen on the other cases. Also, given the dimension m + 1 with m ≡ 2 (mod q), the previous theorem gives us algebras over any field 2 with characteristic p > 3 m, which did not happen in Sections 4.2 and 4.3. On the other hand, both families of algebras from Sections 4.2 and 4.3 contain algebras where the characteristic of the field is smaller than the dimension of the algebra, for example, in Section 4.2 we have an algebra over F3 with dimension 4 and in Section 4.3 we have the algebra in the discussion closing that Section over F34 with dimension 5.

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