CHARACTER TABLES of FINITE GROUPS by Adriana Nenciu August 2006 Chair: Alexandre Turull Major Department: Mathematics

CHARACTER TABLES OF FINITE GROUPS

ADRIANA NENCIU

A DISSERTATION PRESENTED TO THE GRADUATE SCHOOL OF THE UNIVERSITY OF FLORIDA IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF DOCTOR OF PHILOSOPHY UNIVERSITY OF FLORIDA 2006 Copyright 2006 by

Adriana Nenciu ACKNOWLEDGMENTS

My ﬁrst and foremost thanks go to my advisor, Dr. Alexandre Turull. Without his constant guidance and support my doctoral research would never have been possible. The members of my doctoral dissertation committee, Dr. Ion Giviriga, Dr. Jorge Martinez, Dr. Peter Sin, and Dr. Pham Tiep, have been very generous with their time and help. I am grateful to all the mathematics teachers I have had, especially to Ioana Andrei and Dr. Victor Vuletescu.

I wish to thank my friends from Bucharest for their interest in my work and for our discussions on mathematical and non-mathematical topics. Last but not least I want to thank Zia for his patience, understanding, and support.

iii TABLE OF CONTENTS page

ACKNOWLEDGMENTS ...... iii

TABLE ...... vi ABSTRACT ...... vii

CHAPTER 1 INTRODUCTION ...... 1 2 PRELIMINARIES ...... 7

2.1 Classiﬁcation of p-groups with Derived Subgroup of Order p .... 7 2.2 Irreducible Characters of p-groups ...... 12

3 ISOMORPHIC CHARACTER TABLES ...... 18 4 THE NUMBER OF CHARACTER TABLES ...... 22 4.1 Character Tables of p-groups with Derived Subgroup of Order p .. 22 4.2 (m,n,p)-admissible Triples ...... 39 4.3 Equivalent Kernels ...... 41 4.4 Equivalent Images ...... 42 4.5 Double Cosets ...... 45 ¯ ¯ 4.6 Characterizations of Autπ(B) and AutA(B) ...... 49 ¯ ¯ 4.7 The Number of (AutA(B), Autπ(B)) Double Cosets ...... 55 4.8 Combinatorial Character Tables and Admissible Triples ...... 66 4.9 Combinatorial Character Tables and Blackburn Triples ...... 73 4.10 The Preimage of the Map cctn,m ...... 83 4.11 Classiﬁcation of Character Tables ...... 93 5 THE LIMIT ...... 99

5.1 Upper Bounds for the Size of the Preimage of the Map cctn,m ... 99 5.2 The Number of λ’s ...... 106 5.3 The Average Number of Distinct Parts of a Partition of an Integer . 112 5.4 lim NG(n, m)/NCT (n, m) ...... 113 n→∞ 6 NUMERICAL RESULTS ...... 126

REFERENCES ...... 127

iv BIOGRAPHICAL SKETCH ...... 128

v TABLE Table page

6–1 The number of groups and character tables ...... 126

vi Abstract of Dissertation Presented to the Graduate School of the University of Florida in Partial Fulﬁllment of the Requirements for the Degree of Doctor of Philosophy CHARACTER TABLES OF FINITE GROUPS By Adriana Nenciu August 2006 Chair: Alexandre Turull Major Department: Mathematics

Precise formulas and estimates for the number of finite p-groups up to isomorphism are known, but much less is known about the number of non-isomorphic character tables of such groups. Two character tables of finite groups are isomorphic if there exist a bijection for the irreducible characters and a bijection for the conjugacy classes that preserve all the character values. We give necessary and sufficient conditions for two finite groups to have isomorphic character tables. In the case of finite p-groups with derived subgroup of order p, we classify up to isomorphism their irreducible character tables, and estimate their number. The number of such character tables turns out to be considerably fewer than the corresponding number of groups.

vii CHAPTER 1 INTRODUCTION The determination of all isomorphism types of groups of a ﬁxed order has interested mathematicians since Cayley published his paper in 1854. In the last century a lot of eﬀort was put into solving this problem and interesting results have been proved. In the particular case of p-groups, there are results that give an algorithm for constructing p-groups (see Eick and O’Brien [5] and Newman [10]), and results that give estimates for f(n, p), the number of groups of order pn (see

Blackburn [3], Higman [7] and Sims [11]). G. Higman [7] proved that

f(n, p) = pAn3 where A depends on n and p and

2 2 − ε ≤ A ≤ + ε 27 n 15 n and lim εn = 0. Later S. R. Blackburn [3] gave upper and lower bounds for the n→∞ number of p-groups in an isoclinism class:

m Theorem 2 ([3]) Let Φ be an isoclinism class. Deﬁne fΦ(p ) to be the number of groups of order pm in Φ. Suppose that for all P ∈ Φ we have |P/Z(P )P 0| = pa and |P 0 ∩ Z(P )| = pc. Then

µ ¶(a+c)/2 m m fΦ(p ) (a+c)/2 k1 2 ≤ m ≤ k2m (log m) fab(p )

m where k1 and k2 are constants depending only on Φ and fab(p ) denotes the number of abelian groups of order pm.

If progress has been made into classifying finite groups, little is known about the classification up to isomorphism of character tables of finite groups.

1 2

There are results about the uniqueness of character table (see Davydov [4] and

Mattarei [9]). Mattarei’s result gives a suﬃcient condition for two groups to have isomorphic character tables. Using quasi-Hopf algebras methods A. A. Davydov

[4] gives necessary and sufficient conditions for two arbitrary finite groups to have isomorphic character tables. Motivated by Davydov’s result, in Chapter 3, we prove a similar result (see Proposition 3.0.14). Our statement and proof uses only methods from the representation theory of finite groups.

In a paper published in Journal of Algebra in 1999 [2], S. R. Blackburn gave a classiﬁcation up to isomorphism of p-groups with derived subgroup of order p, for any prime p. The classiﬁcation is independent of the prime p. Blackburn describes some combinatorial objects which can easily be enumerated and which are in one–to–one correspondence with the isomorphism classes of such groups.

In this paper, we classify up to isomorphism the character tables of all p-groups whose derived subgroup has order p. We prove that there are two sets that are in bijection with the set of isomorphism classes of character tables of p-groups P with |P 0| = p. One set is algebraic and a priori it depends on the prime p. The second set is combinatorial and is independent of p. Thus, the number of character tables of all groups of order pn and derived subgroup of order p does not depend on the prime p, see Theorem 4.11.3 below.

n Let NG(n) denote the number of non-isomorphic p-groups P with |P | = p

0 and |P | = p and NCT (n) denote the number of non-isomorphic character tables of these p-groups. By our and Blackburn’s results these numbers do not depend on the prime p. We have that NG(n) is much larger than NCT (n). We prove (see Proposition 5.4.12 below) that

NG(n) − NCT (n) ≥ |P(n − 3)| 3 where |P(n − 3)| is the number of partitions of n − 3. It then follows (see Theorem

5.4.13 below) that

NG(n) − NCT (n) lim √ √ = ∞ n→∞ eπ 2/3 n n1+ε for all ε > 0.

It is easily seen that, given a p-group P with |P | = pn and |P 0| = p, we

2m n−1 have [P : Z(P )] = p for some positive integer 1 ≤ m ≤ b 2 c. We denote n 0 by NG(n, m) the number of non-isomorphic p-groups P with |P | = p , |P | = p

2m and [P : Z(P )] = p . We denote by NCT (n, m) the number of non-isomorphic character tables of these p-groups. With this notation, we obtain (see Theorem

5.4.11 below)

NG(n, m) (2m)! lim = m . n→∞ NCT (n, m) 2 m! n−1 Let p be a prime and let m, n be positive integers such that 1 ≤ m ≤ b 2 c. n We denote by Pm,n,p the set of isomorphism classes of p-groups P with |P | = p , |P 0| = p and [P : Z(P )] = p2m. S. .R. Blackburn [2] proves that there exists a set

Sn,m of integer vectors having certain properties (see Deﬁnition 4.2.6 below) and a bijection

Θm,n,p : Pm,n,p → Sn,m.

The set Sn,m is independent of p; thus the number NG(n, m) = |Pm,n,p| of these p-groups is independent of p.

We denote by CT m,n,p the set of isomorphism classes of character tables of p-groups with derived subgroup of order p. We have a natural surjective map

ctm,n,p : Pm,n,p → CT m,n,p that associates to each isomorphism class of p-groups the isomorphism class of its character table. Using Proposition 3.0.14, we prove, in Corollary 4.1.11, that two groups P1,P2 ∈ Pm,n,p have isomorphic character tables if and only if there exist 4

0 0 group isomorphisms α : P1/P1 → P2/P2 and β : Z(P1) → Z(P2) such that the following diagram is commutative

π¯1 / 0 Z(P1) P1/P1

β α π¯2 / 0 Z(P2) P2/P2 whereπ ¯1 andπ ¯2 are the restrictions to Z(P1) and Z(P2) respectively if the

0 0 canonical projections P1 ³ P1/P1 and P2 ³ P2/P2 respectively. This result motivates the introduction, in Section 4.2, of (m, n, p)-admissible triples and their equivalence. We denote by Tm,n,p a full set of non-equivalent (m, n, p)-admissible triples. There is a natural map

tm,n,p : Pm,n,p → Tm,n,p that associates to each isomorphism class of p-groups its corresponding (m, n, p)-admissible triple. The map tm,n,p is always surjective (see Theorem 4.10.17 below). Furthermore, there is a unique map

Υm,n,p : CT m,n,p → Tm,n,p making the diagram

Pm,n,p t II ctm,n,p tt IItm,n,p tt II tt II ty t I$ Υm,n,p CT m,n,p / Tm,n,p commutative, and Υm,n,p is a bijection, see Theorem 4.11.1 below.

In Section 4.8, we construct a combinatorial object CCT n,m (CCT n,m stands for combinatorial character tables) and we show that there is a bijection between

CCT n,m and Tm,n,p, the set of non-equivalent (m, n, p)-admissible triples. Moreover, to every P ∈ Pm,n,p we can associate an (m, n, p)-admissible triple, namely (P/P 0,Z(P ), π¯) and using Blackburn’s classiﬁcation we show, in Theorem 4.9.9, 5 that we have the following commutative diagram

Θm,n,p / Pm,n,p ∼ Sn,m

tm,n,p cctn,m Cp / Tn,m,p ∼ CCT n,m where Sn,m is defined in Definition 4.2.6 and cctn,m : Sn,m → CCT n,m is defined in Definition 4.9.1. Furthermore, in Corollary 4.11.2 we obtain the following commutative diagram

Θm,n,p Pm,n,p / Sn,m t II ctm,n,p tt IItm,n,p tt II cct tt II n,m ty t I$ Υm,n,p Cp CT m,n,p / Tm,n,p / CCT n,m where the horizontal arrows are bijective maps and the vertical arrows are surjective maps. Hence, we obtain two characterizations of the character tables of p-groups with derived subgroup of prime order. One characterization is algebraic, using the (m, n, p)-admissible triples. The second characterization involves the set

CCT n,m and is independent of the prime p. Thus, as in the case of p-groups the classiﬁcation does not depend on the prime p. Furthermore, we can now easily compute the number of non-isomorphic character tables of p groups P of order pn and derived subgroup of order p. The number is given in Theorem 4.11.3.

Let NG(n, m) = |Pm,n,p| and let NCT (n, m) = |CT m,n,p|. In the last chapter we compare NG(n, m) to NCT (n, m). Our main result is Theorem 5.4.11 which says that

NG(n, m) (2m)! lim = m . n→∞ NCT (n, m) 2 m! This can be interpreted in the following way. Let m be any positive integer.

Then if [CT ] ∈ CT m,n,p is a random character table for large n, there are 6 approximately (2m)!/(2m m!) groups whose character tables are isomorphic to [CT ].

We conclude the chapter with Theorem 5.4.13 which gives an estimate for the diﬀerence between the number of non-isomorphic p-groups with derived subgroup of order p and the number of non-isomorphic character tables of these p-groups. Theorem For all ε > 0 we have

NG(n) − NCT (n) lim √ √ = ∞. n→∞ eπ 2/3 n n1+ε Having the classification of p-groups with derived subgroup of order p given by S. R. Blackburn [2] and our classification of character tables of these groups, we wrote a computer program using GAP that computes NG(n), the number of non-isomorphic p-groups with derived subgroup of order p, and NCT (n), the number of non-isomorphic character tables of these groups for any integer n ≥ 3. Some of these numbers are presented in Chapter 6. CHAPTER 2 PRELIMINARIES 2.1 Classification of p-groups with Derived Subgroup of Order p

Let p be a prime and let P be a finite p-group with |P 0| = p. S. R. Blackburn [2] classifies these groups up to isomorphism. We will briefly present his results (see

Blackburn [2] for more details). Let n be a positive integer. We denote by P the set of all isomorphism classes of p-groups with derived subgroup of order p. Let Sn be the set of triples (ρ, e, A) which satisfy the following properties:

(i) ρ = (k1, k2, . . . , kr) is a partition of some integer c where c < n and n − c = 2m. (ii) e is an integer such that ρ has at least one part of size e.

(iii) Let Jt be deﬁned as

2 Jt = {(i, j) ∈ Z : 1 ≤ i ≤ j ≤ t + 1}.

Then A = {αi,j :(i, j) ∈ Jc+1} where the elements αi,j are non-negative integers such that X n − c (a) α = i,j 2 (i,j)∈Jc+1 (b) For all integers i, deﬁne mi to be the number of parts of ρ of size i. Then for all k ∈ {2, 3, . . . , c + 2},    mk−1 if k ≤ e   X X  1 if k = e + 1 αi,k + αk,j ≤  1≤i≤k k≤j≤c+2  me − 1 if k = e + 2   mk−2 if k > e + 2

7 8

Theorem 2.1.1. (Theorem 8 in Blackburn [2])

Let p be a prime and let n be a positive integer. Then there is a bijection Θ

n between the set of isomorphism classes of groups P ∈ P of order p and the set Sn deﬁned above.

Let P ∈ P with derived subgroup of order p. Since P is nilpotent we have [P 0,P ] $ P 0 and as |P 0| = p we have [P 0,P ] = 1; thus P has nilpotency class 2 and P 0 ⊆ Z(P ). Moreover for all g, h ∈ P we have [gp, h] = [g, h]p = 1 (the ﬁrst equality holds as P has nilpotency class 2 and the second follows since |P 0| = p). Hence, the Frattini subgroup of P , Φ(P ) = P 0P p ⊆ Z(P ), where P p = {xp : x ∈ P }. It follows that P/Z(P ) is elementary abelian. If p 6= 2 then the bijection Θ is deﬁned as follows: given P ∈ P with

|P | = pn, |P 0| = p and |Z(P )| = pc we associate the following triple (ρ, e, A).

1) Z(P ) is an abelian group hence Z(P ) ' Z/pk1 Z × Z/pk2 Z × · · · × Z/pkr Z where (k1, k2, . . . , kr) is a partition of c. We take ρ = (k1, k2, . . . , kr). 2) Let e be the largest integer such that P 0 ⊆ Z(P )pe−1 . S. R. Blackburn [2] proves that Z(P ) has a direct factor isomorphic to Z/peZ. 3) The deﬁnition of A is more laborious. For i ∈ {0, 1, . . . , c + 1} deﬁne

Zi ⊆ Z(P ) by   i  {z ∈ Z(P ): zp = 1} if i < e  pe−1 0 Zi = {z ∈ Z(P ): z ∈ P } if i = e    i−1  {z ∈ Z(P ): zp = 1} if i > e We have

Z0 ⊆ Z1 ⊆ ... ⊆ Zc+1.

p Deﬁne Y0,Y1,...,Yc+1 subspaces of Z(P )/Z(P ) by

p p Yi = ZiZ(P ) /Z(P ) . 9

Let π : P/Z(P ) → Z(P )/Z(P )p be the map induced by the p-power map

p x 7−→ x in P . Let V = P/Z(P ). For i ∈ {0, 1, . . . c + 2} deﬁne Vi ⊆ P/Z(P ) by    0 if i = 0 Vi =  −1  π (Yi−1) = {v ∈ V : π(v) ∈ Yi−1} if i > 0

0 0 Let h be a non-identity element of P . We identify P with Fp by identifying

i 0 the element h ∈ P with i ∈ Fp. Using this identification, we may define a mapping Φ : P/Z(P ) × P/Z(P ) → Fp by Φ(xZ(P ), yZ(P )) = [x, y] for all x, y ∈ P . Note that Φ is a non-degenerate alternating form. For all 1 ≤ i ≤ c + 2 define Ri to be the radical of the form Φ when restricted to the subspace Vi. Let

2 Jc+1 = {(i, j) ∈ Z : 1 ≤ i ≤ j ≤ c + 2}. For (i, j) ∈ Jc+1 deﬁne    dim((Ri ∩ Rj−1) + Vi−1) − dim((Ri ∩ Rj) + Vi−1) if i < j  α = i,j  X X  1 (dimV − dimV − α − α ) if i = j  2 i i−1 k,i i,k 1≤k≤i−1 i+1≤k≤c+2

We deﬁne A = {αi,j ∈ Z :(i, j) ∈ Jc+1}. Thus, to every group P ∈ P we associate a triple (ρ, e, A). Conversely, given (ρ, e, A) we will construct a p-group P as follows: we deﬁne for all i ∈ {2, 3, . . . , c + 2} the integer ri by    mi−1 if i ≤ e    1 if i = e + 1 ri =   me − 1 if i = e + 2   mi−2 if i > e + 2

For all 2 ≤ i ≤ c + 2 we deﬁne the non-negative integer di by X X di = ri − αj,i − αi,j. 1≤j≤i i≤j≤c+2 10

For all i ∈ {1, 2, . . . , c + 2} we deﬁne    i − 1 if i ≤ e + 1 s = i   i − 2 if i ≥ e + 2

We have that Xc+2 risi = c. i=2

Let IA be the set of triples of integers deﬁned by

(i, j, k) ∈ IA if and only if (i, j) ∈ Jc+1 and 1 ≤ k ≤ αi,j.

Proposition 2.1.2. (Proposition 9 in Blackburn [2])

Let p be an odd prime and let n be a positive integer. Let P ∈ P have order pn and let (ρ, e, A) ∈ Sn. Deﬁne the integers c, αi,j, di, si and the set IA as above. Then P has type (ρ, e, A) if and only if there exists a generating set X for P, where X is the set

{xa : a ∈ IA} ∪ {ya : a ∈ IA} ∪ {z(i,k) : 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di}, with the property that X together with the following relations forms a presentation for P

psi+1 x(i,j,k) = 1 for all (i, j, k) ∈ IA psj +1 y(i,j,k) = 1 for all (i, j, k) ∈ IA psi z(i,k) = 1 for 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di p [xa, x] = 1 for all x ∈ X and for all a ∈ IA

p [ya, x] = 1 for all x ∈ X and for all a ∈ IA £ ¤ z(i,k), x = 1 for all x ∈ X, where 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di

0 [xa, xa0 ] = 1 for all a, a ∈ IA

0 [ya, ya0 ] = 1 for all a, a ∈ IA

0 0 [xa, ya0 ] = 1 for all a, a ∈ IA, where a 6= a 11

pe−1 [xa, ya] = b for all a ∈ IA where    z(e+1,1) if de+1 = 1  b = xp if α = 1 where j > e + 1  (e+1,j,1) e+1,j   p y(j,e+1,1) if αj,e+1 = 1 where j < e + 1 If p = 2 we summarize the results from Blackburn [2] in the following proposition.

Proposition 2.1.3. Let n be a positive integer and let P ∈ P be a group of order

n 2 . Let (ρ, e, A) ∈ Sn. Deﬁne the integers ri, di and si as usual. Then

a) if e 6= 1 or α2,j = 1 for some 3 ≤ j ≤ c + 2 then P has type (ρ, e, A) if and only if there exists a generating set X for P , where X is as in Proposition 2.1.2

b) if e = 1 and α1,2 = 1 then P has type (ρ, e, A) if and only if there exists a generating set X for P , where X is the set

{xa : a ∈ IA} ∪ {ya : a ∈ IA} ∪ {z(i,k) : 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di} with the property that X together with the following relations form a presentation for P

2si+1 x(i,j,k) = 1 for all (i, j, k) ∈ IA \{(1, 2, 1)}

2 2 x(1,2,1) = y(1,2,1)

2sj +1 y(i,j,k) = 1 for all (i, j, k) ∈ IA

2si z(i,k) = 1 for 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di

2 [xa, x] = 1 for all x ∈ X and for all a ∈ IA

2 [ya, x] = 1 for all x ∈ X and for all a ∈ IA £ ¤ z(i,k), x = 1 for all x ∈ X, where 2 ≤ i ≤ c + 2, 1 ≤ k ≤ di

0 [xa, xa0 ] = 1 for all a, a ∈ IA

0 [ya, ya0 ] = 1 for all a, a ∈ IA

0 0 [xa, ya0 ] = 1 for all a, a ∈ IA, where a 6= a 12

2 [xa, ya] = x(1,2,1) for all a ∈ IA.

2.2 Irreducible Characters of p-groups

Lemma 2.2.1. Let G be a group with G0 ⊆ Z(G). Let χ ∈ Irr(G) and assume that χ is faithful on G0. Then χ(g) = 0 for all g ∈ G \ Z(G).

Proof. Let g ∈ G \ Z(G). Then there exists x ∈ G such that [g, x] 6= 1, where 1 is the identity element of G. Let y = [g, x] and let X denote the representation of G which aﬀords the character χ. Since G0 ⊆ Z(G) we have X(y) is diag(ε, ..., ε) for some ε ∈ C. Moreover, ε 6= 1 as χ is faithful on G0. We have χ(g) = χ(x−1gx) = χ(gy) = χ(g)ε. Since ε 6= 1 we must have χ(g) = 0.

Lemma 2.2.2. (Exercise 2.13 in Isaac [8]) Let G be a group with |G0| = p. Assume that G0 ⊆ Z(G). If χ ∈ Irr(G) with χ(1) > 1 then

χ(1)2 = [G : Z(G)].

Proof. Let χ ∈ Irr(G) such that χ(1) > 1. Then χ is faithful on G0 and by Lemma

2.2.1 we have χ(x) = 0 for all x 6∈ Z(G). By Cliﬀord’s Theorem we have

χ|Z(G) = χ(1)λ for some λ ∈ Irr(Z(G)). We have

1 1 1 =< χ, χ >= χ(1)2 < λ, λ >= χ(1)2. [G : Z(G)] [G : Z(G)]

Hence, χ(1)2 = [G : Z(G)]. 13

Theorem 2.2.3. Let P be an abelian p-group generated by x1, x2, . . . , xk with

µi o(xi) = p for all i = 1, . . . , k and µ1 ≥ µ2 ≥ ... ≥ µk. Then the group of automorphisms of P is isomorphic to the following group of matrices:

   µi   i) 0 ≤ aij < p       ii) for all 1 ≤ i ≤ j ≤ k we have  (aij)1≤i,j≤k : aij ∈ Z satisfy  µi−µj 0 0 µj   aij = p aij where 0 ≤ aij < p      iii) det((aij)1≤i,j≤k) ≡/ 0 (mod p) where the multiplication is matrix multiplication followed by reduction modulo pµi in column i for all 1 ≤ i ≤ k.

Proof. Let Aut(P ) denote the group of automorphisms of P and let φ ∈ Aut(P ).

µi Then for all 1 ≤ i, j ≤ k there exist aij ∈ Z such that 0 ≤ aij < p and

Yk aij φ(xj) = xi . i=1

We will show that (aij)1≤i,j≤k must satisfy conditions ii) and iii). As φ is a group

µj isomorphism, the order of φ(xj) must be p for all j = 1, . . . , k. We have

k µ Y µj p j p aij 1 = φ(xj) = xi . i=1

µj p aij If i is such that 1 ≤ i ≤ j and µi ≥ µj then xi = 1 if and only if

µi−µj 0 0 µj aij = p aij, for some 0 ≤ aij < p . Hence, condition ii) holds. Let Φ(P ) denote the Frattini subgroup of P and let proj : P → P/Φ(P ) denote the canonical projection. We have that P/Φ(P ) is an elementary abelian group and proj ◦ φ is a surjective map. Moreover, settingx ¯i = proj(xi) for all i = 1, . . . , k we have that {x¯1, x¯2,..., x¯k} is a basis of P/Φ(P ) as a vector space over Fp. We obtain Yk a¯ij (proj ◦ φ)(xj) = x¯i i=1 14

wherea ¯ij = aij + pZ for all i, j = 1, . . . , k. Since proj ◦ φ is a surjective map we obtain that det((¯aij)1≤i,j≤k) ≡/ 0 (mod p). Hence, det((aij)1≤i,j≤k) ≡/ 0 (mod p) and condition iii) is satisﬁed.

Conversely, given a matrix (aij)1≤i,j≤k with aij ∈ Z satisfying i), ii) and iii) we Yk aij deﬁne φ : P → P by φ(xj) := xi for all j = 1, . . . , k. It is easy to see that φ is i=1 an well deﬁned group homomorphism. Since det((aij)1≤i,j≤k) ≡/ 0 (mod p) we have that proj ◦ φ is surjective. Hence, P =< Im φ, Φ(P ) > and by Theorem (23.1)(2) in

[1] we have P =< Im φ >, thus φ is surjective. Furthermore, φ is bijective because P is ﬁnite. Hence,    µi   i) 0 ≤ aij < p       ii) for all 1 ≤ i ≤ j ≤ k we have  Γ: Aut(P ) → (aij)1≤i,j≤k : aij ∈ Z satisfy  µi−µj 0 0 µj   aij = p aij where 0 ≤ aij < p      iii) det((aij)1≤i,j≤k) ≡/ 0 (mod p) is onto. Moreover, if φ1 and φ2 are automorphisms of P such that Γ(φ1) = Γ(φ2) =

(aij)1≤i,j≤k then we have Yk aij φ1(xj) = xi = φ2(xj) i=1 for all 1 ≤ j ≤ k. Hence φ1 = φ2 and Γ is one–to–one.

Let φ1, φ2 ∈ Aut(P ) and let (aij)1≤i,j≤k and (bij)1≤i,j≤k be the corresponding matrices. Then Yk Yk aij bij φ1(xj) = xi and φ2(xj) = xi i=1 i=1 for all 1 ≤ j ≤ k. We have:

(φ1 ◦ φ2)(xj) = φ1(φ2(xj)) Yk bij = φ1( xi ) i=1 Yk bij = φ1(xi) i=1 15

Yk Yk alibij = xl i=1 l=1 P Q k k i=1 alibij = l=1 xl P P k k µl where i=1 alibij ≡ i=1 alibij (mod p ) for all 1 ≤ l ≤ k. Hence, Γ(φ1 ◦ φ2) =

Γ(φ1)Γ(φ2). Thus, Γ is a group homomorphism.

Deﬁnition 2.2.4. Let G be a group and let H1 and H2 be subgroups of G. We say that H1 is Aut(G)-equivalent to H2 and we write H1 ∼Aut(G) H2, if there exists an automorphism f of G such that f(H1) = H2. Remark 2.2.5. The relation deﬁned above is an equivalence relation.

n Theorem 2.2.6. Let P be an abelian group of order p generated by a1, a2, . . . , al,

λi with o(ai) = p , for all i = 1, . . . , l and λ = (λ1 ≥ λ2 ≥ ... ≥ λl) a partition of n. Let N be an elementary abelian subgroup of P of order pt, t ≥ 1. Then there exist i1, i2, . . . , it ∈ {1, 2, . . . , l} distinct such that λ −1 λ −1 λ −1 p i1 p i2 p it N ∼Aut(P ) < ai1 , ai2 , . . . , ait >.

Proof. We will prove by induction on t. Suppose that t = 1. Then N is generated

λ −1 λ −1 λ −1 p 1 r1 p 2 r2 p l rl by x = a1 a2 ··· al , with 0 ≤ ri < p for all i = 1, . . . , l. Let s be such rs 6= 0 and ri = 0 for all i > s. We deﬁne

f : P → P by    ai if i 6= s s f(ai) = Y λj −λs  p rj  aj if i = s j=1 Using Theorem 2.2.3 it is easy to see that f is an automorphism of P . Moreover, s s Y λj −λs Y λj −1 pλs−1 p rj pλs−1 p rj f(as ) = ( aj ) = aj = x. j=1 j=1

pλs−1 Hence, N ∼Aut(P )< as >. 16

Assume that t > 1 and that the conclusion holds for t − 1. Let {x1, x2, . . . , xt} denote a set of generators for N. We denote by N¯ the subgroup of P generated ¯ by x1, x2, . . . , xt−1. Then N is an elementary abelian subgroup of P of order

t−1 p . Hence, by the induction hypothesis there exist distinct i1, i2, . . . , it−1 ∈ {1, 2, . . . , l} and φ an automorphism of P such that φ(N¯) is the subgroup

λ −1 λ −1 λ −1 p i1 p i2 p it−1 generated by ai1 , ai2 , . . . , ait−1 . Then

λ −1 λ −1 λ −1 p i1 p i2 p it−1 φ(N) =< ai1 , ai2 , . . . , ait−1 , φ(xt) > .

λ −1 λ −1 λi −1 p i1 p i2 p t−1 t−1 If φ(xt) ∈< ai1 , ai2 , . . . , ait−1 >, then |N| = |φ(N)| = p which is false. Hence, we have

λ −1 λ −1 λ −1 p i1 p i2 p it−1 φ(xt) = ai1 ai2 ··· ait−1 · z

λ −1 λ −1 λ −1 p 1 r1 p 2 r2 p l rl where z = a1 a2 . . . al , with 0 ≤ ri < p for all i = 1, . . . , l z 6= 1, and ri1 = ri2 = ... = rit−1 = 0. We obtain that

λ −1 λi −1 λi −1 it−1 p 1 s1 p 2 s2 p st−1 φ(N) = < ai1 , ai2 , . . . , ait−1 , φ(xt) > λ −1 λ −1 λ −1 p i1 p i2 p it−1 = < ai1 , ai2 , . . . , ait−1 , z > . ¯ Let P be the subgroup of P generated by {aj : j 6= i1, i2, . . . , it−1}. We have that z ∈ P¯ and has order p. Hence, by the case t = 1, there exists ψ¯ an

λ −1 ¯ ¯ p it automorphism of P such that ψ(< z >) =< ait >, for some 1 ≤ it ≤ l with it 6= i1, i2, . . . , it−1. We deﬁne   ai if i 6∈ {i1, . . . , it−1} ψ : P → P by ψ(ai) :=  ¯  ψ(ai) otherwise

It is easy to see that ψ is an automorphism of P . Moreover, since ψ(aij ) = aij for all j = 1, . . . , t − 1 we have that ψ(φ(N)) is the elementary abelian subgroup

λ −1 λ −1 λ −1 p i1 p i2 p it of P generated by ai1 , ai2 , . . . , ait . Since ψ ◦ φ ∈ Aut(P ) we have λ −1 λ −1 λ −1 p i1 p i2 p it N ∼Aut(P )< ai1 , ai2 , . . . , ait >. 17

Lemma 2.2.7. (Exercise 2.7 in Isaac [8]) Let G be an abelian group and write G∗ = Irr(G).

a) G∗ is an abelian group under the pointwise multiplication of characters.

b) If H is a subgroup of G, let H⊥ = {λ ∈ G∗ | H ⊆ Ker λ}. Then ⊥ is a bijection from the set of subgroups of G onto the set of subgroups of G∗. c) G ' G∗

Remark 2.2.8. Let G be an abelian group and assume that G is generated by

∗ a1, . . . , ak with o(ai) = ni, for all i = 1, . . . , k. Then G is generated by χa1 , . . . , χak , where

χa : G → C i    1 if i 6= j χ (a ) = ai j   εi if i = j

εi is a primitive ni-root of unity for all i = 1, . . . , k. We extend χai as follows

Yk Yk rj rj χai ( aj ) = (χai (aj)) . j=1 j=1 CHAPTER 3 ISOMORPHIC CHARACTER TABLES Let G be a ﬁnite group and let Cl(G) denote the set of conjugacy classes of G.

Suppose Cl(G) = {K1,K2,...,Kt}. We denote by Irr(G) the set of irreducible characters of G.

Deﬁnition 3.0.9. Two groups G1 and G2 have isomorphic character tables if there exist two bijections

α : G1 → G2 and β : Irr(G1) → Irr(G2)

such that for all g ∈ G1 and for all χ ∈ Irr(G1) we have

χ(g) = β(χ)(α(g)). (3.0.1)

Deﬁnition 3.0.10. A class function on a group G is a function ϕ : G → C which is constant on conjugacy classes. Let Gb denote the set of class functions on G. Remark 3.0.11. Gb is a C–algebra with point-wise addition and multiplication of functions. Irr(G) is a basis of Gb as a vector space over C.

Remark 3.0.12. Let Cl(G) = {K1,K2,...,Kt} be the set of conjugacy classes of G. For all 1 ≤ i ≤ t we deﬁne

ei : G → C   1 if g ∈ Ki ei(g) =  0 otherwise

Then {e1, e2, . . . , et} is a set of central, primitive, orthogonal idempotents. Hence, there exists an algebra isomorphism

b G ' Ce1 ⊕ Ce2 ⊕ · · · ⊕ Cet.

18 19

Lemma 3.0.13. Let G1 and G2 be two ﬁnite groups. Suppose there exists an b b algebra isomorphism Φ: G1 → G2 such that Φ(Irr(G1)) = Irr(G2). Then G1 and G2 have isomorphic character tables.

Proof. We have to construct a bijection α : G1 → G2 such that for al g ∈ G1 and for all χ ∈ Irr(G1) we have χ(g) = Φ(χ)(α(g)). Since Φ is a bijection such that

Φ(Irr(G1)) = Irr(G2) we have |Irr(G1)| = |Irr(G2)| = t. Hence, G1 and G2 have the same number of conjugacy classes. Let e1, e2, . . . , et and f1, f2, . . . , ft be the central, b b primitive, orthogonal idempotents from Remark 3.0.12 of G1 and G2 respectively. Then, b G1 = Ce1 ⊕ Ce2 ⊕ · · · ⊕ Cet,

b G2 = Cf1 ⊕ Cf2 ⊕ · · · ⊕ Cft. and Φ : Ce1 ⊕ Ce2 ⊕ · · · ⊕ Cet → Cf1 ⊕ Cf2 ⊕ · · · ⊕ Cft is an algebra isomorphism such that Φ({e1, . . . , et}) = {f1, . . . , ft}.

1 2 Letα ¯ : Cl(G1) → Cl(G2) be deﬁned byα ¯(Ki )Kj , where Φ(ei) = fj.

Let χ ∈ Irr(G1) and let ν = Φ(χ). We have

t 1 χ = i=1χ(Ki )ei and Xt 1 ν = χ(Ki )Φ(ei). i=1

Since ν ∈ Irr(G2) we also have

Xt 2 ν = ν(Kj )fj. j=1

1 2 We obtain χ(Ki )ν(Kj ), for all i = 1, . . . , t and j ∈ {1, . . . , t} such that Φ(ei) = fj.

In particular, we obtain χ(1) = ν(1) = Φ(χ)(1). Since, χ ∈ Irr(G1) was chosen arbitrarily we obtain that |G1| = |G2|. 20

1 2 It remains to show that |Ki | = |Kj | for all i, j ∈ {1, . . . , t} such that

1 2 Φ(ei) = fj. Let g1 ∈ Ki and g2 ∈ Kj . we have

1 2 |Ki | = [G1 : CG1 (g1)] and |Kj | = [G2 : CG2 (g2)]

Since |G1| = |G2| it is enough to show that |CG1 (g1)| = |CG2 (g2)|. We have X X X 2 2 2 |CG1 (g1)| = |χ(g1)| = |Φ(χ)(g2)| = |ν(g2)| = |CG2 (g2),

χ∈Irr(G1) χ∈Irr(G1) ν∈Irr(G2)

1 2 where the second equality follows from the fact that χ(Ki ) = Φ(χ)(Kj ).

1 1 Thus, we can deﬁne α : G1 → G2 such that α(Ki ) =α ¯(Ki ). It is easy to see that α is a bijection such that χ(g) = Φ(χ)(α(g)) for all χ ∈ Irr(G1) and for all g ∈ G1.

Proposition 3.0.14. Two groups G1 and G2 have isomorphic character tables b b if and only if there exists an algebra isomorphism Φ: G1 → G2 such that

Φ(Irr(G1)) = Irr(G2).

Proof. Suppose that G1 and G2 have isomorphic character tables. Then there exist two bijections α : G1 → G2 and β : Irr(G1) → Irr(G2) such that for all g ∈ G1 and for all χ ∈ Irr(G1) we have χ(g) = β(χ)(α(g)). (3.0.2)

As in Lemma 3.0.13 we have

b G1 = Ce1 ⊕ Ce2 ⊕ · · · ⊕ Cet and b G2 = Cf1 ⊕ Cf2 ⊕ · · · ⊕ Cft, where e1, . . . , et and f1, . . . , ft are the central, primitive, orthogonal idempotents b b from Remark 3.0.12 of G1 and G2 respectively. Thus, there exists an algebra 21 isomorphism

Φ: Ce1 ⊕ Ce2 ⊕ · · · ⊕ Cet → Cf1 ⊕ Cf2 ⊕ · · · ⊕ Cft

1 2 such that Φ({e1, . . . , et}) = {f1, . . . , ft} and Φ(ei) = fj where α(Ki ) = Kj .

It is enough to show that Φ(Irr(G1)) = Irr(G2). Let χ ∈ Irr(G1). Then we have

Xt 1 χ = χ(Ki )ei i=1 and Xt 1 Φ(χ) = χ(Ki )Φ(ei) i=1 Let ν = β(χ), then

Xt Xt 2 1 ν = ν(Kj )fj = ν(α(Ki ))Φ(ei). j=1 i=1

Since G1 and G2 have isomorphic character tables we have

1 2 χ(Ki ) = ν(Kj )

hence, Φ(χ) = ν = β(χ). Thus, Φ(Irr(G1)) ⊆ Irr(G2). Since β and Φ are bijections we can also prove that Irr(G2) ⊆ Φ(Irr(G1)). Hence, Φ(Irr(G1)) = Irr(G2). The converse is Lemma 3.0.13. CHAPTER 4 THE NUMBER OF CHARACTER TABLES 4.1 Character Tables of p-groups with Derived Subgroup of Order p

In what follows let p be a prime number and let P be a p-group with |P | = pn and derived subgroup P 0 of order p. Then P 0 ⊆ Z(P ) and P/Z(P ) is elementary abelian. By Lemma 2.2.2 we have that [P : Z(P )] = p2m, for some positive integer m and χ(1) = pm for all χ ∈ Irr(P ) with χ(1) > 1. Hence, Irr(P ) contains pn−1 linear characters and the non-linear characters all have degree pm.

n−1 Deﬁnition 4.1.1. Let m and n be two positive integers such that m ≤ b 2 c, and let p be a prime. We denote by CT m,n,p the set of non-isomorphic irreducible character tables of p-groups of order pn with derived subgroup of order p and center of index p2m. Let A := P/P 0,B := Z(P ) and ϕ : B → A be the restriction to Z(P ) of the projection P ³ P/P 0. Since A and B are abelian groups by Lemma 2.2.7 we have that A∗ := Irr(A) and B∗ := Irr(B) are abelian groups with the point-wise multiplication of characters. Moreover, A ' A∗ and B ' B∗ are isomorphic as groups. Let ϕ∗ : A∗ → B∗ denote the dual map, ϕ∗(χ) = χ ◦ ϕ for all χ ∈ A∗.

Deﬁnition 4.1.2. Let P be a p-group with |P 0| = p and let π : P → P/P 0 denote the canonical projection. Let linIrr(P ) denote the irreducible linear characters of P . We deﬁne

Λ : Irr(P/P 0) → linIrr(P ) by Λ(χb) = χb ◦ π.

Lemma 4.1.3. The map Λ from Deﬁnition 4.1.2 is a bijection. Furthermore, linIrr(P ) is closed under the point-wise multiplication of characters and Λ is multiplicative.

22 23

Proof. The fact that Λ is bijective follows immediately from Lemma 2.22 and

Corollary 2.23 in Isaac [8].

The product of two linear characters is a linear character and for all χb1, χb2 ∈ Irr(P/P 0) we have

Λ(χb1 · χb2) = (χb1 · χb2) ◦ π

= (χb1 ◦ π) · (χb2 ◦ π)

= Λ(χb1) · Λ(χb2).

Hence, Λ is multiplicative.

Deﬁnition 4.1.4. Let P be a p-group with |P 0| = p and [P : Z(P )] = p2m. Let nlIrr(P ) denote the set of non-linear irreducible characters of P . We deﬁne

Ψ: {θ ∈ Irr(Z(P )) : θ 6∈ Im ϕ∗} → nlIrr(P ) by    pmθ(x) if x ∈ Z(P ) Ψ(θ)(x) =   0 otherwise

Lemma 4.1.5. The map Ψ from Deﬁnition 4.1.4 is a bijection. X Proof. It is easy to see that Ψ(θ) is a class function. Hence, Ψ(θ) = aχχ, χ∈Irr(P ) where aχ ∈ C. We will show that there exists ν ∈ nlIrr(P ) such that aν = 1 and aχ = 0 for all χ ∈ Irr(P ), χ 6= ν.

Let χ ∈ Irr(P ). Then, by Cliﬀord’s Theorem we have χ|Z(P ) = χ(1)η, for some η ∈ Irr(Z(P )). We have

aχ = < Ψ(θ), χ > 1 X = Ψ(θ)(x)χ(x) |P | x∈P 1 X = pmθ(x)χ(1)η(x) (by Deﬁnition 4.1.4) p2m|Z(P )| x∈Z(P ) 24

χ(1) 1 X = θ(x)η(x) pm |Z(P )| x∈Z(P ) χ(1) = < θ, η > pm  χ(1)  if η = θ = pm   0 otherwise χ(1) Hence, Ψ(θ) = χ, where χ = χ(1)θ. We obtain, pm |Z(P )

χ(1) pm = Ψ(θ)(1) = χ(1). pm

Hence, χ(1) = pm and Ψ(θ) = χ, with χ ∈ nlIrr(P ). Thus, Ψ is well deﬁned.

∗ We will prove that Ψ is one–to–one. Let θ1, θ2 ∈ Irr(Z(P )) \ Im ϕ such that

m m Ψ(θ1) = Ψ(θ2). Then for all x ∈ Z(P ) we have p θ1(x) = p θ2(x), hence θ1 = θ2. Thus, Ψ is one–to–one. We will show that Ψ is onto. Let χ ∈ nlIrr(P ). Then χ(1) = pm > 1 and by Lemma 2.2.1 we have χ(g) = 0 for all g ∈ P \ Z(P ). By Cliﬀord’s Theorem we have that

χ|Z(P ) = χ(1)θ for some θ ∈ Irr(Z(P )). Hence, χ(x) = pmθ(x) if x ∈ Z(P ) and 0 otherwise. It remains to show that θ 6∈ Im ϕ∗. Suppose that θ = ϕ∗(λ) for some λ ∈ Irr(P/P 0). Then

m χ|Z(P ) = p (λ ◦ φ).

Hence, for all x ∈ P 0 we have

χ(x) = pm(λ(φ(x))) = pm = χ(1).

Thus, P 0 ⊆ Ker χ which is impossible as χ is non-linear. Hence, Ψ is onto.

Proposition 4.1.6. Let P be a p-group with |P 0| = p and [P : Z(P )] = p2m. Let Pb denote the algebra of class functions. Then the multiplication on Pb can be described 25 as follows:    Λ(µ1 µ2) if χ1 = Λ(µ1), χ2 = Λ(µ2)    ∗  Ψ(ϕ (µ)θ) if χ1 = Λ(µ), χ2 = Ψ(θ) χ1 · χ2 =  pm Ψ(θ θ ) if χ = Ψ(θ ), χ = Ψ(θ ) and θ θ 6∈ Im ϕ∗  1 2 1 1 2 2 1 2   X  Λ(µ) if χ = Ψ(θ ), χ = Ψ(θ ) and θ θ ∈ Im ϕ∗  1 1 2 2 1 2 ∗ ϕ (µ)=θ1θ2

for all χ1, χ2 ∈ Irr(P ) and we extend by linearity.

Proof. Let χ1, χ2 ∈ Irr(P ). Suppose that χ1 and χ2 are linear. Then, by Lemma

0 4.1.3, there exist µ1, µ2 ∈ Irr(P/P ) such that Λ(µ1) = χ1 and Λ(µ2) = χ2. We have

χ1 · χ2 = Λ(µ1) · Λ(µ2) = Λ(µ1 · µ2) where the second equality follows from Lemma 4.1.3.

If χ1(1) = 1 and χ2(1) > 1 then, since Λ and Ψ are bijections, there exist

0 ∗ µ ∈ Irr(P/P ) and θ ∈ Irr(Z(P )) \ Im ϕ such that Λ(µ) = χ1 and Ψ(θ) = χ2. We will show that

∗ (χ1 · χ2)(x) = Ψ(ϕ (µ)θ)(x) for all x ∈ P .

Let x ∈ P . If x 6∈ Z(P ) then (χ1 · χ2)(x) = χ1(x)χ2(x) = 0 by Lemma 2.2.1 and by Deﬁnition 4.1.4 we have Ψ(ϕ∗(µ)θ)(x) = 0. If x ∈ Z(P ) we have

(χ1 · χ2)(x) = χ1(x)χ2(x) = Λ(µ)(x)Ψ(θ)(x)

= µ(ϕ(x))pmθ(x) (by Deﬁnition 4.1.2 and Deﬁnition 4.1.4) = pm(ϕ∗(µ)θ)(x) = Ψ(ϕ∗(µ)θ)(x). 26

∗ Hence, (χ1 · χ2)(x) = Ψ(ϕ (µ)θ)(x) for all x ∈ P .

If χ1(1) > 1 and χ2(1) > 1 then, by Lemma 4.1.5, there exist θ1, θ2 ∈

∗ Irr(Z(P )) \ Im ϕ such that Ψ(θ1) = χ1 and Ψ(θ2) = χ2.

∗ Suppose that θ1θ2 6∈ Im ϕ . Then for all x ∈ P \ Z(P ) we have

(χ1 · χ2(x)) = 0 = Ψ(θ1θ2)(x).

Let x ∈ Z(P ) then

(χ1 · χ2)(x) = χ1(x)χ2(x)

= Ψ(θ1)(x)Ψ(θ2)(x)

m m = p θ1(x)p θ2(x) (by Deﬁnition 4.1.4)

m m = p (p θ1 · θ2)(x)

m = p Ψ(θ1θ2)(x).

m Hence, (χ1 · χ2)(x) = p Ψ(θ1θ2)(x) for all x ∈ P . ∗ ∗−1 b Assume that θ1θ2 ∈ Im ϕ . Let T = ϕ (θ1θ2). Since Irr(P ) is a basis for P over C we have X χ1 · χ2 = aχχ χ∈Irr(P ) and aχ =< χ1 · χ2, χ > for all χ ∈ Irr(P ). We will prove that aχ = 1 for all

χ ∈ Λ(T ) and aχ = 0 if χ 6∈ Λ(T ). Let χ ∈ Λ(T ). Then χ = Λ(µ) for some µ ∈ T . We have

aχ = < χ1 · χ2, χ > 1 X = χ (x)χ (x)χ(x) |P | 1 2 x∈P 1 X = Ψ(θ )(x)Ψ(θ )(x)Λ(µ)(x) |P | 1 2 x∈P 1 X = Ψ(θ )(x)Ψ(θ )(x)Λ(µ)(x) (by Lemma 2.2.1) |P | 1 2 x∈Z(P ) 1 X = pmθ (x)pmθ (x)µ(ϕ(x)) (by Deﬁnitions 4.1.4 and 4.1.2) |P | 1 2 x∈Z(P ) 27

1 X = p2m(θ · θ )(x)ϕ∗(µ)(x) |P | 1 2 x∈Z(P ) 1 X = p2m(θ · θ )(x)(θ · θ )(x) (since µ ∈ T ) |P | 1 2 1 2 x∈Z(P ) 1 X = (θ · θ )(x)(θ · θ )(x) (since [P : Z(P )] = p2m) |Z(P )| 1 2 1 2 x∈Z(P )

= < θ1 · θ2, θ1 · θ2 > = 1 as θ1 · θ2 ∈ Irr(Z(P )).

Let χ ∈ Irr(P ) \ Λ(T ). Then χ|Z(P ) = χ(1)ν for some ν ∈ Irr(Z(P )), ν 6= θ1θ2. We have

aχ = < χ1 · χ2, χ > 1 X = χ (x)χ (x)χ(x) |P | 1 2 x∈P 1 X = χ (x)χ (x)χ(x) (by Lemma 2.2.1) |P | 1 2 x∈Z(P ) 1 X = pmθ (x)pmθ (x)χ(1)ν(x) (by Deﬁnition 4.1.4) |P | 1 2 x∈Z(P ) 1 X = χ(1) (θ · θ )(x)ν(x) (since [P : Z(P )] = p2m) |Z(P )| 1 2 x∈Z(P )

= χ(1) < θ1 · θ2, ν > = 0 as ν 6= θ1θ2. Thus, X χ1 · χ2 = Λ(µ). ∗ ϕ (µ)=θ1θ2

n Proposition 4.1.7. Let G1 and G2 be two p-groups with |G1| = |G2| = p ,

0 0 2m 0 |G1| = |G2| = p, and [G1 : Z(G1)] = [G2 : Z(G2)] = p . Let ϕ1 : Z(G1) → G1/G1

0 and ϕ2 : Z(G2) → G2/G2 be the restrictions to Z(G1) and Z(G2) respectively of

0 0 ∗ 0 ∗ ∗ the projections G1 ³ G1/G1 and G2 ³ G2/G2. Let ϕ1 :(G1/G1) → Z(G1)

∗ 0 ∗ ∗ and ϕ2 :(G2/G2) → Z(G2) denote the dual maps. Suppose there exist group 28

0 ∗ 0 ∗ b ∗ ∗ isomorphisms αb :(G1/G1) → (G2/G2) and β : Z(G1) → Z(G2) such that the following diagram is commutative

ϕ∗ 0 ∗ 1 / ∗ (G1/G1) Z(G1)

αb βb ϕ∗ 0 ∗ 2 / ∗ (G2/G2) Z(G2)

b ∗ ∗ Then β(Im ϕ1) = Im ϕ2 and there exist bijections

αblin : linIrr(G1) → linIrr(G2) and b βnl : nlIrr(G1) → nlIrr(G2) such that the following diagrams are commutative

0 ∗ Λ1 / ∗ ∗ Ψ1 / (G1/G1) linIrr(G1) Z(G1) \ Im ϕ1 nlIrr(G1)

b b αb αblin β βnl 0 ∗ Λ2 / ∗ ∗ Ψ2 / (G2/G2) linIrr(G2) Z(G2) \ Im ϕ2 nlIrr(G2) where Λ1, Λ2, Ψ1 and Ψ2 are as in Deﬁnition 4.1.2 and Deﬁnition 4.1.4.

−1 Proof. We deﬁne αblin : linIrr(G1) → linIrr(G2) by αblin := Λ2 ◦ αb ◦ Λ1 . Then it is easy to see that αblin is a bijection and αblin ◦ Λ1 = Λ2 ◦ αb. b ∗ ∗ b ∗ ∗ b ∗ Since β ◦ ϕ1 = ϕ2 ◦ αb we have that β(Im ϕ1) = Im ϕ2. Hence, β(Z(G1) \

∗ ∗ ∗ Im ϕ1) = Z(G2) \ Im ϕ2. We deﬁne

−1 b b b ∗ ∗ βnl : nlIrr(G1) → nlIrr(G2) by β = Ψ2 ◦ β|Z(G1) \Im ϕ ◦ Ψ1 .

b b b Then it is easy to see that βnl is a bijection and βnl ◦ Ψ1 = Ψ2 ◦ β.

Remark 4.1.8. Let G be a ﬁnite group and let χ ∈ Irr(G). Then χ is linear if and only if χ · ν ∈ Irr(G) for all ν ∈ Irr(G). 29

n Proposition 4.1.9. Let G1 and G2 be two p-groups with |G1| = |G2| = p ,

0 0 2m |G1| = |G2| = p, and [G1 : Z(G1)] = [G2 : Z(G2)] = p . Suppose that there b b exists an algebra isomorphism F : G1 → G2 such that F (Irr(G1)) = Irr(G2). Then

F (linIrr(G1)) = linIrr(G2) and F (nlIrr(G1)) = nlIrr(G2).

Proof. Let χ ∈ linIrr(G1). We will show that F (χ) ∈ linIrr(G2). By Remark 4.1.8 it is enough to show that F (χ)ν ∈ Irr(G2) for all ν ∈ Irr(G2).

Let ν ∈ Irr(G2). Since F is a bijection there exists µ ∈ Irr(G1) such that F (µ) = ν. We have F (χ)ν = F (χ)F (µ) = F (χµ) where the last equality holds as F is multiplicative. Since χ is linear by Remark

4.1.8 we have χµ ∈ Irr(G1), hence F (χ)ν ∈ F (Irr(G1)) = Irr(G2).

−1 −1 Using a similar argument for F we can prove F (linIrr(G2)) ⊆ linIrr(G1), hence, linIrr(G2) ⊆ F (linIrr(G1)). Thus, F (linIrr(G1)) = linIrr(G2).

Since F is a bijection such that F (Irr(G1)) = Irr(G2) it follows immediately from the ﬁrst part that F (nlIrr(G1)) = nlIrr(G2).

n Theorem 4.1.10. Let G1 and G2 be two p-groups with |G1| = |G2| = p ,

0 0 2m 0 |G1| = |G2| = p, and [G1 : Z(G1)] = [G2 : Z(G2)] = p . Let ϕ1 : Z(G1) → G1/G1

0 and ϕ2 : Z(G2) → G2/G2 be the restrictions to Z(G1) and Z(G2) respectively of the

0 0 projections G1 ³ G1/G1 and G2 ³ G2/G2. Then the following are equivalent:

0 0 i) there exist α : G1/G1 → G2/G2 and β : Z(G1) → Z(G2) group isomorphisms satisfying α ◦ ϕ1 = ϕ2 ◦ β b b ii) there exists F : G1 → G2 an algebra isomorphism such that F (Irr(G1)) =

Irr(G2).

0 Proof. i) ⇒ ii) Let A1 := G1/G1,B1 := Z(G1),A2 := G2/G2,B2 := Z(G2) and let

α : A1 → A2 and β : B1 → B2 be group isomorphisms such that α ◦ ϕ1 = ϕ2 ◦ β.

∗ ∗ ∗ ∗ ∗ ∗ Then α : A2 → A1 and β : B2 → B1 are group isomorphisms satisfying 30

∗ ∗ ∗ ∗ ∗ ∗ ∗−1 b ∗−1 ϕ1 ◦ α = β ◦ ϕ2, where ϕ1 and ϕ2 are the dual maps. Let αb = α and β = β . ∗ b ∗ Then we have ϕ2 ◦ αb = β ◦ ϕ1.

∗ ∗ ∗ ∗ Let Λ1 : A1 → linIrr(G1), Λ2 : A2 → linIrr(G2), Ψ1 : B1 \ Im ϕ1 → nlIrr(G1)

∗ ∗ and Ψ2 : B2 \ Im ϕ2 → nlIrr(G2) be as in Deﬁnition 4.1.2 and Deﬁnition 4.1.4. By Proposition 4.1.7 there exist bijections

b αblin : linIrr(G1) → linIrr(G2) and βnl : nlIrr(G1) → nlIrr(G2) such that b b αblin ◦ Λ1 = Λ2 ◦ αb and βnl ◦ Ψ1 = Ψ2 ◦ β. (4.1.1)

We deﬁne b b F : G1 → G2 by    αblin(χ) if χ ∈ linIrr(G1) F (χ) =   b βnl(χ) if χ ∈ nlIrr(G1) for all χ ∈ Irr(G1) and we extend by linearity.

It is straightforward to see that F is well deﬁned and F (Irr(G1)) = Irr(G2) as b b αblin and βnl are bijections. Furthermore, since Irr(G1) is a basis for G1 over C and b Irr(G2) is basis for G2 over C we obtain that F is a bijection. It remains to show that F is multiplicative. Since F is linear it is enough to show that F (χ1 · χ2) = F (χ1) · F (χ2) for all χ1, χ2 ∈ Irr(G1).

Let χ1, χ2 ∈ Irr(G1). We will consider several cases.

Case 1. Suppose that χ1, χ2 ∈ linIrr(G1). Then χ1 · χ2 ∈ linIrr(G1) hence,

F (χ1 · χ2) = αblin(χ1 · χ2).

∗ By Lemma 4.1.3 there exist µ1, µ2 ∈ A1 such that Λ1(µ1) = χ1 and Λ1(µ2) = χ2. By Proposition 4.1.6 we obtain 31

χ1 · χ2 = Λ1(µ1µ2). (4.1.2)

We compute

F (χ1) · F (χ2) = αblin(χ1) · αblin(χ2)

= αblin(Λ1(µ1)) · αb(Λ1(µ2))

= Λ2(αb(µ1)) · Λ2(αb(µ2)) (αblin ◦ Λ1 = Λ2 ◦ αb)

= Λ2(αb(µ1µ2)) (α,b Λ2 are multiplicative)

= αblin(Λ1(µ1µ2)) (αblin ◦ Λ1 = Λ2 ◦ αb)

= αblin(χ1 · χ2) (by equation (4.1.2))

= F (χ1 · χ2)

Case 2. Suppose that χ1 ∈ linIrr(G1) and χ2 ∈ nlIrr(G1). Then by Lemma

∗ 4.1.3 there exists µ ∈ A1 such that Λ1(µ) = χ1 and by Lemma 4.1.5 there exist

∗ ∗ θ ∈ B1 \ Im ϕ1 such that Ψ1(θ) = χ2. Then by Proposition 4.1.6 we have

∗ χ1 · χ2 = Ψ1(ϕ1(µ)θ) (4.1.3)

b hence, F (χ1 · χ2) = βnl(χ1χ2). We have

b F (χ1) · F (χ2) = αblin(χ1) · βnl(χ2) b = αblin(Λ1(µ)) · βnl(Ψ1(θ)) b = Λ2(αb(µ)) · Ψ2(β(θ)) (by equation 4.1.1) ∗ b = Ψ2(ϕ2(αb(µ)) · β(θ)) (by Proposition 4.1.6) b ∗ b ∗ b ∗ = Ψ2(β(ϕ1(µ))β(θ)) (ϕ2 ◦ αb = β ◦ ϕ1) b ∗ b = Ψ2(β(ϕ1(µ)θ)) (β is multiplicative) b ∗ b b = βnl(Ψ1(ϕ1(µ)θ)) (Ψ2 ◦ β = βnl ◦ Ψ1) b = βnl(χ1 · χ2) (by equation 4.1.3)

= F (χ1 · χ2) 32

Case 3. Suppose that χ1, χ2 ∈ nlIrr(G1). Then, by Lemma 4.1.5, there exist ∗ ∗ b θ1, θ2 ∈ B1 \ Im ϕ1 such that Ψ1(θ1) = χ1 and Ψ1(θ2) = χ2. Note that since β is a b ∗ ∗ ∗ group isomorphism such that β(Im ϕ1) = Im ϕ2 we have that θ1θ2 ∈ Im ϕ1 if and b b ∗ only if β(θ1)β(θ2) ∈ Im ϕ2.

∗ Case 3.1 Suppose that θ1θ2 6∈ Im ϕ1. Then, by Proposition 4.1.6 we have χ1 · χ2 =

m p Ψ1(θ1θ2) hence m b F (χ1 · χ2) = p βnl(Ψ1(θ1θ2)).

We have

b b F (χ1) · F (χ2) = βnl(χ1) · βnl(χ2) b b = βnl(Ψ1(θ1)) · βnl(Ψ1(θ2)) b b b b = Ψ2(β(θ1)) · Ψ2(β(θ2)) (βnl ◦ Ψ1 = Ψ2 ◦ β) m b b = p Ψ2(β(θ1)β(θ2)) (by Proposition 4.1.6) m b b = p Ψ2(β(θ1θ2)) (β is a group homomorphism) m b b b = p βnl(Ψ1(θ1θ2)(βnl ◦ Ψ1 = Ψ2 ◦ β)

= F (χ1 · χ2)

∗ Case 3.2 Suppose that θ1θ2 ∈ Im ϕ1. Then X χ1 · χ2 = Λ1(µ).

∗ µ ∈ A1 ∗ ϕ1 (µ) = θ1θ2

Hence, X F (χ1 · χ2) = αblin(Λ1(µ)).

∗ µ ∈ A1 ∗ ϕ1 (µ) = θ1θ2 We have

b b F (χ1) · F (χ2) = βnl(χ1) · βnl(χ2) b b = βnl(Ψ1(θ1)) · βnl(Ψ1(θ2)) b b b b = Ψ2(β(θ1)) · Ψ2(β(θ2)) (βnl ◦ Ψ1 = Ψ2 ◦ β) 33

X = Λ2(µ2) (by Proposition 4.1.6)

∗ µ2 ∈ A2 ∗ b b ϕ (µ) = β(θ1)β(θ2) 2 X = Λ2(αb(µ1)) (αb is bijective)

∗ µ1 ∈ A1 ∗ b b ϕ (αb(µ1)) = β(θ1)β(θ2) 2 X ∗ b ∗ = Λ2(αb(µ1)) (ϕ2 ◦ αb = β ◦ ϕ1)

∗ µ1 ∈ A1 b ∗ b b β(ϕ (µ1)) = β(θ1)β(θ2) 1 X b = Λ2(αb(µ1)) (β is multiplicative)

∗ µ1 ∈ A1 b ∗ b β(ϕ (µ1)) = β(θ1θ2) 1 X = αblin(Λ1(µ)) (Λ2 ◦ αb = αblin ◦ Λ1)

∗ µ1 ∈ A1 b ∗ b β(ϕ (µ1)) = β(θ1θ2) 1X b = αblin(Λ1(µ)) (β is bijective)

∗ µ1 ∈ A1 ∗ ϕ1 (µ) = θ1θ2

= F (χ1 · χ2). Thus, F is multiplicative. ii) ⇒ i) Suppose that there exists an algebra isomorphism

b b F : G1 → G2

such that F (Irr(G1)) = Irr(G2). We will construct group isomorphisms

0 ∗ 0 ∗ b ∗ ∗ αb :(G1/G1) → (G2/G2) and β : Z(G1) → Z(G2)

∗ b ∗ such that ϕ2 ◦ αb = β ◦ ϕ1.

Since F is an algebra isomorphism such that F (Irr(G1)) = Irr(G2) by

Proposition 4.1.9 we have F (linIrr(G1)) = linIrr(G2) and F (nlIrr(G1)) = nlIrr(G2). We deﬁne

0 ∗ 0 ∗ −1 αb :(G1/G1) → (G2/G2) by αb = Λ2 ◦ F ◦ Λ1 34

where Λ1, Λ2 are as in Deﬁnition 4.1.2. It is easy to see that αb is well deﬁned and is bijective.

0 ∗ We will show that αb is a group homomorphism. Let µ1, µ2 ∈ (G1/G1) and let

χ1 = Λ1(µ1), χ2 = Λ1(µ2). We have

−1 αb(µ1µ2) = Λ2 (F (Λ1(µ1µ2)))

−1 = Λ2 (F (χ1 · χ2)) (by Proposition 4.1.6)

−1 = Λ2 (F (χ1 · χ2))

−1 = Λ2 (F (χ1) · F (χ2)) (F is multiplicative)

−1 −1 −1 = Λ2 (F (χ1)) · Λ2 (F (χ2)) (Λ2 is multiplicative)

= αb(µ1) · αb(µ2).

Hence, αb is a group homomorphism. We define b ∗ ∗ β : Z(G1) → Z(G2) by   −1 ∗  (Ψ2 ◦ F ◦ Ψ1)(θ) if θ 6∈ Im ϕ1 βb(θ) =  ∗ ∗ ϕ2(αb(µ)) θ = ϕ1(µ) b We will prove first that β is well defined. We must show that for any µ1 µ2 ∈

0 ∗ ∗ ∗ ∗ ∗ (G1/G1) such that ϕ1(µ1) = ϕ1(µ2) we have ϕ2(αb(µ1)) = ϕ2(αb(µ2)). Let

0 ∗ ∗ ∗ ∗ µ1, µ2 ∈ (G1/G1) such that ϕ1(µ1) = ϕ1(µ2). Then, as ϕ2 and αb are group

−1 ∗ homomorphisms, it is enough to show that αb(µ1µ2 ) ∈ Ker ϕ2. Since Λ2 is an

−1 ∗ isomorphism it is enough to show that Λ2(αb(µ1µ2 )) ∈ Λ2(Ker ϕ2). We remark

0 ∗ 0 that a linear character µ of G2 is in Λ2(Ker ϕ2) if and only if µ χ2 = χ2 for all

−1 χ2 ∈ nlIrr(G2). Let χ2 ∈ nlIrr(G2). We will show that Λ2(αb(µ1µ2 ))χ2 = χ2. Since

F (nlIrr(G1)) = nlIrr(G2) there exists χ1 ∈ nlIrr(G1) such that F (χ1) = χ2. We 35 have:

−1 −1 Λ2(αb(µ1µ2 ))χ2 = F (Λ1(µ1µ2 )) · F (χ1)

−1 = F (Λ1(µ1µ2 )χ1)(F is multiplicative)

−1 ∗ = F (χ1) (as Λ1(µ1µ2 ) ∈ Λ1(Ker ϕ1))

= χ2.

b b ∗ Thus, β is well deﬁned. It is easy to see that β is bijective and that ϕ2 ◦ αb = b ∗ β ◦ ϕ1.

∗ ∗ We will prove that for all θ1, θ2 ∈ Z(G1) we have θ1θ2 ∈ Im ϕ1 if and only if b b ∗ β(θ1)β(θ2) ∈ Im ϕ2.

∗ ∗ Let θ1, θ2 ∈ Z(G1) and assume that θ1θ2 ∈ Im ϕ1. If both θ1 and θ2 are in ∗ b b ∗ Im ϕ1, then it is easy to see that β(θ1)β(θ2) ∈ Im ϕ2. ∗ ∗ b b ∗ Suppose that θ1, θ2 6∈ Im ϕ1, θ1θ2 ∈ Im ϕ1, and β(θ1)β(θ2) 6∈ Im ϕ2. Let

χ1 = Ψ1(θ1) and χ2 = Ψ1(θ2). Then, by Proposition 4.1.6, we have X χ1 · χ2 = Λ1(µ).

0 ∗ µ ∈ (G1/G1) ∗ ϕ1 (µ) = θ1θ2

Hence, X F (χ1 · χ2) = F (Λ1(µ))

0 ∗ µ ∈ (G1/G1) ∗ ϕ (µ) = θ1θ2 1 X = Λ2(αb(µ)) (F ◦ Λ1 = Λ2 ◦ αb)

0 ∗ µ ∈ (G1/G1) ∗ ϕ (µ) = θ1θ2 1 X = Λ2(αb(µ)) (αb is bijective)

0 ∗ µ2 ∈ (G2/G2) ∗ −1 ϕ1 (αb (µ2)) = θ1θ2 36 and

F (χ1) · F (χ2) = F (Ψ1(θ1)) · F (Ψ1(θ2)) b b b = Ψ2(β(θ1)) · Ψ2(β(θ2)) (F ◦ Ψ1 = Ψ2 ◦ β) m b b = p Ψ2(β(θ1)β(θ2)) (by Proposition 4.1.6) = pmν

b b where ν = Ψ2(β(θ1)β(θ2)) ∈ nlIrr(G2). Since F is multiplicative we obtain

1 X ν = Λ (αb(µ)) pm 2 0 ∗ µ2 ∈ (G2/G2) ∗ −1 ϕ1 (αb (µ2)) = θ1θ2

b b ∗ which is impossible since ν is an irreducible character. Hence, β(θ1)β(θ2) ∈ Im ϕ2. b b ∗ Using a similar argument we can prove that β(θ1)β(θ2) ∈ Im ϕ2 implies

∗ θ1θ2 ∈ Im ϕ1. b ∗ We will show that β is a group homomorphism. Let θ1, θ2 ∈ Z(G1) . We will prove that b b b β(θ1θ2) = β(θ1)β(θ2).

We have to consider several cases.

∗ Case 1. Suppose that θ1, θ2, θ1θ2 6∈ Im ϕ1. Let χ1 = Ψ1(θ1) and let χ2 = Ψ1(θ2). b b Then F (χ1) = Ψ2(β(θ1)) and F (χ2) = Ψ2(β(θ2)). We have

b −1 β(θ1θ2) = Ψ2 (F (Ψ1(θ1θ2)))

−1 1 = Ψ2 (F ( pm χ1 · χ2)) (by Proposition 4.1.6) −1 1 = Ψ2 ( pm F (χ1) · F (χ2)) (F is an algebra homomorphism) −1 1 m b b = Ψ2 ( pm p Ψ2(β(θ1))Ψ2(β(θ2))) (by Proposition 4.1.6) −1 b b = Ψ2 (Ψ2(β(θ1)β(θ2))) (by Proposition 4.1.6) b b = β(θ1)β(θ2) 37

∗ ∗ Case 2. Suppose that θ1, θ2 6∈ Im ϕ1 but θ1θ2 ∈ Im ϕ1. Then, we have b b ∗ b b ∗ β(θ1), β(θ2) 6∈ Im ϕ2 but β(θ1)β(θ2) ∈ Im ϕ2. Moreover, we have

b ∗ β(θ1θ2) = ϕ2(αb(µ))

0 ∗ ∗ for some µ ∈ (G1/G1) such that θ1θ2 = ϕ1(µ) and

b b ∗ 0 β(θ1)β(θ2) = ϕ2(µ )

0 0 ∗ 0 ∗ for some µ ∈ (G2/G2) . Since αb is onto there existsµ ¯ ∈ (G1/G1) such that αb(¯µ) = µ0. Hence, b b ∗ b ∗ β(θ1)β(θ2) = ϕ2(αb(¯µ)) = β(ϕ1(¯µ)).

It is enough to show that

∗ ϕ1(¯µ) = θ1θ2.

Let χ1 = Ψ1(θ1) and χ2 = Ψ1(θ2). Then by Proposition 4.1.6 we have X X F (χ1χ2) = F (Λ1(µ)) = Λ2(αb(µ))

0 ∗ 0 ∗ µ ∈ (G1/G1) µ ∈ (G1/G1) ∗ ∗ ϕ1(µ) = θ1θ2 ϕ1(µ) = θ1θ2 and

X X 0 F (χ1)F (χ2) = Λ2(µ ) = Λ2(αb(µ)).

0 0 ∗ 0 ∗ µ ∈ (G2/G2) µ ∈ (G1/G1) ∗ 0 b b ∗ b b ϕ2(µ ) = β(θ1)β(θ2) ϕ2(αb(µ)) = β(θ1)β(θ2)

Since F (χ1χ2) = F (χ1)F (χ2) we must have that

0 ∗ ∗ 0 ∗ ∗ b b { µ˜ ∈ (G1/G1) : ϕ1(˜µ) = θ1θ2} = { µˆ ∈ (G1/G1) : ϕ2(αb(ˆµ)) = β(θ1)β(θ2)}.

∗ b b 0 ∗ ∗ Since ϕ2(αb(¯µ)) = β(θ1)β(θ2) we haveµ ¯ ∈ { µˆ ∈ (G1/G1) : ϕ2(αb(ˆµ)) = b b 0 ∗ ∗ ∗ β(θ1)β(θ2)}, henceµ ¯ ∈ { µ˜ ∈ (G1/G1) : ϕ1(˜µ) = θ1θ2}. Thus, ϕ1(¯µ) = θ1θ2. This b b b proves that in this case β(θ1θ2) = β(θ1)β(θ2). 38

∗ ∗ ∗ Case 3. Assume that θ1 ∈ Im ϕ1 but θ2 6∈ Im ϕ1. Then θ1θ2 6∈ Im ϕ1 and b b ∗ 0 ∗ ∗ β(θ1)β(θ2) 6∈ Im ϕ2. Let µ ∈ (G1/G1) be such that θ1 = ϕ1(µ) and let χ1 = Λ1(µ) b ∗ and χ2 = Ψ1(θ2). We have β(θ1) = ϕ2(αb(µ)) and

b −1 β(θ1θ2) = Ψ2 (F (Ψ1(θ1θ2)))

−1 ∗ = Ψ2 (F (Ψ1(ϕ1(µ)θ)))

−1 = Ψ2 (F (χ1 · χ2)) (by Proposition 4.1.6)

−1 = Ψ2 (F (χ1) · F (χ2)) (F is multiplicative)

−1 = Ψ2 (F (Λ1(µ))F (Ψ1(θ2))) −1 b b = Ψ2 (Λ2(αb(µ))Ψ2(β(θ2))) (using the deﬁnition of αb and β) −1 ∗ b = Ψ2 (Ψ2(ϕ2(αb(µ))β(θ2))) (by Proposition 4.1.6) b b = β(θ1)β(θ2).

∗ 0 ∗ Case 4. Suppose that θ1, θ2 ∈ Im ϕ1. Let µ1, µ2 ∈ (G1/G1) be such that

∗ ∗ ∗ θ1 = ϕ1(µ1) and θ2 = ϕ1(µ2). Then θ1θ2 = ϕ1(µ1µ2) and we have

b ∗ β(θ1θ2) = (ϕ2 ◦ αb)(µ1µ2)

∗ ∗ ∗ = (ϕ2 ◦ αb)(µ1)(ϕ2 ◦ αb)(µ2)(ϕ2 and αb are group homomorphisms) b b = β(θ1)β(θ2).

Thus, βb is a group homomorphism.

n 0 0 Corollary 4.1.11. Two groups G1 and G2 with |G1| = |G2| = p , |G1| = |G2| = p,

2m and [G1 : Z(G1)] = p = [G2 : Z(G2)] have isomorphic irreducible character

0 0 tables if and only if there exist α : G1/G1 → G2/G2 and β : Z(G1) → Z(G2) group isomorphisms such that the following diagram is commutative

ϕ1 / 0 Z(G1) G1/G1

β α ϕ2 / 0 Z(G2) G2/G2 39

where ϕ1 and ϕ2 are the restrictions to Z(G1) and Z(G2) respectively of the

0 0 projections G1 ³ G1/G1 and G2 ³ G2/G2.

Proof. It follows immediately from Theorem 4.1.10 and Proposition 3.0.14.

4.2 (m,n,p)-admissible Triples

n−1 Deﬁnition 4.2.1. Let m and n be two positive integers with m ≤ b 2 c, and let p be a prime. An (m, n, p)-admissible triple is a triple (A, B, ϕ) such that

i) A is an abelian group of order pn−1, ii) B is an abelian group of order pn−2m, and iii) ϕ : B → A is a group homomorphism such that |Ker ϕ| = p and A/ϕ(B) is an elementary abelian group of order p2m.

Deﬁnition 4.2.2. Two (m, n, p)-admissible triples (A1,B1, ϕ1) and (A2,B2, ϕ2) are said to be equivalent if there exist group isomorphisms α : A1 → A2 and

β : B1 → B2 such that α ◦ ϕ1 = ϕ2 ◦ β. We denote this equivalence by

(A1,B1, ϕ1) ∼ (A2,B2, ϕ2). Remark 4.2.3. The above relation is an equivalence relation.

n−1 Deﬁnition 4.2.4. Let m and n be two positive integers such that m ≤ b 2 c, and let p be a prime. Let Tm,n,p denote a full set of non-equivalent (m, n, p)-admissible triples (A, B, ϕ).

Deﬁnition 4.2.5. Let p be a prime and let m, n be positive integers such that n − 2m − 1 ≥ 0. We deﬁne Pm,n,p to be a full set of non-isomorphic p-groups P with |P | = pn, |P 0| = p and [P : Z(P )] = p2m.

Deﬁnition 4.2.6. Let m, n be positive integers such that n−2m−1 ≥ 0. We deﬁne

Sn,m to be the set of triples (ρ, e, (αi,j)1≤i≤j≤n−2m+2) that satisfy the following: 1. ρ is a partition of n − 2m, 2. ρ has at least one part of size e,

3. αi,j are non-negative integers satisfying: 40

X a) αi,j = m i,j b) for all k ∈ {2, 3, . . . , n − 2m + 2} we have    mk−1 if k ≤ e   X X  1 if k = e + 1 αi,k + αk,j ≤  1≤i≤k k≤j≤n−2m+2  me − 1 if k = e + 2   mk−2 if k > e + 2 where mk denotes the number of parts of µ of size k. Theorem 4.2.7. (Theorem 8 in Blackburn [2]) Let p be a prime and let m, n be positive integers with n − 2m − 1 ≥ 0. Then there exists a bijection Θm,n,p between the set Pm,n,p and the set Sn,m. Deﬁnition 4.2.8. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. We deﬁne

tm,n,p : Pm,n,p → Tm,n,p by tm,n,p(P ) = (A, B, ϕ)

0 where (A, B, ϕ) ∈ Tm,n,p is such that (A, B, ϕ) ∼ (P/P ,Z(P ), π¯) andπ ¯ is the restriction to Z(P ) of the canonical projection π : P ³ P/P 0.

Proposition 4.2.9. The map tm,n,p is well deﬁned.

n 0 0 If P1,P2 are isomorphic p-groups with |P1| = |P2| = p , |P1| = |P2| = p

2m and [P1 : Z(P1)] = [P2 : Z(P2)] = p then it is straightforward to see that

0 0 (P1/P1,Z(P1), π¯1) ∼ (P2/P2,Z(P2), π¯2). Thus, tm,n,p is well deﬁned. 41

4.3 Equivalent Kernels

In what follows m and n will be positive integers such that n − 2m − 1 ≥ 0 and p will be a prime.

Theorem 4.3.1. Let (A, B, ϕ) be in Tm,n,p. Suppose that B is generated by

µi b1, b2, . . . , bk with o(bi) = p for all i = 1, . . . , k. Then, there exist 1 ≤ i0 ≤ k and β an automorphism of B such that β(Ker ϕ) is the subgroup of B generated by

µ −1 p i0 −1 bt . Moreover, (A, B, ϕ) and (A, B, ϕ◦β ) are equivalent (m, n, p)-admissible triples.

Proof. We have that B is an abelian p-group and Ker ϕ is an elementary abelian subgroup of B of order p. Hence, by Theorem 2.2.6 there exist 1 ≤ i0 ≤ k and β an µ −1 p i0 automorphism of B such that β(Ker ϕ) is the subgroup of B generated by bt . It is straightforward to see that (A, B, ϕ) ∼ (A, B, ϕ ◦ β−1).

Proposition 4.3.2. Let (A1, B, ϕ1) and (A2, B, ϕ2) be (m, n, p)-admissible triples.

µi Suppose that B is generated by b1, b2, . . . , bk with o(bi) = p for all i = 1, . . . , k

pµs−1 and µ1 ≥ ... ≥ µk. Assume that Ker ϕ1 is generated by bs and that Ker ϕ2 is

pµt−1 generated by bt with s 6= t. Then Ker ϕ1 ∼Aut(B) Ker ϕ2 if and only if µs = µt.

Proof. Suppose Ker ϕ1 ∼Aut(B) Ker ϕ2. We have to prove that µs = µt. We will argue by contradiction. Assume µs 6= µt. Without loss of generality we can assume

pµs−1 pµt−1 that µs > µt. Let β be an automorphism of B such that β(< bs >) =< bt >. Then

µ −1 µs−1 µ −1 µ −1 µ −1 p s p a1s p s ass p t ats p k aks β(bs ) = b1 . . . bs . . . bt . . . bk .

µs−1 µt−1 µt p ats p Since µs > µt and the order of bt is p we have that bt = 1. Thus, bt 6∈

pµs−1 β(< bs >), which is a contradiction. Hence, µs = µt.

Conversely, suppose µs = µt. We deﬁne β : B → B by 42

   bi if i 6= s, t  β(bi) := b if i = s  t   bs if i = t

pµs−1 pµt−1 It is easy to see that β is an automorphism of B and β(< bs >) =< bt >.

Hence, Ker ϕ1 ∼Aut(B) Ker ϕ2.

4.4 Equivalent Images

Proposition 4.4.1. Let A be an abelian p-group and let D be a subgroup of A such

t that A/D is elementary abelian of order p . Suppose A is generated by a1, a2, . . . , al

λi with o(ai) = p for all i = 1, . . . , l. Then there exists 1 ≤ i1 < i2 < ··· < it ≤ l such

p p that D ∼Aut(A)< a1, a2, . . . , ai1 , . . . , ait , . . . , al >.

Proof. Let A∗ := Irr(A) and let D⊥ := {χ ∈ A∗ : D ⊆ Ker χ}. By Lemma

2.2.7 we have that D⊥ is a subgroup of A∗ and D⊥ ' Irr(A/D). Moreover, Irr(A/D) ' A/D, hence we obtain D⊥ is elementary abelian of order pt. Hence, by

⊥ Proposition 2.2.6 there exist 1 ≤ i1 < i2 < . . . < it ≤ l such that D ∼Aut(A∗) <

λ −1 λ −1 p i1 p it χ , . . . , χ >, where χa are as in Remark 2.2.8. Thus, using again Lemma ai1 ait j p p 2.2.7 we obtain that D ∼Aut(A)< a1, a2, . . . , ai1 , . . . , ait , . . . , al >.

Theorem 4.4.2. Let (A, B, ϕ) ∈ Tm,n,p and suppose that A is generated by

λi a1, . . . , al with o(ai) = p for all i = 1, . . . , l. Then there exist 1 ≤ i1 < i2 < ··· <

p p i2m ≤ l and α ∈ Aut(A) such that α(ϕ(B)) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >. Moreover, (A, B, ϕ) ∼ (A, B, α ◦ ϕ).

Proof. Since (A, B, ϕ) ∈ Tm,n,p we have that ϕ(B) is a subgroup of A such that A/ϕ(B) is elementary abelian. The conclusion follows from Proposition 4.4.1.

Theorem 4.4.3. Let (A, B, ϕ) and (A, B, ϕ0) be (m, n, p)-admissible triples.

λi Suppose that A is generated by a1, . . . , al with o(ai) = p for all i = 1, . . . , l.

Assume that there exist 1 ≤ i1 < . . . < i2m ≤ l such that ϕ(B) =< 43

p p a1, . . . , ai1 , . . . , ai2m , . . . , al > and 1 ≤ j1 < . . . < j2m ≤ l such that 0 p p 0 ϕ (B) =< a1, . . . , aj1 , . . . , aj2m , . . . , al >. Then ϕ(B) ∼Aut(A) ϕ (B) if and only if for all r ∈ {1, 2,..., 2m} we have λir = λjr .

Proof. We have

λ λ −1 λ −1 λ ϕ(B) ' Z/p 1 Z × · · · × Z/p i1 Z × · · · × Z/p i2m Z × · · · × Z/p l Z and

0 λ λ −1 λ −1 λ ϕ (B) ' Z/p 1 Z × · · · × Z/p j1 Z × · · · × Z/p j2m Z × · · · × Z/p l Z.

0 Suppose that ϕ(B) ∼Aut(A) ϕ (B). Then there exists α ∈ Aut(A) such that α(ϕ(B)) = ϕ0(B). Hence,

λ λ −1 λ −1 λ Z/p 1 Z × · · · × Z/p i1 Z × · · · × Z/p i2m Z × · · · × Z/p l Z

λ λ −1 λ −1 λ ' Z/p 1 Z × · · · × Z/p j1 Z × · · · × Z/p j2m Z × · · · × Z/p l Z.

¯ ˜ ¯ By Krull-Schimdt Theorem we have λi = λi for all 1 ≤ i ≤ l, where λi = λi if ¯ ˜ i 6= i1, . . . i2m and λi = λi − 1 otherwise, and similarly λi = λi if i 6= j1, . . . , j2m and ˜ λi = λi − 1 otherwise. Thus, we must have λir = λjr for all r = 1,..., 2m.

Conversely, if λir = λjr for all 1 ≤ r ≤ 2m we deﬁne

α : A → A by   ai if i 6∈ {i1, . . . , i2m} α(ai) =  ajr if i = ir, r ∈ {1,..., 2m} It is easy to see that α is an automorphism of A and α(ϕ(B)) = ϕ0(B). Hence,

0 ϕ(B) ∼Aut(A) ϕ (B). 44

Theorem 4.4.4. Let (A, B, ϕ) ∈ Tm,n,p. Assume that B is generated by b1, . . . , bk

µi with o(bi) = p for all 1 ≤ i ≤ k and that A is generated by a1, . . . , al with o(ai) =

λi p for all 1 ≤ i ≤ l. Then, there exist 1 ≤ i0 ≤ k and 1 ≤ i1 < . . . < i2m ≤ l and µ −1 0 0 p i0 0 p p ϕ : B → A such that Ker ϕ =< bi0 >, ϕ (B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al > and (A, B, ϕ) ∼ (A, B, ϕ0).

Proof. By Theorem 4.3.1 there exist 1 ≤ i0 ≤ k and β ∈ Aut(B) such that

µ −1 p i0 β(Ker ϕ) =< bi0 > .

Moreover, (A, B, ϕ) ∼ (A, B, ϕ ◦ β−1). Similarly, by Theorem 4.4.2 there exist

1 ≤ i1 < . . . < i2m ≤ l and α ∈ Aut(A) such that

−1 p p α(ϕ ◦ β (B)) =< a1, . . . , ai1 , . . . , ai2m , . . . , al > .

Furthermore, (A, B, ϕ ◦ β−1) ∼ (A, B, α ◦ ϕ ◦ β−1). Let ϕ0 = α ◦ ϕ ◦ β−1.

µ −1 0 p i0 0 p p Then, Ker ϕ =< bi0 > and ϕ (B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >. Since (A, B, ϕ) ∼ (A, B, ϕ ◦ β−1) and (A, B, ϕ ◦ β−1) ∼ (A, B, α ◦ ϕ ◦ β−1) we obtain (A, B, ϕ) ∼ (A, B, ϕ0).

Remark 4.4.5. Let (A, B, ϕ) be an (m, n, p)-admissible triple. Assume that B is

µi generated by b1, . . . , bk with o(bi) = p for all 1 ≤ i ≤ k and that A is generated by

λi a1, . . . , al with o(ai) = p for all 1 ≤ i ≤ l. By Theorem 4.4.4 we can assume that µ −1 p i0 there exist 1 ≤ i0 ≤ k and 1 ≤ i1 < . . . < i2m ≤ l such that Ker ϕ =< bi0 > p p and ϕ(B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >. By Proposition 4.3.2 we can assume that µi0 > µi0+1 and by Theorem 4.4.3 we can assume that it+1 = it + 1 whenever λ = λ and λ > λ whenever λ 6= λ for all t ∈ {1,..., 2m}. Let it  it+1 it it+1 it it+1  k if µi > 1 k¯ = 0  k − 1 otherwise 45

For all 1 ≤ i ≤ k¯ let      µ if i 6= i  λ if i 6= i , . . . , i i 0 ¯ i 1 2m µ¯i = and λi =   µi0 − 1 if i = i0, µi0 > 1 λi − 1 if i = i1, . . . , i2m, λi > 1

We obtain

B/Ker ϕ ' Z/pµ¯1 Z × · · · × Z/pµ¯k¯ Z and

¯ ¯ ϕ(B) ' Z/pλ1 Z × · · · × Z/pλk¯ .

By the First Isomorphism Theorem we have

B/Ker ϕ ' ϕ(B) hence, by Krull-Schmidt Theorem we have µ¯ = λ¯. Thus, λ is obtained from µ¯ by increasing by 1 at r ≤ 2m parts and adding 2m − r parts equal to 1 at the end. 4.5 Double Cosets

Deﬁnition 4.5.1. Let m, n be positive integers such that n − 2m − 1 ≥ 0. We deﬁne Q(m, n) to be the set of triples (µ, e, λ) where

1. µ is a partition of n − 2m. We write µ = (µ1 ≥ ... ≥ µk)

2. µ has a part of size e. We deﬁne i0 to be the largest integer such that

µi0 = e and letµ ¯ be the partition of n − 2m − 1 whose parts are    µi if i 6= i0, i ≤ k 

µ¯i = µi − 1 if i = i0    0 if i > k

3. λ is a partition of n − 1 obtained fromµ ¯ by increasing by 1 at 2m parts

µ¯i1 ,..., µ¯i2m

Remark 4.5.2. Given (µ, e, λ) ∈ Q(m, n) we also have µ,¯ i0, i1, . . . , i2m. Further- more, λi =µ ¯i if i 6∈ {i1, . . . , i2m} and λi =µ ¯i + 1 otherwise. 46

In what follows m and n will be positive integers such that n − 2m − 1 ≥ 0 and p will be a prime.

Notation 4.5.3. Let (µ, e, λ) ∈ Q(m, n). Let (A, B, K, I) be as follows:

n−1 λi 1. A is an abelian group of order p generated by a1, . . . , al with o(ai) = p for all 1 ≤ i ≤ l.

p p 2. I is the subgroup of A generated by a1, a2, . . . , ai1 , . . . , ai2m , . . . , al. n−2m 3. B is an abelian group of order p generated by b1, b2, . . . , bk with

µi o(bi) = p for all i = 1, . . . , k. µ −1 p i0 4. K is the subgroup of B generated by bi0 . Proposition 4.5.4. Let (µ, e, λ) ∈ Q(m, n) and let (A, B, K, I) be as in Notation 4.5.3. Then B/K ' I.

˜ Proof. Let π : B → B/K be the canonical projection and let bi = π(bi) for all ˜ ˜ ¯ ¯ 1 ≤ i ≤ k. Then, B/K is generated by b1,..., bk¯ where k = k if e 6= 1 and k = k − 1 otherwise.

Let      a if i 6∈ {i , . . . , i }  λ if i 6∈ {i , . . . , i } i 1 2m ¯ i 1 2m a˜i = and λi =  p  ai if i ∈ {i1, . . . , i2m}, λi > 1 λi − 1 otherwise

Since λ is obtained fromµ ¯ by increasing by 1 at 2m parts we have o(˜ai) =

λ¯i µ¯i p = p and onlya ˜1,..., a˜k¯ are non-trivial. Hence, we can deﬁne

˜ f : B/K → I by f(bi) =a ˜i for all 1 ≤ i ≤ k¯. Obviously f is a group isomorphism.

Proposition 4.5.5. Let (A, B, K, I) be as in Notation 4.5.3. Then there exist

0 0 0 0 0 (A ,B , ϕ ) ∈ Tm,n,p and group isomorphisms α : A → A and β : B → B such that β(K) = Ker ϕ and α(I) = Im ϕ. 47

Proof. We have that A is an abelian group of order pn−1 and B is an abelian group of order pn−2m. It is easy to see that A/I is elementary abelian of order p2m. By

Proposition 4.5.4 we have that there exists a group isomorphism ϕ1 : B/K ' I. Let

ϕ := i ◦ ϕ1 ◦ π where π : B → B/K is the canonical projection and i : I → A is the canonical injection. Then it is easy to see that (A, B, ϕ) is an (m, n, p)-admissible

0 0 0 0 0 0 triple. Thus, there exists (A ,B , ϕ ) ∈ Tm,n,p such that (A, B, ϕ) ∼ (A ,B , ϕ ). The conclusion follows now immediately.

0 0 0 Proposition 4.5.6. Let (A ,B , ϕ ) ∈ Tm,n,p. Then there exists (A, B, ϕ) an (m, n, p)-admissible triple such that (A, B, Ker ϕ, Im ϕ) = (A, B, K, I) as in

Notation 4.5.3 and (A0,B0, ϕ0) ∼ (A, B, ϕ).

Proof. It follows immediately from Theorem 4.4.4.

Notation 4.5.7. Let (µ, e, λ) ∈ Q(m, n) and let (A, B, K, I) be as in Notation ¯ 4.5.3. We fix (B, θ1, θ2) to be such that i) B¯ is an abelian p-group isomorphic to B/K and I, ¯ ¯ ii) θ1 : B/K → B and θ2 : I → B are group isomorphisms. Definition 4.5.8. Let (µ, e, λ) ∈ Q(m, n) and let (A, B, K, I) be as in Notation ¯ 4.5.3. Let (B, θ1, θ2) be as in Notation 4.5.7. We define the following subsets of Aut(B¯)

¯ ¯ ¯ ¯ Autπ(B) := {β ∈ Aut(B) : there exists β ∈ Aut(B) such that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β}

¯ ¯ −1 −1 AutA(B) := {α¯ ∈ Aut(B) : there exists α ∈ Aut(A) such that α◦i◦θ2 = i◦θ2 ◦α¯}.

¯ ¯ ¯ Proposition 4.5.9. Autπ(B) and AutA(B) are subgroups of Aut(B).

¯ ¯ ¯ ¯ Proof. Let β1, β2 ∈ Autπ(B). Then there exist β1 β2 ∈ Aut(B) such that ¯ βi ◦ θ1 ◦ π = θ1 ◦ π ◦ βi for all i = 1, 2. We have

¯ ¯ ¯ (β1 ◦ β2) ◦ θ1 ◦ π = β1 ◦ θ1 ◦ π ◦ β2 = θ1 ◦ π ◦ (β1 ◦ β2) 48

¯ ¯ ¯ ¯ hence, β1 ◦ β2 ∈ Autπ(B). Obviously idB¯ ◦ θ1 ◦ π = θ1 ◦ π ◦ idB. Thus, Autπ(B) is a subgroup of Aut(B¯). ¯ ¯ Similarly, AutA(B) is a subgroup of Aut(B).

Theorem 4.5.10. Let (A, B, ϕ1) and (A, B, ϕ2) be (m, n, p)-admissible triples.

Assume that Ker ϕ1 = Ker ϕ2 = K and Im ϕ1 = Im ϕ2 = I. Let ϕ˜1 :

B/Ker ϕ1 → ϕ1(B) and ϕ˜2 : B/Ker ϕ2 → ϕ2(B) be the isomorphism induced ¯ by the First Isomorphism Theorem. Let (B, θ1, θ2) be as in Notation 4.5.7 and let

−1 −1 ϕ¯1 = θ2 ◦ ϕ˜1 ◦ θ1 and ϕ¯2 = θ2 ◦ ϕ˜2 ◦ θ1 . Then (A, B, ϕ1) is equivalent to (A, B, ϕ2) ¯ ¯ if and only if ϕ¯1 and ϕ¯2 belong to the same (AutA(B), Autπ(B))–double coset.

Proof. Suppose that (A, B, ϕ1) ∼ (A, B, ϕ2). Then, there exist β ∈ Aut(B) and ˜ α ∈ Aut(A) such that α ◦ ϕ1 = ϕ2 ◦ β. It follows that there exist β ∈ Aut(B/K) andα ˜ ∈ Aut(I) such that the following diagram is commutative

π / ϕ˜1 / i / B B/K ∼ I A

β β˜ α˜ α π / ϕ˜2 / i / B B/K2 ∼ I A

¯ ˜ −1 ¯ ¯ Let β = θ1 ◦ β ◦ θ1 . Then β ∈ Aut(B) and

¯ ˜ β ◦ θ1 ◦ π = θ1 ◦ β ◦ π = θ1 ◦ π ◦ β.

¯ ¯ Hence, β ∈ Autπ(B). −1 ¯ Letα ¯ = θ2 ◦ α˜ ◦ θ2 . Thenα ¯ ∈ Aut(B) and

−1 −1 −1 i ◦ θ2 ◦ α¯ = i ◦ α˜ ◦ θ2 = α ◦ i ◦ θ2

¯ Hence,α ¯ ∈ AutA(B).

−1 −1 Furthermore, sinceϕ ¯1 = θ2 ◦ ϕ˜1 ◦ θ1 andϕ ¯2 = θ2 ◦ ϕ˜2 ◦ θ1 we obtain 49

−1 ϕ¯1 = θ2 ◦ ϕ˜1 ◦ θ1 −1 ˜ −1 = θ2 ◦ α˜ ◦ ϕ˜2 ◦ β ◦ θ1 −1 −1 −1 ˜ −1 = (θ2 ◦ α˜ ◦ θ2 ) ◦ (θ2 ◦ ϕ˜2 ◦ θ1 ) ◦ (θ1 ◦ β ◦ θ1 ) ¯ =α ¯ ◦ ϕ¯2 ◦ β. ¯ ¯ Thus, we haveϕ ¯1 ∈ AutA(B)ϕ ¯2Autπ(B). Similarly we can prove that ¯ ¯ ϕ¯2 ∈ AutA(B)ϕ ¯1Autπ(B). ¯ ¯ Conversely, suppose thatϕ ¯1 andϕ ¯2 are in the same (AutA(B), Autπ(B))–double ¯ ¯ ¯ ¯ coset. Then there existα ¯ ∈ AutA(B) and β ∈ Autπ(B) such thatα ¯ ◦ ϕ¯1 =ϕ ¯2 ◦ β. ¯ −1 −1 Sinceα ¯ ∈ AutA(B) there exists α ∈ Aut(A) such that α ◦ i ◦ θ2 = i ◦ θ2 ◦ α¯. ¯ Similarly, there exists β ∈ Aut(B) such that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. We have

α ◦ ϕ1 = α ◦ (i ◦ ϕ˜1 ◦ π)

−1 −1 = α ◦ i ◦ θ2 ◦ ϕ¯1 ◦ θ1 ◦ π (asϕ ˜1 = θ2 ◦ ϕ¯1 ◦ θ1)

−1 −1 −1 = i ◦ θ2 ◦ α¯ ◦ ϕ¯1 ◦ θ1 ◦ π (as α ◦ i ◦ θ2 = i ◦ θ2 ◦ α¯) −1 ¯ ¯ = i ◦ θ2 ◦ ϕ¯2 ◦ β ◦ θ1 ◦ π (asα ¯ ◦ ϕ¯1 =ϕ ¯2 ◦ β) −1 ¯ = i ◦ θ2 ◦ ϕ¯2 ◦ θ1 ◦ π ◦ β (as β ◦ θ1 ◦ π = θ1 ◦ π ◦ β)

−1 = i ◦ ϕ˜2 ◦ π ◦ β (asϕ ˜2 = θ2 ◦ ϕ¯2 ◦ θ1)

= ϕ2 ◦ β.

Thus, (A, B, ϕ1) ∼ (A, B, ϕ2).

¯ ¯ Our goal is to compute the number of double cosets (AutA(B), Autπ(B)) in Aut(B¯). ¯ ¯ 4.6 Characterizations of Autπ(B) and AutA(B) Hypothesis 4.6.1. Let m, n be positive integers such that n−2m−1 ≥ 0 and let p be a prime. Let (µ, e, λ) ∈ Q(m, n) and let (A, B, K, I) be as in Notation 4.5.3. Let π : B → B/K and i : I → A be the canonical projection and the canonical injection ¯ ˜ respectively. For all 1 ≤ i ≤ k let bi = π(bi) and leta ˜i = ai if i 6= i1, . . . , i2m and 50

p ˜ ˜ a˜i = ai if i ∈ {i1, . . . , i2m} with λi > 1. Then B/K is generated by b1,..., bk¯ with ¯ ˜ µ¯i λi µ¯i o(bi) = p and I is generated bya ˜1,..., a˜k¯ with o(˜ai) = p = p . Let B¯ be an abelian p-group such that B¯ ' B/K and B¯ ' I. Assume that B¯

¯ ¯ ¯ µ¯i ¯ is generated by b1,..., bk¯ with o(bi) = p for all 1 ≤ i ≤ k. We deﬁne

¯ ˜ ¯ θ1 : B/K → B by θ1(bi) = bi and ¯ ¯ θ2 : I → B by θ2(˜ai) = bi

¯ ¯ for all 1 ≤ i ≤ k. Then (B, θ1, θ2) is as in Notation 4.5.7. In what follows we will use the above assumptions and notations. We will give ¯ ¯ characterizations of Autπ(B) and AutA(B) using matrices. ¯ ¯ Theorem 4.6.2. Autπ(B) = Aut(B) if and only if µi0 = µk.

¯ ¯ Proof. Assume that µi0 = µk. Let β ∈ Aut(B). We have to construct β ∈ Aut(B) ¯ such that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. By Theorem 2.2.3 we have that

Yk¯ ¯ ¯ ¯aij β(bj) = bi i=1 for all 1 ≤ j ≤ k¯. ¯ Suppose that µi0 = 1. Then k = k − 1 and we deﬁne   kY−1  aij bi if 1 ≤ j ≤ k − 1 β : B → B by β(b ) = j  i=1  bk if j = k

¯ It is easy to see that β ∈ Aut(B). Furthermore, we also have β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. ¯ ¯ Hence, β ∈ Autπ(B). ¯ Suppose that µi0 > 1. Then k = k and for all 1 ≤ i, j ≤ k we deﬁne 51

   aij j 6= k  cij = pa if i < k, j = k  ik   akk if i = j = k

It is easy to see that (cij)1≤i,j≤k satisﬁes the conditions of Theorem 2.2.3. Hence,

Qk cij β : B → B deﬁned by β(bj) = i=1 bi for all 1 ≤ j ≤ k is an automorphism of B. ¯ ¯ ¯ Moreover, we have β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. Hence, β ∈ Autπ(B). ¯ ¯ Conversely, suppose that Autπ(B) = Aut(B). We will show that µi0 = µk. We ¯ will argue by contradiction. Suppose that µi0 > µk. Then µi0 > 1 and k = k. Let β¯ : B¯ → B¯ be deﬁned by    ¯b if i 6= k ¯ ¯ i β(bi) = µ −1−µ  ¯p i0 k ¯ bi0 · bk if i = k for all 1 ≤ i ≤ k. It is easy to see that β¯ is an automorphism of B¯. Since ¯ ¯ ¯ Autπ(B) = Aut(B) there exists β and automorphism of B such that β ◦ θ1 ◦ π =

θ1 ◦ π ◦ β. For all 1 ≤ j ≤ k we have

Yk aij β(bj) = bi i=1 where (aij)1≤i,j≤k satisﬁes the conditions of Theorem 2.2.3. In particular, since

µi0 > µk we have

µi0 −µk 0 ai0k = p ai0k where 0 ≤ a0 < pµk . We compute i0k µ −1−µ ¯ ¯p i0 k ¯ (β ◦ θ1 ◦ π)(bk) = bi0 · bk and µ −µ p i0 k a0 ¯a1k ¯ i0k ¯akk (θ1 ◦ π ◦ β)(bk) = b1 ··· bi0 ··· bk .

¯ ¯ µi0 −1 Since β ◦ θ1 ◦ π = θ1 ◦ π ◦ β and o(bi0 ) = p we must have

µi0 −1−µk µi0 −µk 0 µi0 −1 p ≡ p ai0k (mod p ) 52

which is impossible. Thus, µi0 = µk.

¯ ¯ ¯ ¯ Theorem 4.6.3. If Autπ(B) $ Aut(B) then k = k and Autπ(B) is isomorphic to the following group of matrices  ¯  ¯  ¯  ¯ 1. (aij)1≤i,j≤k satisfy the conditions of Theorem 2.2.3 (a ) ¯  ij 1≤i,j≤k ¯   µi0 −µt  ¯ 2. ai0t = p a¯i0t for all t > i0 where the multiplication is matrix multiplication followed by reduction modulo pµ¯i in column i for all i = 1, . . . , k.

¯ ¯ Proof. Since Autπ(B) $ Aut(B) by Theorem 4.6.2 we have µi0 > µk ≥ 1. Hence, ¯ ¯ ¯ k = k. Let β ∈ Autπ(B). Then for all 1 ≤ j ≤ k we have

Yk ¯ ¯ ¯aij β(bj) = bi i=1 where (aij)1≤i,j≤k satisfy the conditions of Theorem 2.2.3. In particular for all t > i0

µi0 −1−µt 0 we have ai0t = p ai0t. 0 ¯ ¯ Let t > i0. We will show that ai0t ≡ 0 (mod p). Since β ∈ Autπ(B) there exists ¯ β ∈ Aut(B) such that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. We have

Yk cij β(bj) = bi i=1 for all 1 ≤ j ≤ i where (cij)1≤i,j≤k satisfy the conditions of Theorem 2.2.3. In

µi0 −µt 0 0 µt particular, we have ci0t = p ci0t, 0 ≤ ci0t < p . We compute a ¯ ¯a1t ¯ i0t ¯akt (β ◦ θ1 ◦ π)(bt) = b1 ··· bi0 ··· bk and

c ¯c1t ¯ i0t ¯ckt (θ1 ◦ π ◦ β)(bt) = b1 ··· bi0 ··· bk .

¯ ¯ µi0 −1 Since β ◦ θ1 ◦ π = θ1 ◦ π ◦ β and o(bi0 ) = p we must have

µi0 −1 ai0t ≡ ci0t (mod p ) 53 which implies

µi0 −1−µt 0 µi0 −µt 0 µi0 −1 p ai0t ≡ p ci0t (mod p )

0 µi0 −µt Hence, ai0t ≡ 0 (mod p). Thus, ai0t = p a¯i0t for all t > i0.

Conversely, let (aij)1≤i,j≤k be such that it satisﬁes the conditions of Theorem

µi0 −µt 2.2.3 and for all t > i0 we have ai0t = p a¯i0t. Let

Yk ¯ ¯ ¯ ¯ ¯ ¯aij β : B → B be deﬁned by β(bj) = bi . i=1 ¯ ¯ ¯ ¯ Then β is an automorphism of B. We will show that β ∈ Autπ(B). We will ¯ construct β an automorphism of B such that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. We deﬁne

Yk aij β : B → B by β(bj) = bi . i=1

µi0 −µt Since ai0t = p a¯i0t for all t > i0 we have, by Theorem 2.2.3, that β is an ¯ automorphism of B. Furthermore, it is easy to see that β ◦ θ1 ◦ π = θ1 ◦ π ◦ β. Thus, ¯ ¯ β ∈ Autπ(B).

¯ Theorem 4.6.4. The group AutA(B) is isomorphic to the following group of matrices  ¯  ¯  ¯   ¯ 1. (aij)1≤i,j≤k satisfy the conditions of Theorem 2.2.3   ¯  ¯ (aij)1≤i,j≤k¯ 2. for all 1 ≤ t ≤ 2m and for all j > i , j 6= i , . . . , i  ¯ t t+1 2m   ¯   ¯ 0  ¯ we have ajit = p ajit where the multiplication is the matrix multiplication followed by reduction modulo pµ¯i in column i for all 1 ≤ i ≤ k¯.

¯ Proof. Letα ¯ ∈ AutA(B) then by Theorem 2.2.3 we have

Yk¯ ¯ ¯aij α¯(bj) = bi i=1 54

µ¯i µ¯i−µ¯j 0 ¯ where aij ∈ Z with 0 ≤ aij < p , aij = p aij for all 1 ≤ i ≤ j ≤ k and

0 µ¯j 0 ≤ aij < p and det (aij) 6≡ 0 (mod p). ¯ −1 −1 Sinceα ¯ ∈ AutA(B) there exist α ∈ Aut(A) such that α ◦ i ◦ θ2 = i ◦ θ2 ◦ α¯.

Ql cij Suppose that α(aj) = i=1 ai for all 1 ≤ j ≤ l, where (cij)1≤i,j≤l satisfy the conditions of Theorem 2.2.3. Let 1 ≤ t ≤ 2m and let j > it, j 6= it+1, . . . , i2m. We

have to show that ajit ≡ 0 (mod p). We compute pc pc −1 ¯ 1it lit (α ◦ i ◦ θ2 )(bit ) = a1 . . . al and

a p a p a a −1 ¯ 1it i1it i2mit kit (i ◦ θ2 ◦ α¯)(bit ) = a1 . . . ai1 . . . ai2m . . . ak .

Hence, for j > it, j 6= it+1, . . . , i2m we obtain p cjit = ajit .

Conversely, let (aij)1≤i,j≤k¯ be a matrix satisfying the above conditions. Then

Yk¯ ¯ ¯ ¯ ¯aij α¯ : B → B deﬁned byα ¯(bj) := bi i=1 is an automorphism of B¯. For all 1 ≤ i, j ≤ k¯ we deﬁne   aiit  if j = it, i 6= it+1, . . . , i2m  p c := ij  and for all 1 ≤ t ≤ 2m   aij otherwise

Then, (cij)1≤i,j≤k satisfy the conditions of Theorem 2.2.3, hence the map

 ¯  Yk  cij ¯ ai if 1 ≤ j ≤ k α : A → A deﬁned by α(aj) = i=1   ¯ aj for all k < j ≤ l

−1 −1 is an automorphism of A. It is straightforward to see that α ◦ i ◦ θ2 = i ◦ θ2 ◦ α¯, ¯ thus,α ¯ ∈ AutA(B).

¯ ¯ Corollary 4.6.5. AutA(B) = Aut(B) if and only if i1 = l − 2m + 1. 55

Proof. Suppose i1 6= l − 2m + 1. Since 1 ≤ i1 < i2 < ··· < i2m ≤ l we have i1 < l − 2m + 1 and there exists i1 < t ≤ l, t 6= i1, i2, . . . , i2m. We deﬁne aij = δij if

i 6= t, j 6= i1 and ati1 = 1. Then (aij)1≤i,j≤k¯ satisfy the conditions of Theorem 2.2.3 but since ati1 6≡ 0 (mod p) the matrix (aij)1≤i,j≤k¯ does not satisfy the conditions Yk ¯ ¯ ¯ ¯aij of Theorem 4.6.4. Hence, the mapα ¯ : B → B deﬁned byα ¯(bj) := bi is an i=1 ¯ ¯ element of Aut(B) but is not in AutA(B) which is a contradiction.

Conversely, if i1 = l − 2m + 1 then i2 = l − 2m + 2, . . . , i2m = l, since

1 ≤ i1 < i2 < ··· < i2m ≤ l. Hence, the situation of the second condition of ¯ ¯ Theorem 4.6.4 is not a restriction. Thus, Aut(B) = AutA(B).

¯ ¯ ¯ ¯ Remark 4.6.6. If Autπ(B) = Aut(B) or AutA(B) = Aut(B) then we have only one double coset. ¯ ¯ 4.7 The Number of (AutA(B), Autπ(B)) Double Cosets Proposition 4.7.1. Let f : G → G¯ be a surjective group homomorphism. Let H and K be subgroups of G and assume that either Ker f ⊆ H or Ker f ⊆ K. Then, the number of (H,K)-double cosets in G equals the number of (f(H), f(K))- double cosets in G¯.

Proof. Suppose that Ker f ⊆ H. Let {Hg1K, . . . , HgtK} be a set of representatives for (H,K)-double cosets. We have:

G = Hg1K ∪ Hg2K ∪ · · · ∪ HgtK.

Since f is a surjective group homomorphism we have

[t [t ¯ G = f(G) = f( HgiK) = f(H)f(gi)f(K). i=1 i=1

We have to show that f(gi) 6= f(gj) and f(H)f(gi)f(K) ∩ f(H)f(gj)f(K) = ∅ whenever i 6= j. 56

−1 If i 6= j but f(gi) = f(gj) then we have gigj ∈ Ker f ⊆ H. Thus, there exists h ∈ H such that gi = hgj and this implies HgiK ∩ HgjK 6= ∅, which is a contradiction. Hence, the f(gi)’s are all distinct.

Suppose there exists y ∈ f(H)f(gi)f(K) ∩ f(H)f(gj)f(K). Then, there exist h1, h2 ∈ H and k1, k2 ∈ K such that y = f(h1gik1) = f(h2gjk2). −1 ˜ It follows that (h1gik1)(h2gjk2) ∈ Ker f ⊆ H. Hence, there exists h ∈ H ˜ ˜ such that h1gik1 = hh2gjk2. But h1gik1 ∈ HgiK and hh2gjk2 ∈ HgjK, hence [t ¯ HgiK ∩ HgjK 6= ∅ which is a contradiction. Thus, G = f(H)f(gi)f(K) i=1 is a decomposition into disjoint double cosets and we have that the number of (H,K)-double cosets in G equals the number of (f(H), f(K))-double cosets in G¯. Similarly if Ker f ⊆ K. Hence, the number of (H,K)-double cosets in G equals the number of (f(H), f(K))-double cosets in G¯.

Theorem 4.7.2. Let V be a vector space over a ﬁeld L and let U and W be subspaces of V with U a subspace of codimension 1. Let

H = {f ∈ GL(V ): f(W ) = W } and K = {f ∈ GL(V ): f(U) = U}. i) If W 6= 0,V then

GL(V ) = Hf1K ∪ Hf2K where f1, f2 ∈ GL(V ) such that W ⊆ f1(U) and W 6⊆ f2(U), and the union is disjoint. ii) If W = 0 then GL(V ) = Hf1K, for some f1 ∈ GL(V ). iii) If W = V then GL(V ) = Hf2K, for some f2 ∈ GL(V ). 57

Proof. i). Suppose that 0 6= W 6= V . We will show ﬁrst that

Hf1K = {ψ ∈ GL(V ): W ⊆ ψ(U)} and

Hf2K = {ψ ∈ GL(V ): W 6⊆ ψ(U)} where f1, f2 ∈ GL(V ) such that W ⊆ f1(U) and W 6⊆ f2(U).

Since K = {f ∈ GL(V ): f(U) = U} it is easy to see that f1K = {g ∈

GL(V ): g(U) = f1(U)}. We will show that

Hf1K = {ψ ∈ GL(V ): W ⊆ ψ(U)}.

Let g1 ∈ H and g2 ∈ K. Then g2(U) = U and f1(g2(U)) = f1(U) k W .

Since g1 ∈ H we have g1(W ) = W hence, g1(f1(g2(U))) k g2(W ) = W . Thus, g1f1g2 ∈ {ψ ∈ GL(V ): W ⊆ ψ(U)}. Conversely, let ψ ∈ GL(V ) such that W ⊆ ψ(U). We will show that

ψ ∈ Hf1K. Since W ⊆ ψ(U) and W ⊆ f1(U) there exists g ∈ H such that

−1 g(ψ(U)) = f1(U). Hence, gψ ∈ f1K. Thus, ψ ∈ g f1K ⊆ Hf1K.

Let f2 ∈ GL(V ) be such that W 6⊆ f2(U). We will prove that

Hf2K = {ψ ∈ GL(V ): W * ψ(U)}.

Let αf2β ∈ Hf2K. Then α(f2(β(U))) = α(f2(U)) and if W ⊆ α(f2(U)) we obtain

W ⊆ f2(U) as α(W ) = W . Hence, W 6⊆ α(f2(β(U))). Thus, αf2β ∈ {ψ ∈ GL(V ): W * ψ(U)}.

Conversely, let ψ ∈ GL(V ) such that W * ψ(U). Since U is a subspace of codimension 1 and ψ is an automorphism of V we have

V = ψ(U)⊕ < w1 > 58

for some 0 6= w1 ∈ W \ ψ(U). Similarly we can write

V = f2(U)⊕ < w2 >

for some 0 6= w2 ∈ W \ f2(U). Hence,

W =< w1 > ⊕(W ∩ ψ(U)) =< w2 > ⊕(W ∩ f2(U))

and there exist α ∈ GL(V ) such that α(W ) = W and α(ψ(U)) = f2(U).

−1 −1 Let β := f2 ◦ α ◦ ψ. Then β ∈ GL(V ) and β(U) = f2 (α(ψ(U))) =

−1 f2 (f2(U)) = U. Thus, β ∈ K and ψ = αf2β ∈ Hf2K. Since

GL(V ) = {ψ ∈ GL(V ): W ⊆ ψ(U)} ∪ {ψ ∈ GL(V ): W * ψ(U)} we obtain

GL(V ) = Hf1K ∪ Hf2K.

Furthermore, the above union is disjoint. ii) Suppose that W = 0. Then f(W ) = W for all f ∈ GL(V ) hence, H = GL(V ).

Thus, GL(V ) = Hf1K, for some f1 ∈ GL(V ). iii) Suppose that W = V . Then f(W ) = W and W 6⊆ f(U) for all f ∈ GL(V ).

Hence, H = GL(V ) and GL(V ) = Hf2K for some f2 ∈ GL(V ).

¯ Let (B, θ1, θ2) be as in Hypothesis 4.6.1. Suppose that

µ¯1 = ... =µ ¯j1 > µ¯j1+1 = ... = µj2 > . . . > µ¯js−1+1 = ... =µ ¯js

and let ni be the multiplicity of partµ ¯ji for all 1 ≤ i ≤ s. ¯ ¯ If µi0 = 1 then by Theorem 4.6.2 we have Autπ(B) = Aut(B). Hence, by Remark 4.6.6 we have only one double coset. ¯ ¯ Similarly if i1 = l −2m+1 then by Corollary 4.6.5 we have AutA(B) = Aut(B). Hence by Remark 4.6.6 we have one double coset. 59

In what follows we assume that µi0 > 1 and i1 < l − 2m − 1. We have

µ¯i0 6= 0, and not all of λi1 , . . . , λi2m are 1. Henceµ ¯i0 =µ ¯jq for some q ∈ {1, . . . , s}.

Moreover, i0 = jq−1 + 1. Suppose that

¯ ¯ ¯ ¯ ¯ ¯ λi = ... = λi > λi = ... = λi > . . . > λi = ... = λi > 0 1 f1 f1+1 f2 fh−1+1 fh

¯ ¯ ¯ and either if = i2m or λi +1 = ... = λi = 0. Since λ =µ ¯ we have h fh 2m

¯ ¯ λi =µ ¯j ,..., λi =µ ¯j . f1 e1 fh eh

¯ ¯ ¯ ¯ µ¯i Recall that B is an abelian group generated by b1,..., bk with o(bi) = p for all 1 ≤ i ≤ k. Let Φ(B¯) be the Frattini subgroup of B¯ and let

proj : B¯ → B/¯ Φ(B¯)

¯ be the canonical projection. For all 1 ≤ j ≤ k let xj := proj(bj). Then {x1, . . . , xk} ¯ ¯ is a basis of B/Φ(B) as a vector space over Fp. Deﬁnition 4.7.3. We deﬁne

¯ ¯ ¯ Modp : Aut(B) → Aut(B/Φ(B)) by ¯ Modp(ψ)(xj) = proj(ψ(bj)) for all ψ ∈ Aut(B¯) and for all 1 ≤ j ≤ k, and we extend by linearity.

Remark 4.7.4. The map Modp is well deﬁned and is a group homomorphism. ¯ ¯ ¯ For all 1 ≤ r ≤ s let Vr be the subspace of B/Φ(B) generated by {xt : o(bt) =

µ¯jr p }. Then dimFp Vr = nr. Deﬁnition 4.7.5. We deﬁne

Ys ¯ Π : Modp(Aut(B)) → GL(Vr) r=1 60 by ¯ ¯ ¯ Π(ψ) = (ψ1,..., ψs) Yk Yjr ¯ ait ¯ ait where ψ(xt) = xi and ψr(xt) = xi for all 1 ≤ r ≤ s and for all i=1 i=jr−1+1 jr−1 + 1 ≤ t ≤ jr and we extend by linearity. Remark 4.7.6. Π is well deﬁned and is a group homomorphism. Ys ¯ Lemma 4.7.7. The map Π ◦ Modp : Aut(B) → GL(Vr) is onto. r=1 Q ¯ ¯ s ¯ ¯ Proof. Let (ψ1,..., ψs) ∈ r=1 GL(Vr). Then we deﬁne f : B → B by

Yjr ¯ ¯ait f(bt) = bi i=jr−1+1

Yjr ¯ ait whereµ ¯t =µ ¯jr for some r ∈ {1, . . . , s} and ψr(xt) = xi . It is easy to see i=jr−1+1 ¯ ¯ ¯ that f ∈ Aut(B) and Π(Modp(f)) = (ψ1,..., ψs). Thus, Π ◦ Modp is onto.

Qs Deﬁnition 4.7.8. We deﬁne Πq : r=1 GL(Vr) → GL(Vq) to be the projection onto the qth coordinate. ¯ Lemma 4.7.9. The map Πq ◦ Π ◦ Modp : Aut(B) → GL(Vq) is onto.

Proof. Straightforward.

¯ Lemma 4.7.10. Ker (Πq ◦ Π ◦ Modp) ⊆ Autπ(B).

Proof. Let f ∈ Ker (Πq ◦ Π ◦ Modp). Then ψ = Πq(Π(Modp(f))) = idVq . Since f ∈ Aut(B¯) by Theorem 2.2.3 we have that

Yk ¯ ¯aij f(bj) = bi . i=1 ¯ In order to prove that f ∈ Autπ(B) it is enough, by Theorem 4.6.3, to show that

ai0t ≡ 0 (mod p) for all t > i0 withµ ¯t =µ ¯i0 .

Let t > i0 such thatµ ¯t =µ ¯i0 . Recall, i0 = jq−1 + 1. Then xt ∈ Vq and we have

ψ(xt) = xt. 61

On the other hand, since ψ = Πq(Π(Modp(f))) we have

a¯jq−1+1t a¯jqt ψ(xt) = xjq−1+1 ··· xjq

wherea ¯ut = aut + pZ for all jq−1 + 1 ≤ u ≤ jq. Hence,a ¯jq−1+1t = 0 so ai0t ≡ 0 ¯ (mod p) for all t > i0 withµ ¯t =µ ¯i0 . Thus, f ∈ Autπ(B) and Ker (Πq ◦ Π ◦ Modp) ⊆ ¯ Autπ(B).

¯ Proposition 4.7.11. Let K := (Πq ◦ Π ◦ Modp)(Autπ(B)). Then

K = {ψ ∈ GL(Vq): ψ(U) = U}

where U is the subspace of Vq generated by {xjq−1+1, . . . , xjq }.

¯ Proof. Let f ∈ Autπ(B) and let ψ = (Πq(Π(Modp(f)))). Then

Yk ¯ ¯aij f(bj) = bi i=1 where (aij)1≤i,j≤k satisfy the conditions of Theorem 4.6.3. Hence, for all t > i0 we

have ai0t ≡ 0 (mod p). Let jq−1 + 1 ≤ t ≤ jq then, sinceµ ¯i0 =µ ¯jq−1+1 we have

a¯jq−1+2 t a¯jq t ψ(xt) = xjq−1+2 ··· xjq

wherea ¯ut = aut + pZ for all jq−1 + 2 ≤ u ≤ jq. Since det((aij)1≤i,j≤k) 6≡ 0 (mod p) we obtain that ψ(U) = U.

Conversely, let ψ ∈ GL(Vq) such that ψ(U) = U. We deﬁne

f : B¯ → B¯ by    bt if t 6= jq−1 + 2, . . . , jq ¯ jq f(bt) = Y  ¯bait otherwise  i i=jq−1+2 62

Q jq a¯it where ψ(xt) = i=jq−1+2 xi for all t ∈ {jq−1 + 2, . . . , xjq }. Since ψ ∈ GL(Vq) we ¯ ¯ immediately have that f ∈ Aut(B) and since ψ(U) = U we have f ∈ Autπ(B).

Moreover, Πq(Π(Modp(f))) = ψ. ¯ Thus, (Πq ◦ Π ◦ Modp)(Autπ(B)) = {ψ ∈ GL(Vq): ψ(U) = U}.

¯ Proposition 4.7.12. Let H := (Πq ◦ Π ◦ Modp)(AutA(B)) and let he := |{1 ≤ t ≤ ¯ 2m : λit =µ ¯i0 }|. Then   GL(Vq) he ∈ {0, nq} H =   {ψ ∈ GL(Vq): ψ(W ) = W } if 0 < he < nq

where W is the subspace of Vq generated by {xjq−1+1, . . . , xjq−1+he }.

Proof. Suppose that he = 0. We will show that H = GL(Vq). ¯ Let f ∈ AutA(B) and let ψ = Πq(Π(Modp(f))). Then for all jq−1 + 1 ≤ t ≤ jq we have Yjq a¯ij ψ(xt) = xi i=jq−1+1 Q ¯ t ¯ait where f(bt) = i=1 bi ,(ait)1≤i,t≤k satisfy the conditions of Theorem 4.6.4 and a¯ut = aut + pZ, for all jq−1 + 1 ≤ u ≤ jq. Since det((ait)) 6≡ 0 (mod p) we have

det((¯aut)jq−1+1≤u,t≤jq ) 6= 0. Thus, ψ ∈ GL(Vq).

Conversely, let ψ ∈ GL(Vq). Then we deﬁne

f : B¯ → B¯ by    ¯b if t 6∈ {j + 1, . . . , j } ¯ t q−1 q f(bt) = Q  jq ¯ait i=jq−1+1 bt otherwise Q jq a¯it where ψ(xt) = i=jq−1+1 xt anda ¯ut = aut + pZ for all jq−1 + 1 ≤ t ≤ jq. It is ¯ easy to see that f ∈ AutA(B) and Πq(Π(Modp(f))) = ψ. Thus, in this case we have ¯ (Πq ◦ Π ◦ Modp)(AutA(B)) = GL(Vq) 63

Suppose that 0 < he ≤ nq. Let W be the subspace of Vq generated by

{xjq−1+1, . . . , xjq−1+he }. We will show that

H = {ψ ∈ GL(Vq): ψ(W ) = W }.

¯ Let f ∈ AutA(B) and let ψ = Πq(Π(Modp(f))). Let jq−1 + 1 ≤ t ≤ jq. By Theorem 4.6.4 we have that Yk ¯ ¯ait f(bt) = bi i=1 with aut ≡ 0 (mod p) for all jq−1 + he < u ≤ jq. Hence,

Yhe a¯jq−1+r t ψ(xt) = xjq−1+r r=1

wherea ¯jq−1+r t = ajq−1+r t +pZ for all 1 ≤ r ≤ he. Since det((aij)1≤i,j≤k) 6≡ 0 (mod p) we obtain det((auv)jq−1+1≤u,v≤jq−1+he ) 6= 0 and we have ψ(W ) = W . ¯ ¯ Conversely, let ψ ∈ GL(Vq) such that ψ(W ) = W . Then we deﬁne f : B → B by   ¯ bt if t 6∈ {jq−1 + 1, . . . , jq−1 + he} f(¯b ) = t Q a  he ¯ jq−1+r t r=1 bjq−1+r otherwise

a¯ Qhe jq−1+r t where ψ(xt) = r=1 xjq−1+r ,a ¯ut = aut + pZ for all jq−1 + 1 ≤ u, t ≤ jq−1 + he. ¯ It is easy to see that f ∈ AutA(B) and Πq(Π(Modp(f))) = ψ. Thus, in this case

H = {ψ ∈ GL(Vq): ψ(W ) = W }.

If he = nq then W = Vq and {ψ ∈ GL(Vq): ψ(W ) = W } = GL(Vq).

¯ ¯ ¯ Theorem 4.7.13. The number of (AutA(B), Autπ(B)) double cosets in Aut(B) is at most two.

Proof. We have to consider several cases. ¯ ¯ Case 1. Suppose that µi0 = 1. Then, by Theorem 4.6.2 we have Autπ(B) = Aut(B) hence, by Remark 4.6.6 we have only one double coset. 64

Suppose that µi0 > 1 and let

Ys Modp / Π / Πq / Aut(B¯) Aut(B/¯ Φ(B¯)) GL(Vr) GL(Vq) r=1 be the maps from Definition 4.7.3, Definition 4.7.5, and Definition 4.7.8 respectively.

Then, by Lemma 4.7.9, we have Πq ◦ Π ◦ Modp is onto and by Lemma 4.7.10 we ¯ have Ker (Πq ◦ Π ◦ Modp) ⊆ Autπ(B). Hence, by Proposition 4.7.1 we have that the ¯ ¯ ¯ number of (AutA(B), Autπ(B)) double cosets in Aut(B) is the same as the number of (H,K) double cosets in GL(Vq), where H was deﬁned in Proposition 4.7.12 and K was deﬁned in Proposition 4.7.11. ¯ Let he = |{1 ≤ t ≤ 2m : λit =µ ¯i0 }|. Then 0 ≤ he ≤ nq.

Case 2. Suppose that he ∈ {0, nq}. Then, by Proposition 4.7.12 we have H =

GL(Vq), hence, we have only one double coset.

Case 3. Suppose that 0 < he < nq. Then by Proposition 4.7.12 we have

H = {ψ ∈ GL(Vq): ψ(W ) = W } and by Proposition 4.7.11 we have that

K = {ψ ∈ GL(Vq): ψ(U) = U}

where U is the subspace of Vq generated by {xjq−1+2, . . . , xjq } and W is the subspace of Vq generated by {xjq−1+1, . . . , xjq−1+he }. Since U has codimension 1 we can apply Theorem 4.7.2 and we obtain that we have two double cosets. ¯ ¯ ¯ Hence, the number of (AutA(B), Autπ(B)) double cosets in Aut(B) is at most two.

Corollary 4.7.14. We have exactly two double cosets if and only if µi0 6= 1 and

0 < |{1 ≤ t ≤ 2m : λit = µi0 }| < nq. 65

Proof. Suppose that we have two double cosets. If µi0 = 1 then as in the proof of Theorem 4.7.13 we have only one double coset, which is a contradiction. Hence,

µi0 6= 1. If |{1 ≤ t ≤ 2m : λit = µi0 }| ∈ {0, nq}, then we have only one double coset

(see the proof of Theorem 4.7.13 Case 2). Thus, 0 < |{1 ≤ t ≤ 2m : λit = µi0 }| < nq.

Conversely if µi0 6= 1 and 0 < |{1 ≤ t ≤ 2m : λit = µi0 }| < nq then as in Case 3 of Theorem 4.7.13 we have two double cosets.

¯ ¯ Assume that there are two (AutA(B), Autπ(B)) double cosets. Then, by

Corollary 4.7.14 we have µi0 6= 1 and 0 < he := |{1 ≤ t ≤ 2m : λit = µi0 }| < nq. Givenϕ ¯ ∈ Aut(B¯) we want to have necessary and suﬃcient conditions forϕ ¯ to be ¯ ¯ in the trivial (AutA(B), Autπ(B)) double coset. Recall, ¯ Πq ◦ Π ◦ Modp : Aut(B) → GL(Vq)

is onto, Vq is a vector space over Fp of dimension nq generated by {xjq−1+1, . . . , xjq }, ¯ ¯ H = (Πq ◦ Π ◦ Modp)(AutA(B)) and K = (Πq ◦ Π ◦ Modp)(Autπ(B)). ¯ Lemma 4.7.15. Let ϕ¯ ∈ Aut(B) and let ψ = Πq(Π(Modp(ϕ ¯))). Then ϕ¯ ∈ ¯ ¯ AutA(B)Autπ(B) if and only if ψ ∈ HK.

Proof. Straightforward.

¯ Corollary 4.7.16. Let ϕ¯ ∈ Aut(B) and assume that nq ≥ 2. Let ψ = ¯ ¯ Πq(Π(Modp(ϕ ¯))). Then ϕ¯ ∈ AutA(B)Autπ(B) if and only if W * ψ(U), where

U is the subspace of V generated by {xjq−1+2, . . . , xjq } and W is the subspace of V generated by {xjq−1+1, . . . , xjq−1+he }.

¯ ¯ Proof. By Lemma 4.7.15 we haveϕ ¯ ∈ AutA(B)Autπ(B) if and only if ψ ∈ HK. Since U has codimension 1 we can apply Theorem 4.7.2 and we have

GL(Vq) = {ψ ∈ GL(Vq): W * ψ(U)} ∪ {ψ ∈ GL(Vq): W ⊆ ψ(U)}. 66

Since W * U we have idVq ∈ {ψ ∈ GL(Vq): W * ψ(U)} and we obtain ¯ ¯ ψ ∈ HK if and only if W * ψ(U). Hence,ϕ ¯ ∈ AutA(B)Autπ(B) if and only if W * ψ(U).

¯ The result of the above Corollary depends on the choice of (B, θ1, θ2). To prove this we consider the following counterexample:

Example 4.7.17. Let p be a prime. Let B be an abelian p-group generated by b1, b2

2 with o(b1) = p and o(b2) = p. Let A be an abelian p-group generated by a1, a2, a3

2 with o(a1) = p and o(a2) = o(a3) = p. Let ϕ : B → A be deﬁned by

p ϕ(b1) = a1, ϕ(b2) = a2.

It is easy to see that (A, B, ϕ) is an (2, 5, p)-admissible triple. ¯ ¯ ¯ ¯ ¯ Let B be an abelian p-group generated by b1, b2 with o(b1) = o(b2) = p. We have that B¯ ' B/Ker ϕ and B¯ ' ϕ(B). Let π : B → B/Ker ϕ be the canonical projection and let i : ϕ(B) → A be the canonical injection. Then B/Ker ϕ is

p generated by π(b1), π(b2) and ϕ(B) is generated by a1, a2. ¯ We will show that the choice of the isomorphisms θ1 : B/Ker ϕ → B and ¯ ¯ ¯ θ2 : ϕ(B) → B can change the (AutA(B)Autπ(B)) double cosets. ¯ ¯ ¯ We deﬁne θ1 : B/Ker ϕ → B by θ1(π(b1)) = b1 and θ1(π(b2)) = b2 and ¯ p ¯ ¯ −1 θ2 : ϕ(B) → B by θ2(a1) = b1 and θ2(a2) = b2. Thenϕ ¯ = θ2 ◦ ϕ˜ ◦ θ1 and we have ¯ ¯ ¯ ¯ ¯ ¯ ϕ¯(b1) = b1 andϕ ¯(b2) = b2. Thus,ϕ ¯ = idB¯ and in this caseϕ ¯ ∈ AutA(B)Autπ(B). ¯ ¯ ¯ If we deﬁne θ1 : B/Ker ϕ → B by θ1(π(b1)) = b2 and θ1(π(b2)) = b1 and ¯ p ¯ ¯ ¯ ¯ ¯ ¯ θ2 : ϕ(B) → B by θ2(a1) = b1 and θ2(a2) = b2 we obtainϕ ¯(b1) = b2 andϕ ¯(b2) = b1. ¯ ¯ Thus, in this caseϕ ¯ 6∈ AutA(B)Autπ(B). 4.8 Combinatorial Character Tables and Admissible Triples

Deﬁnition 4.8.1. Let m, n be positive integers with n − 2m − 1 ≥ 0. We deﬁne

CCT n,m to be the set of quadruples (µ, e, λ, φ) satisfying

1) µ is a partition of n − 2m and we write µ = (µ1 ≥ µ2 ≥ ... ≥ µk) 67

2) µ has at least one part of size e. We let i0 be the largest integer such that

µi0 = e. 3) We deﬁneµ ¯ to be the partition of n − 2m − 1 with parts    µi if i 6= i0, i ≤ k 

µ¯i = µi − 1 if i = i0    0 if i > k

4) λ is a partition of n − 1 which is obtained fromµ ¯ as follows: There exist

1 ≤ i1 < i2 < . . . < i2m, then λ is the partition of n − 1 whose parts are   µ¯i + 1 if i ∈ {i1, i2, . . . , i2m} λ := i   µ¯i otherwise   {1, 2} if e 6= 1 and 0 < |{1 ≤ t ≤ 2m, λi = e}| 6 me−1 5. φ ∈ t  {1} otherwise where me−1 denotes the number of parts of µ of size e − 1.

Remark 4.8.2. Given (µ, e, λ, φ) ∈ CCT n,m we also have uniquely determined µ¯ and i0, i1, . . . , i2m.

Remark 4.8.3. CCT n,m does not depend on the prime p.

We will prove that there exists a bijection between Tm,n,p and CCT n,m. Deﬁnition 4.8.4. Let m, n be positive integers with n − 2m − 1 ≥ 0. We deﬁne

Cp : Tm,n,p → CCT n,m

as follows: given (C, D, ψ) ∈ Tm,n,p by Theorem 4.4.4 we have that there exists (A, B, ϕ) an (m, n, p)-admissible triple such that (A, B, Ker ϕ, ϕ(B)) = (A, B, K, I) from Notation 4.5.3 and (A, B, ϕ) ∼ (C, D, ψ). Then we deﬁne

Cp((C, D, ψ)) = (µ, e, λ, φ) 68 where

1. µ is the partition of n − 2m such that B ' Z/pµ1 Z × · · · × Z/pµk Z.

2. e is the smallest integer such that Ker ϕ ⊆ Bpe−1 .

3. λ is the partition of n − 1 such that A ' Z/pλ1 Z × · · · × Z/pλl Z. ¯ −1 4. Let (B, θ1, θ2) be as in Hypothesis 4.6.1 and letϕ ¯ = θ2 ◦ ϕ˜ ◦ θ1 , where ϕ˜ : B/Ker ϕ → ϕ(B) is the isomorphism induced by the First Isomorphism ¯ ¯ Theorem. Let H = AutA(B) and K = Autπ(B) be as in Deﬁnition 4.5.8. Then φ = 1 ifϕ ¯ ∈ HK and φ = 2 otherwise.

Theorem 4.8.5. The map Cp is well deﬁned.

Proof. Let (C, D, ψ) ∈ Tm,n,p and let Cp((C, D, ψ)) = (µ, e, λ, φ). We will show that (µ, e, λ, φ) satisfies the conditions of Definition 4.8.1. We have that µ is a partition of n − 2m. We will show µ has a part of size e. By definition e is the smallest integer such that Ker ϕ ⊆ Bpe−1 . Recall, B

µi is generated by b1, b2, . . . , bk with o(bi) = p for all i = 1, . . . , k and Ker ϕ =< µ −1 p i0 pe−1 bi0 >. Since Ker ϕ ⊆ B we have that e is the smallest integer such that µ −1 p i0 pe−1 < bi0 >⊆ B . Thus, e = µi0 and µ has a part of size e. From Notation 4.5.3 we have that λ is obtained fromµ ¯ by increasing by 1 at 2m parts, whereµ ¯ is the partition of n − 2m − 1 whose parts are    µi if i 6= i0, i < k 

µ¯i = µi − 1 if i = i0  0   0 if i > k

Furthermore, we have that there exist 1 ≤ i1 < . . . < i2m such that λi =µ ¯i + 1 if i ∈ {i1, . . . , i2m} and λi =µ ¯i otherwise. It remains to show that φ ∈ Φ with |Φ| = 2 if and only if |{1 ≤ t ≤ 2m :

λit = e}| ≤ me−1. By Corollary 4.7.14 we have 2 double cosets when e 6= 1 69

and |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1 and 1 double coset otherwise. Thus

(µ, e, λ, φ) ∈ CCT n,m.

We will show that the deﬁnition of Cp is independent of the choice of a representative for an (m, n, p)-admissible triple. Let (C, D, ψ) ∈ Tm,n,p. Then, by

Theorem 4.4.4 we have determined µ, µi0 and λ. Therefore, (A, B, Ker ϕ, ϕ(B)) = ¯ (A, B, K, I) are ﬁxed by Notation 4.5.3 and (B, θ1, θ2) are ﬁxed by Notation 4.5.7. Suppose that there exist (A, B, ϕ) and (A, B, ϕ0) two (m, n, p)-admissible triples such that 1) they both satisfy Notation 4.5.3,

2) (A, B, Ker ϕ, ϕ(B)) = (A, B, Ker ϕ0, ϕ0(B)), and 3) (A, B, ϕ) ∼ (C, D, ψ) ∼ (A, B, ϕ0).

0 Then Cp((C, D, ψ)) = (µ, e, λ, φ) and Cp((C, D, ψ)) = (µ, e, λ, φ ). We have to show that φ = φ0. Since (A, B, ϕ) ∼ (A, B, ϕ0) we have by Theorem 4.5.10 that ϕ¯ and ϕ¯0 are in the same double coset. Thus, φ = 1 if and only if φ0 = 1. Hence,

0 φ = φ and Cp is well deﬁned.

Theorem 4.8.6. Let m,n be positive integers with n − 2m − 1 ≥ 0. Then there exists a bijection Cp : Tm,n,p → CCT n,m given by the map deﬁned in 4.8.4.

0 0 0 Proof. We will show ﬁrst that Cp is one–to–one. Let (C, D, ψ), (C ,D , ψ ) ∈ Tm,n,p

0 0 0 be such that Cp((C, D, ψ)) = Cp((C ,D , ψ )) = (µ, e, λ, φ). Since (µ, e, λ) ∈ Q(m, n) we have, from Notation 4.5.3,(A, B, K, I) associated with (µ, e, λ) and, by ¯ Hypothesis 4.6.1, we have (B, θ1, θ2).

Let ϕ1 : B → A and ϕ2 : B → A be such that:

0 0 0 1) (A, B, ϕ1) ∼ (C, D, ψ) and (A, B, φ2) ∼ (C ,D , ψ ),

2) Ker ϕ1 = Ker ϕ2 = K, and

3) ϕ1(B) = ϕ2(B) = I.

By the deﬁnition of Cp we have thatϕ ¯1 andϕ ¯2 are in the same double coset.

Hence, by Theorem 4.5.10 we have (A, B, ϕ1) ∼ (A, B, ϕ2). Thus, Cp is one–to–one. 70

We will show C is onto. Let (µ, e, λ, φ) ∈ CCT n,m. We have µ = (µ1 ≥ ... ≥

µi0 = e > µi0+1 ≥ ... ≥ µk) and there exist 1 ≤ i1 < . . . < i2m ≤ l such that

λi =µ ¯i + 1 if i ∈ {i1, . . . , i2m} and λi =µ ¯i otherwise, whereµ ¯ = (µ1 ≥ . . . >

µi0 − 1 = e − 1 ≥ µi0+1 ≥ ... ≥ µk, 0,...). We have that (µ, e, λ) ∈ Q(m, n), so we have (A, B, K, I) as in Notation 4.5.3. Recall, B is generated by b1, b2, . . . , bk

µi with o(bi) = p for all i = 1, . . . , k and A is generated by a1, a2, . . . , al with

λi o(ai) = p for all i = 1, . . . , l. We have to deﬁne ϕ : B → A such that (A, B, ϕ) is an (m, n, p)-admissible triple, Ker ϕ = K, ϕ(B) = I and Cp((A, B, ϕ)) = (µ, e, λ, φ). We will consider several cases.

Case 1. Assume that e = 1. Then we consider

Case 1.1 Suppose that {1 ≤ t ≤ 2m : λit = e} = ∅. Then we deﬁne ϕ : B → A by    ai if i 6= i1, i2, . . . , i2m, k  p ϕ(bi) = a if i = i1, . . . , i2m  i   1 if i = k

We have that ϕ(bk) = 1 so < bk >⊆ Ker ϕ. Let x ∈ Ker ϕ and suppose that pr r1 rk µi r1 i1 x = b1 ··· bk , with 0 ≤ ri < p . Then we have 1 = ϕ(x) = a1 ··· ai1 ··· 1. Hence

λi for all i 6∈ {i1, . . . , i2m} we have ri ≡ 0 (mod p ) and since in this case λi = µi we obtain ri = 0 for all i 6∈ {i1, . . . , i2m}. Similarly if i ∈ {i1, . . . , . . . i2m} we obtain

rk ri = 0¿ Thus, x = bk , 0 ≤ rk < p. Hence, Ker ϕ =< bk >. Furthermore, we have p p ϕ(B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >. Thus, (A, B, ϕ) is an (m, n, p)-admissible triple. Since e = 1 we have φ = 1 and since we have only one double coset we obtain Cp((A, B, ϕ)) = (µ, e, λ, φ).

Case 1.2 Suppose that {1 ≤ t ≤ 2m : λit = e} 6= ∅. In this case we deﬁne

ϕ : B → A by   ai if i 6= i1, i2, . . . , i2m ϕ(bi) =  p ai if i = i1, . . . , i2m 71

Note that in this case ϕ(b ) = ap = 1 for some j ∈ {1,..., 2m}. As before we k ij p p can show that Ker ϕ =< bk > and ϕ(B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >. Thus, (A, B, ϕ) is an (m, n, p)-admissible triple. Since e = 1 we have φ = 1 and since we have only one double coset we obtain Cp((A, B, ϕ)) = (µ, e, λ, φ). In the following cases we assume that e > 1.

Case 2. Suppose |{1 ≤ t ≤ 2m : λit = e}| = 0 Then we deﬁne ϕ : B → A by   ai if i 6= i1, . . . , i2m ϕ(bi) :=  p ai otherwise

µ µ p i0 p i0 µi0 −1 In this case we have o(ai0 ) = p , hence ϕ(bi0 ) = ai0 = 1. Hence, < µ µ p i0 p i0 bi0 >⊆ Ker ϕ. We will prove that Ker ϕ =< bi0 >. Let x ∈ Ker ϕ,

r1 rk µi x 6= 1. Then x = b1 . . . bk with 0 ≤ ri < p for all i = 1, . . . , k. We have

r1 pri1 ri0 rl 1 = ϕ(x) = a1 . . . ai1 . . . ai0 . . . al . Hence, if i 6= i1, . . . , i2m we must have

λi λi−1 ri = p si and if i ∈ {i1, . . . , i2m} we have ri = p si with 0 ≤ si < p.

Since for i 6= i0, i1, . . . , i2m we have λi = µi and for i = i1, . . . , i2m we have µ −1 i0 µi −1 p si0 p 0 λi − 1 = µi. We obtain x = bi0 , thus Ker ϕ =< bi0 >. Moreover, p p ϕ(B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al >, hence, A/ϕ(B) is elementary abelian of order p2m. This proves that (A, B, ϕ) is an (m, n, p)-admissible triple. Furthermore

Cp((A, B, ϕ)) = (µ, µi0 , λ, φ) since by Corollary 4.7.14 we have only one double coset.

Case 3. Suppose |{1 ≤ t ≤ 2m : λit = e}| = me−1 + 1, where me−1 denotes the number of parts of µ of size e − 1. In this case we deﬁne ϕ : B → A by   ai if i 6= i1, . . . , i2m ϕ(bi) :=  p ai otherwise

µi −1 µ −1 µi p 0 p i0 p 0 µi0 p We have o(ai0 ) = p and ϕ(bi0 ) = (ai0 ) = ai0 = 1 and as in µ −1 p i0 the previous case we can show that Ker ϕ =< bi0 >. We also have that 72

p p ϕ(B) =< a1, . . . , ai1 , . . . , ai2m , . . . , al > and A/ϕ(B) is elementary abelian of order p2m. Thus, (A, B, ϕ) is an (m, n, p)-admissible triple. Furthermore, in this case we

have only one double coset, hence Cp((A, B, ϕ)) = (µ, µi0 , λ, φ).

Case 4. Suppose 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1. In this case φ ∈ {1, 2}.

Case 4.1 Suppose that φ = 1. Then i0 ∈ {i1, . . . , i2m} and we deﬁne ϕ : B → A by   ai if i 6= i1, . . . , i2m ϕ(bi) :=  p ai otherwise

µ −1 p i0 As before we have Ker ϕ =< bi0 > and (A, B, ϕ) is an (m, n, p)-admissible triple. In this case using Corollary 4.7.16 we have thatϕ ¯ belongs to the double ¯ ¯ coset AutA(B)Autπ(B), hence Cp((A, B, ϕ)) = (µ, µi0 , λ, 1) = (µ, µi0 , λ, φ).

Case 4.2 Suppose that φ = 2. Then let he = |{1 ≤ t ≤ 2m : λit = e}|. We deﬁne ϕ : B → A by   ap if i ∈ {i , . . . , i }\{i , i + 1, . . . , i + h }  i 1 2m 0 0 0 e   p a if i ∈ {i0 + 1, . . . i0 + he} ϕ(b ) := i−1 i   a if i = i  i0+he 0   ai otherwise

µ −1 µ −1 µ −1 p i0 p i0 In this case we have o(a ) = p i0 and ϕ(b ) = a = 1. As in i0+he i0 i0+he µ −1 p i0 the previous cases we can prove Ker ϕ =< bi0 > and (A, B, ϕ) is an (m, n, p)-admissible triple. Furthermore, using Corollary 4.7.16 we have thatϕ ¯ ¯ ¯ does not belong to the Autπ(B)AutA(B) double coset. Hence, Cp((A, B, ϕ)) =

(µ, µi0 , λ, 2) = (µ, µi0 , λ, φ).

Thus, Tm,n,p is in bijection with CCT n,m. 73

4.9 Combinatorial Character Tables and Blackburn Triples

Deﬁnition 4.9.1. Let m, n be positive integers such that n − 2m − 1 ≥ 0. We deﬁne

cctn,m : Sn,m → CCT n,m

cctn,m((ρ, e, (αi,j))) = (ρ, e, λ, φ) as follows

1. let ρ = (ρ1 ≥ ... ≥ ρi0 = e > ρi0+1 ≥ ... ≥ ρk) and letρ ¯ = (ρ1 ≥ . . . >

ρi0 − 1 = e − 1 ≥ ρi0+1 ≥ ... ≥ ρk, 0,..., 0). Then λ is the partition of n − 1 that is obtained fromρ ¯ in the following way: for each αi,j 6= 0 and for each k ∈ {i, j}, we successively increase by 1 αi,j-distinct parts ofρ ¯ of size k − 1 if k ≤ e, and of size k − 2 if k ≥ e + 1.

2. φ is deﬁned as follows   2 if e 6= 1, he+1 = 0 and 0 < he ≤ me−1 φ =  1 otherwise where me−1 denotes the number of parts of ρ of size e − 1 and X X ha := αi,a + αa,j 1≤i≤a a≤j≤n−2m+2 for all a ∈ {2, 3, . . . , n − 2m + 2}.

Remark 4.9.2. The deﬁnition of the map cctn,m does not depend on the prime p.

Remark 4.9.3. Since ha ≤ ma−1 if a ≤ e and ha ≤ ma−2 if a > e + 2 we have that the construction of λ is possible.

Lemma 4.9.4. Let (ρ, e, (αi,j)) ∈ Sn,m and let cctn,m((ρ, e, (αi,j)) = (ρ, e, λ, φ) be as in Deﬁnition 4.9.1. Then λ is obtained from ρ¯ by increasing by 1 at 2m distinct parts.

Proof. Let (ρ, e, (αi,j)) ∈ Sn,m and let cctn,m((ρ, e, (αi,j)) = (ρ, e, λ, φ) be as in

Deﬁnition 4.9.1. Letρ ¯ = (ρ1 ≥ . . . > ρi0 − 1 = e − 1 ≥ ρi0+1 ≥ ... ≥ ρk, 0,..., 0). We 74 will show that λ is a partition of n − 1 obtained fromρ ¯ by increasing by 1 at 2m parts. The conditions on the (αi,j) imply that we can successively increase distinct parts ofρ ¯ to obtain λ. Since λ is obtained fromρ ¯ by increasing by 1 at αi,j-parts, for all αi,j 6= 0 and for all k ∈ {i, j} we have that

Xl X λi = n − 2m − 1 + 2 αi,j i=1 1≤i≤j≤n−2m+2 = n − 2m − 1 + 2m

= n − 1

Thus, λ is a partition of n − 1 obtained fromρ ¯ by increasing by 1 at 2 × X αi,j = 2m parts. 1≤i≤j≤n−2m+2

Remark 4.9.5. Let (ρ, e, (αi,j) ∈ Sn,m and let cctn,m((ρ, e, (αi,j)) = (ρ, e, λ, φ).

Since for each αi,j 6= 0 and for each k ∈ {i, j} we successively increase by 1 at

αi,j-parts we have that we do not increase by at 1 at the same part twice. Hence, if we denote by i1, . . . , i2m the positions were we increase we have that [ ¡ 1 2 ¢ {i1, . . . , i2m} = Ii,j ∪ Ii,j

αi,j 6=0

1 2 where Ii,j is the set of αi,j positions equal to i − 1 if i ≤ e or i − 2 if i > e, and Ii,j is the set of αi,j positions equal to j − 1 if j ≤ e or j − 2 if j > e. Furthermore, we have

1 2 i) Ii,j ∩ Ii,j = ∅

1 1 1 2 ii) Ii,j ∩ Ik,l = ∅ and Ii,j ∩ Ik,l = ∅, if (i, j) 6= (k, l)

1 2 iii) |Ii,j| = |Ii,j| = αi,j.

Lemma 4.9.6. Let (ρ, e, (αi,j)) ∈ Sn,m with e 6= 1 and let cctn,m((ρ, e, (αi,j))) =

(ρ, e, λ, φ) be as in Deﬁnition 4.9.1. Let ρ¯ = (ρ1 ≥ . . . > ρi0 − 1 = e − 1 ≥ ρi0+1 ≥

... ≥ ρk, 0,..., 0) and let 1 ≤ i1 < . . . < i2m be such that λij =ρ ¯ij + 1 for all j = 1,..., 2m. Then we have: 75

a) |{1 ≤ t ≤ 2m : λit = e}| = 0 if and only if he = he+1 = 0;

b) |{1 ≤ t ≤ 2m : λit = e}| = me−1 + 1 if and only if he+1 = 1 and he = me−1;

c) 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1 if and only if either (he+1 = 0 and he 6= 0) or (he+1 = 1 and 0 ≤ he < me−1); where me−1 denotes the number of parts of ρ of size e − 1 and for k ∈ {e, e + 1} we X X denote by hk := αi,k + αk,j. i≤k k≤j

Proof. a) Assume that |{1 ≤ t ≤ 2m : λit = e}| = 0. We will show that he = he+1 = 0. Suppose that he 6= 0. Then there exists i ≤ e or j ≥ e such that

αi,e 6= 0 or αe,j 6= 0. In either case we will increase by 1 at a part of size e − 1,

hence {1 ≤ t ≤ 2m : λit = e} 6= ∅ which is a contradiction. Similarly we obtain a contradiction if we assume he+1 6= 0. Thus, he = he+1 = 0.

Conversely, suppose he = he+1 = 0 and |{1 ≤ t ≤ 2m : λit = e}|= 6 0. Then we increase by 1 at a part of size e − 1. We can do this only if there exist i ≤ j ∈ {1, . . . , n − 2m + 2} such that either e ∈ {i, j} or e + 1 ∈ {i, j} and αi,j 6= 0.

If e ∈ {i, j} then he 6= 0 and if e + 1 ∈ {i, j} then he+1 6= 0. This contradicts the

fact that he = he+1 = 0. Hence, |{1 ≤ t ≤ 2m : λit = e}| = 0.

b) Suppose |{1 ≤ t ≤ 2m : λit = e}| = me−1 + 1. Then, we increase by 1 at all the parts of ρ of size e − 1, hence he = me−1. Since |{1 ≤ t ≤ 2m : λit = e}| = me−1 + 1 we increase by 1 at a part of size e − 1 which was not chosen above, so he+1 = 1.

Conversely, suppose that he = me−1. Then the number of parts of ρ of size e − 1 is at most the number of parts ofρ ¯ of size e − 1 where we increase by 1.

Since he+1 = 1 we increase by 1 at a part of size e − 1 which was not chosen above.

Hence, |{1 ≤ t ≤ 2m : λit = e}| = me−1 + 1. c) Follows immediately from parts a) and b). 76

Proposition 4.9.7. The map cctn,m : Sn,m → CCT n,m from Deﬁnition 4.9.1 is well deﬁned.

Proof. Let (ρ, e, (αi,j)) ∈ Sn,m and let cctn,m((ρ, e, (αi,j)) = (ρ, e, λ, φ) be as in

Deﬁnition 4.9.1. Letρ ¯ = (ρ1 ≥ . . . > ρi0 − 1 = e − 1 ≥ ρi0+1 ≥ ... ≥ ρk, 0,..., 0). By Proposition 4.9.4, we have that λ is obtained fromρ ¯ by increasing by 1 at

2m parts. Hence, there exist 1 ≤ i1 < . . . < i2m such that λij =ρ ¯ij + 1 for all j = 1,..., 2m and λj =ρ ¯j if j 6= i1, . . . , i2m.

It remains to show that if e 6= 1 and 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1 then it is possible to have φ = 2. Suppose that e 6= 1 and 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1. Then, by Lemma 4.9.6 c) we have that either (he+1 = 1, he < me−1) or

(he+1 = 0, he 6= 0). Hence, by the deﬁnition of the map cctn,m, we have that either φ = 1 or φ = 2.

Thus, the map cctn,m : Sn,m → CCT n,m is well deﬁned.

Lemma 4.9.8. Let (ρ, e, (αi,j)) ∈ Sn,m. Suppose that ρ = (ρ1 ≥ ... ≥ ρi0 = e >

ρi0+1 ≥ ... ≥ ρk). Let ρ¯ be the partition of n − 2m − 1 whose parts are    ρi if i 6= i0 

ρ¯i := ρi − 1 if i = i0    0 if i > k

−1 Let p be a prime and let P = Θm,n,p((ρ, e, (αi,j)). Then

P/P 0 ' Z/pλ1 Z × · · · × Z/pλl Z, where λ is a partition of n-1 obtained from ρ¯ as follows: for each αi,j 6= 0 and for   k − 1 if k ≤ e each k ∈ {i, j} we increase by 1 αi,j-parts of ρ¯ of size .  k − 2 if k ≥ e + 1 77

Proof. Suppose p is odd. Then using the notation from Blackburn [2], see also earlier in this paper Proposition 2.1.2, we have that P is generated by

{xa : a ∈ I} ∪ {ya : a ∈ I} ∪ {zb : b ∈ J} where

I = {(i, j, k) : 1 ≤ i ≤ j ≤ n − 2m + 2, 1 ≤ k ≤ αi,j} and

J = {(i, k) : 2 ≤ i ≤ n − 2m + 2, 1 ≤ k ≤ di} We also have P 0 =< bpe−1 >, where    z(e+1,1) if de+1 = 1  b = xp if α = 1  (e+1,j,1) e+1,j   p y(i,e+1,1) if αi,e+1 = 1

0 pe−1 Suppose that P =< z(e+1,1) >. Then we have: Y Y Y 0 p P/P ' < xa > × < ya > × < zb > × < z(e+1,1) > a∈I a∈I b∈J b6=(e+1,1) Y Y Y p p Z(P ) ' < xa > × < ya > × < zb > × < z(e+1,1) > . a∈I a∈I b∈J b6=(e+1,1) Hence

Y Y Y 0 p p p Z(P )/P ' < xa > × < ya > × < zb > × < z(e+1,1) > a∈I a∈I b∈J b6=(e+1,1)

ρ1 ρk e Since Z(P ) ' Z/p Z × · · · × Z/p Z and o(z(e+1,1)) = p we have

Z(P )/P 0 ' Z/pρ¯1 Z × · · · × Zpρ¯k Z

P P Moreover, |{xa : a ∈ I}| + |{ya : a ∈ I}| = i,j αi,j + i,j αi,j = 2m. We obtain that P/P 0 ' Z/pλ1 Z × · · · × Z/pλl Z, where λ is a partition of n − 1 obtained fromρ ¯ by increasing by 1 at 2m parts.

Let αi,j 6= 0 and let 1 ≤ k ≤ αi,j. Then a = (i, j, k) ∈ I and 78

   pi if i ≤ e + 1 si+1 o(xa) = p =  pi−1 if i > e + 1    pj if j ≤ e + 1 o(y ) = psj +1 = a   pj−1 if j > e + 1 X X If i ≤ e + 1 we have 0 6= hi := αu,i + αi,v ≤ mi−1. Hence, ρ has at least u≤i i≤v i αi,j parts of size i − 1. For all 1 ≤ k ≤ αi,j we have that x(i,j,k) has order p , hence λ has αi,j-parts of size i obtained by increasing by 1 at αi,j-parts ofρ ¯ of size i − 1.

Similarly if i > e + 1 we have that λ has αi,j-parts of size i − 1 obtained by increasing by 1 at αi,j-parts ofρ ¯ of size i − 2.

The same argument works if we ﬁx i and work with j. Thus, for all αi,j 6= 0 and for all k ∈ {i, j} we have that λ has αi,j parts of size    k if k ≤ e + 1   k − 1 if k > e + 1 obtained by increasing by 1 at αi,j-parts ofρ ¯ of size    k − 1 if k ≤ e + 1  k − 2 if k > e + 1

0 p 0 p The cases P =< x(e+1,j,1) > and P =< y(i,e+1,1) > are treated similarly. Assume that p = 2. Then using Proposition 2.1.3 we have to consider only the case p = 2, e = 1 and α1,2 = 1. In this case we have P is generated by the same set X as above, but the generators satisfy diﬀerent relations. In this case we have

0 2 P =< x(1,2,1) > and Y Y ¡ 2 2 ¢ 2 Z(P ) = < xa > × < ya > × < x(1,2,1) > × < zb > . a∈I a6=(1,2,1) b∈J

Thus, Y Y 0 ¡ 2 2 ¢ Z(P )/P ' < xa > × < ya > × < zb > a∈I a6=(1,2,1) b∈J 79 and

Y Y 0 ¡ 2 2 ¢ P/P ' < xa > × < ya > × < x¯(1,2,1) > × < y¯(1,2,1) > < zb > . a∈I a6=(1,2,1) b∈J

The conclusion follows now as before.

Theorem 4.9.9. Let p be a prime and let m, n be positive integers such that n − 2m − 1 ≥ 0. Then the following diagram is commutative

Θm,n,p / Pm,n,p ∼ Sn,m

tm,n,p cctn,m Cp / Tm,n,p ∼ CCT n,m

Proof. Let (ρ, e, (αi,j)) ∈ Sn,m we will prove

−1 (Cp ◦ tm,n,p ◦ Θm,n,p)((ρ, e, (αi,j))) = cctn,m((ρ, e, (αi,j)).

−1 0 Let P = Θm,n,p((ρ, e, αi,j)) then tm,n,p(P ) ∼ (P/P ,Z(P ), π¯) whereπ ¯ is the restriction to Z(P ) of the canonical projection π : P → P/P 0. By Lemma 4.9.8 we have

P/P 0 ' Z/pλ1 Z × · · · × Z/pλl Z

where λ is the partition of n − 1 obtained fromρ ¯ = (ρ1 ≥ . . . > ρi0 − 1 = e − 1 ≥

ρi0+1 ≥ ... ≥ ρk, 0,..., 0) by increasing by 1 at αi,j parts equal to k − 1 if k ≤ e and equal to k − 2 if k > e, for all αi,j 6= 0 and for all k ∈ {i, j}. Using Blackburn’s results we have

Z(P ) ' Z/pρ1 Z × · · · × Z/pρk Z and e is the smallest integer such that Kerπ ¯ = P 0 ⊆ Z(P )pe−1 . We obtain

Cp(tm,n,p(P )) = (ρ, e, λ, φ). 80

0 0 0 Suppose that cctn,m((ρ, e, (αi,j)) = (ρ, e, λ , φ ). Then we have that λ is obtained fromρ ¯ in the same way λ is obtained (see Deﬁnition 4.9.1). Thus, λ = λ0.

It remains to show φ = φ0. We will show φ0 = 2 if and only if φ = 2.

0 Suppose φ = 2. Then, by Deﬁnition 4.9.1, we have e 6= 1, he+1 = 0 and

0 < he ≤ me−1. In this case using Proposition 2.1.2 we have that P is generated by

0 pe−1 {xa, ya : a ∈ I} ∪ {zb : b ∈ J}, P =< z(e+1,1) > and Y Y p p Z(P ) ' (< xa > × < ya >) × < zb > a∈I b∈J Y Y 0 p P/P ' (< xa > × < ya >) × < zb > × < z(e+1,1) > .

a∈I b ∈ J b 6= (e + 1, 1)

In order to deﬁne Cp(tm,n,p(P )) we have ﬁxed A and B abelian p-groups and ¯ (B, θ1, θ2) as in Hypothesis 4.6.1. Recall that B is an abelian group generated by

ρi b1, b2, . . . , bk with o(bi) = p for all i = 1, . . . , k and A is an abelian group generated

λi by a1, a2, . . . , al with o(ai) = p for all i = 1, . . . , l. We deﬁne a group isomorphism

fZ : Z(P ) → B

p p such that fZ sends the generators {xa, ya : a ∈ I} ∪ {zb : b ∈ J} to {b1, . . . , bk} and such that fZ (z(e+1,1)) = bi0 .

Since he+1 = 0 and he > 0 we have that

he = |{1 ≤ t ≤ 2m : λit = e}|.

Hence, i0 ∈ {i1, . . . , i2m} and since when we increase by 1 we always increase

λi0 e at the ﬁrst parts of size e − 1 we have that o(ai0 ) = p = p . Furthermore, since he is strictly less than the number of parts ofρ ¯ of size e − 1 we obtain that

e−1 o(ai0+he ) = p . 81

We deﬁne a group isomorphism

0 fP/P 0 : P/P → A

such that fP/P 0 sends the generators {xa, ya : a ∈ I} ∪ {zb : b ∈ J \{(e +

p p 0 1, 1)}} ∪ {z(e+1,1)} to the generators of A and such that fP/P (z(e+1,1)) = ai0+he and e 0 0 f({xa, ya : o(xa) = o(ya ) = p }) = {ai0 , . . . , ai0+he−1}.

−1 0 Let ϕ : B → A, ϕ := fP/P ◦ π¯ ◦ fZ . Then ϕ(bi0 ) = ai0+he and ϕ(bi0+1) = ai0 , . . . , ϕ(bi0+he ) = ai0+he−1. Furthermore, (A, B, Ker ϕ, ϕ(B)) are as in Notation ¯ −1 4.5.3. Let (B, θ1, θ2, ) be as in Hypothesis 4.6.1 and letϕ ¯ = θ2 ◦ ϕ˜ ◦ θ1 , where ϕ˜ : B/Ker ϕ → ϕ(B) is the isomorphism given by the First Isomorphism Theorem. ¯ By Deﬁnition 4.8.4 we have that φ = 2 ifϕ ¯ 6∈ HK, where H = AutA(B) and ¯ K = Autπ(B). By Corollary 4.7.16 it is enough to show that W 6⊆ ψ(U), where

ψ = Πq(Π(Modp(ϕ ¯))),

Ys Modp / Π / Πq / Aut(B¯) Aut(B/¯ Φ(B¯)) GL(Vr) GL(Vq), r=1

Vq is an Fp vector space with a basis given by {xi0 , xi0+1, . . . , xi0+he , . . . , xi0+me−1 } and U and W are subspaces of Vq such that W is generated by {xi0 , . . . , xi0+he−1} and U is generated by {xi0+1, . . . , xi0+he , . . . , xi0+me−1 }. Then, we have

ψ(xi0 ) = xi0+he , ψ(xi0+1) = xi0 , . . . , ψ(xi0+he ) = xi0+he−1.

We obtain ψ(U) ⊇< xi0 , . . . , xi0+he−1 >= W . Hence, by Corollary 4.7.16 we have ϕ¯ 6∈ HK. Thus, φ = 2. Conversely, if φ = 2 then e 6= 1 and we have two double cosets. By Corollary

4.7.14 we obtain that we increase by 1 at parts ofρ ¯ of size e − 1 and we do not

increase at all the parts ofρ ¯ of size e − 1. Thus, 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1. By Deﬁnition 4.9.1 we have to show he+1 = 0 and he 6= 0. By Lemma 4.9.6 82

we have that either (he+1 = 0 and he 6= 0) or (he+1 = 1 and 0 ≤ he < me−1). We will show that the case (he+1 = 1, 0 ≤ he < me−1) implies φ = 1.

Suppose that he+1 = 1 and 0 ≤ he < me−1. Then either αi,e+1 = 1 for some i ≤ e + 1 or αe+1,j = 1 for some j ≥ e + 1. Suppose αi,e+1 = 1. Then

0 pe P =< y(i,e+1,1) > and we have

Y Y p p p p Z(P ) ' (< xa > × < ya >) × < x(i,e+1,1) > × < y(i,e+1,1) > × < zb >

a ∈ I b∈J a 6= (i, e + 1, 1) Y Y 0 P/P ' (< xa > × < ya >) × < x(i,e+1,1) > × < y(i,e+1,1) > × < zb >

a ∈ I b∈J a 6= (i, e + 1, 1) Let A and B be abelian groups as in the ﬁrst part. We deﬁne a group isomorphism

gZ : Z(P ) → B

p p such that it sends the generators {xa, ya : a ∈ I} ∪ {zb : b ∈ J} to {b1, . . . , bk} and p such that gZ (y(i,e+1,1)) = bi0 .

Since 0 < |{1 ≤ t ≤ 2m : λit = e}| we have that i0 ∈ {i1, . . . , i2m} as we always

e increase by 1 at the ﬁrst parts of size e − 1. Hence, o(ai0 ) = p . We deﬁne a group isomorphism

0 gP/P 0 : P/P → A such that

p gP/P 0 ({xa, ya : a ∈ I \{(i, e+1, 1)}}∪{x(i,e+1,1), y(i,e+1,1)}∪{zb : b ∈ J}) = {a1, . . . , al}

p 0 and gP/P (y(i,e+1,1)) = ai0 . −1 0 Let ϕ : B → A, ϕ := gP/P ◦ π¯ ◦ gZ . Then ϕ(bi0 ) = ai0 and (A, B, Ker ϕ, ϕ(B)) ¯ are as in Notation 4.5.3. Let (B, θ1, θ2, ) be as in Hypothesis 4.6.1 and letϕ ¯ = 83

−1 θ2 ◦ ϕ˜ ◦ θ1 as before. We will apply again Corollary 4.7.16. In this case Vq, U and

W are the same as before. We have ψ(xi0 ) = xi0 and ψ(U) = U. Hence, W 6⊆ ψ(U) and by Corollary 4.7.16 we haveϕ ¯ ∈ HK. Thus, by Deﬁnition 4.8.4 we have φ = 1 which is a contradiction.

Similarly, if αe+1,j = 1 for some j > e + 1 we obtain a contradiction. Thus,

Cp(tm,n,p((ρ, e, (αi,j)))) = cctn,m((ρ, e, (αi,j))) for all (ρ, e, (αi,j)) ∈ Sn,m.

4.10 The Preimage of the Map cctn,m Deﬁnition 4.10.1. Let I be a set with 2m elements. We deﬁne      |S| = m, |s| = 2 for all s ∈ S    P2(I) := S ∈ P(P(I)) : and for all s , s ∈ S with .  1 2      s1 6= s2 we have s1 ∩ s2 = ∅

Remark 4.10.2. Let I = {a1, . . . , a2m}. Then S = {{a1, a2},..., {a2m−1, a2m}} ∈

P2(I). Thus, the set P2(I) is non-empty. Deﬁnition 4.10.3. Let I and I¯ be two sets such that |I|¯ ≤ |I| = 2m. Let ¯ f : I → I be a surjective map and let S1,S2 ∈ P2(I). We say S1 is f-equivalent to

S2 and we write S1 ∼f S2 if there exists a bijection ψ : I → I such that f ◦ ψ = f and ψ(S1) = S2. Remark 4.10.4. The relation deﬁned above is an equivalence relation.

We want to describe the preimage of the map cctn,m : Sn,m → CCT n,m. Let

(µ, e, λ, φ) ∈ CCT n,m. Then we have

µ = (µ1 ≥ ... ≥ µi0 = e > µi0+1 ≥ ... ≥ µk)

µ¯ = (µ1 ≥ . . . > µi0 − 1 = e − 1 ≥ µi0+1 ≥ ... ≥ µk, 0,..., 0)

λ = (λ1 ≥ λ2 ≥ ... ≥ λl) 84

and there exists 1 ≤ i1 < . . . < i2m ≤ l such that λi =µ ¯i + 1 if i ∈ {i1, . . . , i2m}

and λi =µ ¯i otherwise. Let E := {1 ≤ t ≤ 2m : λit = e}. If E 6= ∅ then assume E = {s + 1, . . . , s + r} for some 0 ≤ s ≤ 2m − r.

Deﬁnition 4.10.5. Let (µ, e, λ, φ) ∈ CCT n,m and letµ, ¯ i0, i1, . . . , i2m and E be as

0 above. For all i ∈ {i1, . . . , i2m} we deﬁne f1 and fi as follows: i) if φ = 1 then    µ¯i + 1 ifµ ¯i ≤ e − 1, i 6= is+1  fi = e + 1 if i = i  s+1   µ¯i + 2 ifµ ¯i ≥ e

ii) if φ = 2 then   µ¯i + 1 ifµ ¯i ≤ e − 1 f = i   µ¯i + 2 ifµ ¯i ≥ e iii) if e = 1 and E 6= ∅ we further deﬁne   1 ifµ ¯i = 0 f 0 = i   µ¯i + 2 ifµ ¯i ≥ 1

Deﬁnition 4.10.6. Let (µ, e, λ, φ) ∈ CCT n,m, I = {1, 2,..., 2m} and let

0 0 fi1 , . . . , fi2m and fi1 , . . . , fi2m be as in Deﬁnition 4.10.5. We deﬁne

F : I → {fi1 , . . . , fi2m }

F(j) = fij and

0 0 0 F : I → {fi1 , . . . , fi2m }

0 0 F (j) = fij . 85

Remark 4.10.7. The maps F and F 0 from Deﬁnition 4.10.6 are always surjective.

0 0 They are injective if and only if fi1 , . . . , fi2m are all distinct and fi1 , . . . , fi2m are all distinct respectively .

Deﬁnition 4.10.8. Let I = {1, 2,..., 2m} and let F and F 0 be the maps from

Deﬁnition 4.10.6. Let S ∈ P2(I). We deﬁne

FS : S → P({fi1 , . . . , fi2m })

FS({a, b}) = {fia , fib } and

0 0 0 FS : S → P({fi1 , . . . , fi2m })

F 0 ({a, b}) = {f 0 , f 0 }. S ia ib

Deﬁnition 4.10.9. Let (µ, e, λ, φ) ∈ CCT n,m, I = {1, 2,..., 2m} and let

S ∈ P2(I). We deﬁne α((µ, e, λ, φ),S) = (αi,j)1≤i≤j≤n−2m+2 by

−1 αi,j := |FS ({i, j})| for all 1 ≤ i ≤ j ≤ n − 2m + 2.

0 0 If e = 1 and E 6= ∅ we further deﬁne α ((µ, 1, λ, 1),S) = (αi,j)1≤i≤j≤n−2m+2 by

0 0 −1 αi,j := |FS ({i, j})|

0 for all 1 ≤ i ≤ j ≤ n − 2m + 2, where FS and FS are the maps from in Deﬁnition 4.10.8.

Lemma 4.10.10. Let (µ, e, λ, φ) ∈ CCT n,m and let I = {1, 2,..., 2m}. Let S1,S2 ∈

1 P2(I) and let α((µ, e, λ, φ),S1) = (αi,j)1≤i≤j≤n−2m+2 and α((µ, e, λ, φ),S2) =

2 (αi,j)1≤i≤j≤n−2m+2 be as in Deﬁnition 4.10.9. Then we have α((µ, e, λ, φ),S1) =

α((µ, e, λ, φ),S2) if and only if S1 ∼F S2 where F : I → {fi1 . . . , fi2m }, F(j) = fij . 86

Proof. Suppose that S1 ∼F S2. Then there exists a bijection ψ : I → I such that

F ◦ ψ = F and ψ(S1) = S2. Let i, j be such that 1 ≤ i ≤ j ≤ n − 2m + 2. We will

1 2 prove that αi,j = αi,j. We have

α1 = |F −1({i, j})| and α2 = |F −1({i, j})|. i,j S1 i,j S2

Assume that F −1({i, j}) = {{r , s },... {r , s }}. Then {r , s },... {r , s } ∈ S S1 1 1 t t 1 1 t t 1 and F(r1) = ... = F(rt) = i and F(s1) = ... = F(st) = j. Since ψ is a bijection such that ψ(S1) = S2 we obtain {ψ(r1), ψ(s1)},..., {ψ(rt), ψ(st)} ∈ S2 and since

F ◦ ψ = F we have F(ψ(rk)) = F(rk) = i and F(ψ(sk)) = F(sk) = j for all k = 1, . . . , t. Thus, {{ψ(r ), ψ(s )},..., {ψ(r ), ψ(s )}} ⊆ F −1({i, j}). Hence, 1 1 t t S2 α2 = |F −1({i, j})| ≥ α1 . i,j S2 i,j 1 2 1 2 Similarly we can prove αi,j ≥ αi,j. Thus, αi,j = αi,j for all 1 ≤ i ≤ j ≤ n − 2m + 2.

Conversely, suppose that α((µ, e, λ, φ),S1) = α((µ, e, λ, φ),S2). We will construct ψ : I → I such that F ◦ ψ = F and ψ(S1) = S2. Assume α1 , . . . , α1 6= 0 and all the other α1 ’s are zero. Then for all a1,b1 at,bt i,j 1 ≤ k ≤ t we have |F −1({a , b })| = |F −1({a , b })|. S1 k k S2 k k Let S := F −1({a , b }) and S := F −1({a , b }). We have S = ∪t S 1,k S1 k k 2,k S2 k k 1 k=1 1,k t and S2 = ∪k=1S2,k. Since |S1,k| = |S2,k| for all k = 1, . . . , t there exists a bijection

ψk : S1,k → S2,k. Furthermore, FS2 (ψk({r, s})) = FS1 ({r, s}) = {ak, bk} for all

{r, s} ∈ S1,k and we have a bijection [ [ ¯ ψ : s1 → s2

s1∈S1,k s2∈S2,k ¯ ¯ such that ψk({r, s}) = {ψk(r), ψk(s)} for all {r, s} ∈ S1,k. We have

[ [t [ I = s¯1 = s¯1

s¯1∈S1 k=1 s¯1∈S1,k 87

[ [t [ I = s¯2 = s¯2

s¯2∈S2 k=1 s¯2∈S2,k and for all r ∈ I there exist a unique k and a unique s ∈ I such that {r, s} ∈ S1,k. We can deﬁne ¯ ψ : I → I by ψ(r) = ψk(r).

Since ψk is a bijection for all k = 1, . . . , t we have that ψ is a bijection. Moreover,

ψ(S1) = S2 and F ◦ ψ = F. Thus, S1 ∼F S2.

Similarly we can prove the following

Lemma 4.10.11. Let (µ, 1, λ, 1) ∈ CCT n,m such that E 6= ∅ and let I =

0 01 {1, 2,..., 2m}. Let S1,S2 ∈ P2(I) and let α ((µ, 1, λ, 1),S1) = (αi,j)1≤i≤j≤n−2m+2

0 02 and α ((µ, 1, λ, 1),S2) = (αi,j)1≤i≤j≤n−2m+2 be as in Deﬁnition 4.10.9. Then

0 0 0 α ((µ, 1, λ, 1),S1) = α ((µ, 1, λ, 1),S2) if and only if S1 ∼F 0 S2 where F : I →

0 0 0 0 {fi1 , . . . , fi2m }, F (j) = fij .

Lemma 4.10.12. Let (µ, e, λ, φ) ∈ CCT n,m and let I = {1, 2,..., 2m}. Let

S ∈ P2(I) and α((µ, e, λ, φ),S) = (αi,j)1≤i≤j≤n−2m+2 be as in Deﬁnition 4.10.9. Then

(αi,j)1≤i≤j≤n−2m+2 satisfy the following:

1) αi,j are non-negative integers for all 1 ≤ i ≤ j ≤ n − 2m + 2 X 2) αi,j = m i,j 3) for all k ∈ {2, 3, . . . , n − 2m + 2} we have    mk−1 if k ≤ e   X X  1 if k = e + 1 αi,k + αk,j ≤  1≤i≤k k≤j≤n−2m+2  me − 1 if k = e + 2   mk−2 if k > e + 2 where mk denotes the number of parts of µ of size k.

−1 Proof. 1) Since αi,j = |FS ({i, j})| we have that the ﬁrst assertion holds. 88

X X −1 2) We have αi,j = |FS ({i, j})| = |S| = m i,j i,j 3) Let k ∈ {2, 3, . . . , n − 2m + 2}. We have X X X X −1 −1 αi,k + αk,j = |F ({i, k})| + |F ({k, j})| 1≤i≤k k≤j≤n−2m+2 1≤i≤k k≤j≤n−2m+2 .

≤ |{1 ≤ t ≤ 2m : fit = k}|

Suppose that k 6= e. Then we have:

   |{1 ≤ t ≤ 2m :µ ¯it + 1 = k}| if k < e    1 if k = e + 1 |{1 ≤ t ≤ 2m : f = k}| ≤ it   |{1 ≤ t ≤ 2m :µ ¯ + 2 = e + 2}| if k = e + 2  it   |{1 ≤ t ≤ 2m :µ ¯it + 2 = k}| if k > e + 2    |{1 ≤ t ≤ 2m :µ ¯it = k − 1}| if k < e    1 if k = e + 1 =   |{1 ≤ t ≤ 2m :µ ¯ = e}| if k = e + 2  it   |{1 ≤ t ≤ 2m :µ ¯it = k − 2}| if k > e + 2    mk−1(¯µ) if k < e    1 if k = e + 1 ≤   me(¯µ) k = e + 2   mk−2(¯µ) k > e + 2 where mk(¯µ) denotes the number of parts ofµ ¯ of size k. If k < e − 1 or k > e we have mk(¯µ) = mk(µ) and me(¯µ) = me(µ) − 1. We obtain:    mk−1 if k < e    1 if k = e + 1 |{1 ≤ t ≤ 2m : fit = k}| ≤   me − 1 if k = e + 2   mk−2 if k > e + 2 89

If k = e then |{1 ≤ t ≤ 2m : fit = e}| ≤ |{1 ≤ t ≤ 2m :µ ¯it + 1 = e}| = |E| ≤

me−1 + 1. If |E| = me−1 + 1 then fis+1 = e + 1 hence, |{1 ≤ t ≤ 2m : fit = e}| <

|E| ≤ me−1 + 1. Thus, the conclusion holds for k = e also.

Similarly we can prove the following:

Lemma 4.10.13. Let (µ, 1, λ, 1) ∈ CCT n,m such that E 6= ∅ and let I =

0 0 {1, 2,..., 2m}. Let S ∈ P2(I) and α ((µ, e, λ, φ),S1) = (αi,j)1≤i≤j≤n−2m+2 be as in

0 Deﬁnition 4.10.9. Then (αi,j)1≤i≤j≤n−2m+2 satisfy the following

0 1) αi,j are non-negative integers for all 1 ≤ i ≤ j ≤ n − 2m + 2 X 0 2) αi,j = m i,j 3) for all k ∈ {2, 3, . . . , n − 2m + 2} we have    mk−1 if k ≤ e   X X  1 if k = e + 1 0 0 αi,k + αk,j ≤  1≤i≤k k≤j≤n−2m+2  me − 1 if k = e + 2   mk−2 if k > e + 2 where mk denotes the number of parts of µ of size k.

Theorem 4.10.14. Let (µ, e, λ, φ) ∈ CCT n,m and assume either e 6= 1 or E = ∅.

Let I = {1, 2,..., 2m} and let F = {fi1 , . . . , fi2m } be as in Deﬁnition 4.10.5. Then

−1 cctn,m((µ, e, λ, φ)) = {(µ, e, α((µ, e, λ, φ),S)) : [S] ∈ P2(I)/ ∼F }

where F : I → {fi1 , . . . , fi2m }, F(j) = fij and α((µ, e, λ, φ),S) is as in Deﬁnition 4.10.9.

Proof. Let P reim := {(µ, e, α((µ, e, λ, φ),S)) : [S] ∈ P2(I)/ ∼F }. Let

0 0 −1 0 0 (µ , e , (αi,j)) ∈ cctn,m((µ, e, λ, φ)). We will prove that (µ , e , (αi,j)) ∈ P reim. We

0 0 0 0 0 0 0 0 −1 have cctn,m((µ , e , (αi,j))) = (µ , e , λ , φ ) and since (µ , e , (αi,j)) ∈ cctn,m((µ, e, λ, φ))

0 0 0 0 0 we also have cctn,m((µ , e , (αi,j))) = (µ, e, λ, φ). Hence, µ = µ , e = e , λ = λ and

0 φ = φ . It remains to show that (αi,j) = α((µ, e, λ, φ),S) for some [S] ∈ P2(I)/ ∼F . 90

We will construct S ∈ P2(I) such that

−1 αi,j = |FS ({i, j})| for all 1 ≤ i ≤ j ≤ n − 2m + 2. By Remark 4.9.5 we can write

[ ¡ 1 2 ¢ I = Ii,j ∪ Ii,j

αi,j 6=0

1 2 where all the sets in the above union are disjoint and |Ii,j| = |Ii,j| = αi,j. Assume that

1 2 Ii,j = {a1, . . . , aαi,j } and Ii,j = {b1, . . . , bαi,j }.

Let   {{a1, b1},..., {aα , bα }} if αi,j 6= 0 S := i,j i,j i,j   ∅ otherwise

We have that Si,j is well deﬁned, |Si,j| = αi,j and for all s1, s2 ∈ Si,j with s1 6= s2 we have s1 ∩ s2 = ∅. Furthermore, Si,j ∩ Sk,l = ∅ for all (i, j) 6= (k, l). Hence, X X |Si,j| = 2 αi,j = 2m. i,j i,j

Thus, S := ∪i,jSi,j ∈ P2(I). It remains to show that

−1 αi,j = |FS ({i, j})|.

−1 1 2 Note that {a, b} ∈ FS ({i, j}) if and only if a ∈ Ii,j and b ∈ Ii,j. Thus, −1 FS ({i, j}) = Si,j. Hence,

−1 |FS ({i, j})| = |Si,j| = αi,j.

−1 Thus, cctn,m((µ, e, λ, φ)) ∈ P reim. 91

Conversely, let (µ, e, λ, α((µ, e, λ, φ),S)) ∈ P reim. We have to show that

cctn,m((µ, e, λ, α((µ, e, λ, φ),S))) = (µ, e, λ, φ).

Suppose that

0 0 cctn,m((µ, e, λ, α((µ, e, λ, φ),S))) = (µ, e, λ , φ ).

We have to show that λ = λ0 and φ = φ0.

From the deﬁnition of α((µ, e, λ, φ),S) = (αi,j) we have [ [ [ {1, 2,..., 2m} = s = s.

s∈S αi,j 6=0 −1 s∈FS ({i,j})

0 Suppose that αa,b = q 6= 0, with a, b ≤ e. Then, in order to obtain λ we increase by 1 at q parts ofµ ¯ of size a − 1 and at q parts ofµ ¯ of size b − 1 not chose before. On

−1 the other hand we have αa,b = |FS ({a, b})|= 6 0 and from the deﬁnition of FS and Deﬁnition 4.10.5 we have that λ is obtained fromµ ¯ by increasing by 1 at q parts of size a − 1 and by increasing by 1 at q parts of size b − 1. Since we do not increase at the same part twice we can conclude that we increase by 1 at the same parts of µ¯ in order to obtain λ and λ0. Same conclusion holds if a ≤ e < b or e < a, b. Thus, λ = λ0.

In order to prove φ = φ0 recall from Deﬁnition 4.9.1 that we have   2 ife 6= 1, he+1 = 0, 0 < he ≤ me−1 φ0 =  1 otherwise

If e = 1 then φ = 1 = φ0. Suppose that e 6= 1. We will show that φ = 2 if and only if φ0 = 2.

Suppose that φ = 2. Then, by Deﬁnition 4.10.5 we have that fi 6= e + 1 for all i ∈ {i1, . . . , i2m}. Hence, αi,e+1 = 0 = αe+1,j for all 1 ≤ i < e + 1 < j ≤ n − 2m + 2.

Hence, he+1 = 0. 92

Since φ = 2 then by Deﬁnition 4.8.1 we have 0 < |{1 ≤ t ≤ 2m : λit = e}| ≤ me−1. Let ge := |{1 ≤ t ≤ 2m : λit = e}|. Then there exist is+1, . . . , is+ge ∈

{i1, . . . , i2m} such that fis+1 = ... = fis+ge = e. We obtain he = ge, hence

0 0 < he ≤ me−1. Thus, φ = 2.

0 If φ = 2 then 0 < he ≤ me−1 and the set {1 ≤ t ≤ 2m : λit = e} 6= ∅. If φ = 1

then by Deﬁnition 4.10.5 there exists is+1 ∈ {i1, . . . , i2m} such that fis+1 = e + 1.

0 0 This implies he+1 = 1 which is impossible as φ = 2. Hence, φ = 2 and φ = φ .

Similarly we can prove:

Theorem 4.10.15. Let (µ, 1, λ, 1) ∈ CCT n,m and assume |E| 6= 0. Let I =

0 0 0 {1, 2,..., 2m} and let F = {fi1 , . . . , fi2m } and F = {fi1 , . . . , fi2m } be as in Deﬁnition 4.10.5. Then

−1 cctn,m((µ, 1, λ, 1)) = {(µ, 1, α((µ, 1, λ, 1),S)) : [S] ∈ P2(I)/ ∼F }

0 0 0 ∪ {(µ, 1, α ((µ, 1, λ, 1),S )) : [S ] ∈ P2(I)/ ∼F 0 }

0 0 0 where F : I → F, F(j) = fij and F : I → F , F(j) = fij and α((µ, 1, λ, 1),S), α0((µ, 1, λ, 1),S0) are as in Deﬁnition 4.10.9. Corollary 4.10.16. In the hypothesis of Theorem 4.10.15 we have

{(µ, 1, α((µ, 1, λ, 1),S)) : [S] ∈ P2(I)/ ∼F }

0 0 0 ∩ {(µ, 1, α ((µ, 1, λ, 1),S )) : [S] ∈ P2(I)/ ∼F 0 } = ∅.

Proof. Suppose that {1 ≤ t ≤ 2m : λit = e} = {s + 1, . . . , s + r} for some

0 0 ≤ s ≤ 2m − r. For all 1 ≤ t ≤ s we have fit =µ ¯it + 2 = fit and for

0 all s + 2 ≤ t ≤ 2m we have fit =µ ¯it + 1 = fit . For t = s + 1 we have

0 fis+1 = 2 and fis+1 = 1. Let S ∈ P2(I) and let α((µ, 1, λ, 1),S) = (αi,j) and 0 0 α ((µ, 1, λ, 1),S) = (αi,j). Since e = 1 and (αi,j) satisfy condition 3) of Lemma 93

4.10.12 we have X α1,2 + αj,2 ≤ 1 j≥2 and similarly we have X 0 0 α1,2 + αj,2 ≤ 1. j≥2

Then, since fis+1 = 2 either α1,2 = 1 or there exists j ∈ {3, 4, . . . , n − 2m + 2}

0 such that α2,j = 1. Since for all 1 ≤ t ≤ 2m we have fit 6= 2 we obtain that

0 0 0 α1,2 = α2,j = 0 for all 3 ≤ j ≤ n−2m+2. Hence, α((µ, 1, λ, 1),S) 6= α ((µ, 1, λ, 1),S) for all S ∈ P2(I). Thus,

{(µ, 1, α((µ, 1, λ, 1),S)) : [S] ∈ P2(I)/ ∼F }

0 0 0 ∩ {(µ, 1, α ((µ, 1, λ, 1),S )) : [S] ∈ P2(I)/ ∼F 0 } = ∅.

Theorem 4.10.17. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. The maps cctn,m : Sn,m → CCT n,m and tm,n,p : Pm,n,p → Tm,n,p are surjective.

Proof. The fact that cctn,m : Sn,m → CCT n,m is surjective follows immediately from Theorem 4.10.14, Theorem 4.10.15 and Remark 4.10.2. In Theorem 4.9.9 we have proved that the following diagram is commutative

Θm,n,p / Pm,n,p ∼ Sn,m

tm,n,p cctn,m C / Tm,n,p ∼ CCT n,m

Since Θm,n,p and C are bijections we have that tm,n,p is onto as cctn,m is.

4.11 Classiﬁcation of Character Tables

n−1 Theorem 4.11.1. For any positive integers n and m such that m ≤ b 2 c there exists a bijection Υm,n,p between the set CT m,n,p and the set Tm,n,p such that the 94 following diagram is commutative

Pm,n,p t II ctm,n,p tt IItm,n,p tt II tt II ty t I$ Υm,n,p CT m,n,p / Tm,n,p where ctm,n,p : Pm,n,p → CT m,n,p is the map that associates to every group its character table and tm,n,p is the map from Deﬁnition 4.2.8.

Proof. Let [CT ] be an isomorphism class in CT m,n,p. Then there exists a p-group P such that the character table of P is in the isomorphism class [CT ]. We deﬁne

Υn,m([CT ]) := (A, B, ϕ) where (A, B, ϕ) is an (m, n, p)-admissible triple equivalent to (P/P 0,Z(P ), π¯) andπ ¯ is the restriction to Z(P ) of the projection map P → P/P 0. Note that

Υm,n,p([CT ]) = tm,n,p(P ), so Υm,n,p([CT ]) ∈ Tm,n,p.

We have to show that Υm,n,p is well deﬁned. Suppose that there exist two p-groups P1 and P2 such that their character tables are both in [CT ]. Then P1 and P2 have isomorphic character tables and by Corollary 4.1.11 there exist

0 0 group isomorphisms α : P1/P1 → P2/P2 and β : Z(P1) → Z(P2) such that

0 α ◦ π¯1 =π ¯2 ◦ β. Hence, the two (m, n, p)-admissible triples (P1/P1,Z(P1), π¯1) and

0 (P2/P2,,Z(P2), π¯2) are equivalent. Thus, Υm,n,p is well deﬁned.

We will show that Υm,n,p is one–to–one. Suppose there exist two isomorphism classes [CT ]1 and [CT ]2 in CT m,n,p such that Υm,n,p([CT ]1) = Υm,n,p([CT ]2).

Then, there exist two p-groups P1 and P2 such that the character table of P1 is in the isomorphism class [CT ]1 and the character table of P2 is in the isomorphism

0 class [CT ]2. Then Υm,n,p([CT ]1) ∼ (P1/P1,,Z(P1), π¯1) and Υm,n,p([CT ]2) ∼

0 0 0 (P2/P2,Z(P2), π¯2). We obtain that (P1/P1,Z(P1), π¯1) and (P2/P2,Z(P2), π¯2) are equivalent (m, n, p)-admissible triples. Hence, by Corollary 4.1.11 we have that 95

P1 and P2 have isomorphic irreducible character tables. Thus, [CT ]1 = [CT ]2 and

Υm,n,p is one–to–one.

We will show Υm,n,p is onto. Let (A, B, ϕ) ∈ Tm,n,p. Since tm,n,p is onto there

0 exists P ∈ Pn,m such that (P/P ,Z(P ), π¯) ∼ (A, B, ϕ). Let [CT ] ∈ CT m,n,p be such that the character table of P is in the isomorphism class [CT ]. Then

0 Υm,n,p([CT ]) = (P/P ,Z(P ), π¯) = (A, B, ϕ). This proves Υm,n,p is onto.

Thus, Υm,n,p is a bijection. Furthermore, Υm,n,p ◦ ctm,n,p = tm,n,p.

Corollary 4.11.2. Let m, n be positive integers such that n − 2m − 1 ≥ 0. Then there exists a bijection between the set CT m,n,p and the set CCT n,m. Furthermore, the following diagram is commutative

Θm,n,p Pm,n,p / Sn,m t II ctm,n,p tt IItm,n,p tt II cct tt II n,m ty t I$ Υm,n,p Cp CT m,n,p / Tm,n,p / CCT n,m

Proof. We have

Υm,n,p / Cp / CT m,n,p ∼ Tm,n,p ∼ CCT n,m.

e Hence, Cm,n,p := Cp ◦ Υm,n,p is a bijection. Furthermore, since by Theorem 4.9.9 we have cctn,m ◦ Θm,n,p = Cp ◦ tm,n,p and by Theorem 4.11.1 we have Υm,n,p ◦ ctm,n,p = e tm,n,p we immediately obtain that cctn,m ◦ Θm,n,p = Cm,n,p ◦ ctm,n,p.

Theorem 4.11.3. Let n ≥ 3 and let p be a prime. Let NCT (n) denote the number of non-isomorphic character tables of p-groups P with |P | = pn and |P 0| = p. Then

NCT (n) does not depend on p and we have

b n−1 c X2 X X X NCT (n) = f(µ, e, λ) m=1 µ∈P(n−2m) e∈D(µ) λ∈P(µ,e) where D(µ) denotes the set of distinct parts of µ and P(µ, e) is the set of distinct partitions λ of n − 1 that are obtained from µ as follows: 96

1. Suppose µ = (µ1 ≥ µ2 ≥ ... ≥ µk) and µi0 = e > µi0+1. Let µ¯ ∈   µi if i 6= i0, i ≤ k 

P(n − 2m − 1) with parts µ¯i = µi − 1 if i = i0    0 if i > k

2. Choose 1 ≤ i1 < i2 < . . . < i2m. We deﬁne λ ∈ P(n − 1) whose parts are   µ¯i + 1 if i ∈ {i1, i2, . . . , i2m} λ := i   µ¯i otherwise    2 if e 6= 1, and 

and f(µ, e, λ) = 0 < |{1 ≤ t ≤ 2m, λi = e}| ≤ me−1  t   1 otherwise where me−1 denotes the number of parts of µ of size e − 1.

Proof. By Corollary 4.11.2 we have:

b n−1 c b n−1 c X2 X2 NCT (n) = |CT m,n,p| = |CCT n,m|. m=1 m=1 Recall,

CCT n,m = {(µ, e, λ, φ): µ ∈ P(n − 2m), e ∈ D(µ), λ ∈ P(µ, e), φ ∈ Φ} where P(µ, e) is the set of distinct partitions of n − 1 that are obtained from µ by ﬁrst decreasing by 1 at a part of size e and then increasing by 1 at 2m parts including parts of size 0 and   {1, 2} if e 6= 1, 0 < |{1 ≤ t ≤ 2m : λi = e}| ≤ me−1 Φ = t   {1} otherwise

(see Deﬁnition 4.8.1). Hence,

X X X |CCT n,m| = f(µ, e, λ) µ∈P(n−2m) e∈D(µ) λ∈P(µ,e) 97

  2 if e 6= 1, 0 < |{1 ≤ t ≤ 2m : λi = e}| ≤ me−1 where f(µ, e, λ) = t  1 otherwise Thus, b n−1 c X2 X X X NCT (n) = f(µ, e, λ). m=1 µ∈P(n−2m) e∈D(µ) λ∈P(µ,e)

Proposition 4.11.4. Let n be a positive integer and let Pn = {(µ, e): µ ∈ P(n) and e is a positive part of µ}. Then there exits a bijection

ψ : Pn → Pn−1 ∪ P(n − 1).

Proof. We deﬁne

ψ : Pn → Pn−1 ∪ P(n − 1) by    (¯µ, e − 1) if e 6= 1 ψ((µ, e)) :=  µ¯ if e = 1 whereµ ¯ is the partition of n − 1 obtained from µ by decreasing by 1 at its last part ¯ ¯ of size e. We deﬁne ψ : Pn−1 ∪ P(n − 1) → Pn by ψ((¯µ, e¯)) = (µ, e¯ + 1) and ψ¯(¯µ) = ((¯µ, 1), 1), where µ is the partition of n obtained fromµ ¯ by increasing by 1 at its ﬁrst part of sizee ¯. It is easy to see that ψ and ψ¯ are inverse functions. Thus, ψ is a bijection.

Corollary 4.11.5. Let n ≥ 3 and let p be a prime. Let NCT (n) denote the number of non-isomorphic character tables of p-groups P with |P | = pn and |P 0| = p. Then,

b n−1 c X2 X X X ¯ NCT (n) = f(¯µ, e,¯ λ)

m=1 µ¯∈P(n−2m−1) e¯∈D(¯µ)∪{0} λ∈Pµ¯ where D(¯µ) denotes the set of distinct parts of µ¯ and Pµ¯ is the set of partitions of

n − 1 obtained from µ¯ by increasing by 1 at 2m parts, µ¯i1 ,..., µ¯i2m which can be 98 zero, and   1e ¯ = 0 or |{1 ≤ t ≤ 2m :µ ¯i =e ¯}| ∈ {0, me¯} f¯(¯µ, e,¯ λ) = t  2 otherwise where me¯ denotes the number of parts of µ¯ of size e¯.

Proof. By Theorem 4.11.3 we have

b n−1 c X2 X X X NCT (n) = f(µ, e, λ) m=1 µ∈P(n−2m) e∈D(µ) λ∈P(µ,e) where λ is obtained by ﬁrst decreasing by 1 at a part of size e and then increasing by 1 at 2m-parts which can be zero. By Proposition 4.11.4 we have {(µ, e): µ ∈ P(n − 2m), e ∈ D(µ)}'{(¯µ, e¯) :µ ¯ ∈ P(n − 2m − 1), e¯ ∈ D(¯µ)} ∪ P(n − 2m − 1). Moreover, given µ ∈ P(n − 2m) and e ∈ D(µ) the set of partitions λ of n − 1 obtained from µ by ﬁrst decreasing by 1 at a part of size e and then increasing by 1 at 2m-parts is the same as the set of partitions on n − 1 obtained fromµ ¯ by increasing by 1 at 2m-parts.

We have f(µ, e, λ) = 2 if and only if e 6= 1 and 0 < |{1 ≤ t ≤ 2m :µ ¯it = e − 1}| < me−1(¯µ). But e 6= 1 if and only ife ¯ 6= 0 and {1 ≤ t ≤ 2m :µ ¯it = e − 1} = ¯ {1 ≤ t ≤ 2m :µ ¯it =e ¯}. Hence, f(µ, e, λ) = 2 if and only if f(¯µ, e,¯ λ) = 2. Thus, f(µ, e, λ) = f¯(¯µ, e,¯ λ) and

b n−1 c X2 X X X ¯ NCT (n) = f(¯µ, e,¯ λ). m=1 µ¯∈P(n−2m−1) e¯∈D(¯µ)∪{0} λ∈P(¯µ,e¯) CHAPTER 5 THE LIMIT

Let I := {1, 2,..., 2m} and let M = {m1, m2, . . . , mt} be a set with at most 2m elements. Let f : I → M be a surjective map. For all 1 ≤ i ≤ t let

−1 ni = |f (mi)| and let G := Sn1 × Sn2 × ... × Snt , where Sni denotes the symmetric group in ni letters. Then G acts on P2(I) and |P2(I)/ ∼f | equals the number of orbits under the action of G. By Cauchy-Frobenius Lemma we have

1 X |P (I)/ ∼ | = |Fix (g)| 2 f |G| P2(I) g∈G

where FixP2(I)(g) = {S ∈ P2(I): g.S = S for all g ∈ G}.

5.1 Upper Bounds for the Size of the Preimage of the Map cctn,m

Theorem 5.1.1. Let I := {1, 2,..., 2m} and G := Sn1 × Sn2 × ... × Snt with n1 + ... + nt = 2m. Let g ∈ G and view G as a subgroup of S2m. For all

1 ≤ r ≤ 2m let kr denote the number of disjoint cycles of length r that appear in a decomposition of g into disjoint cycles. Let  kr  b 2 c µ ¶µ ¶ µ ¶  X rl k k − 2 k − 2l + 2  1 + r r ··· r if r is even  l! 2 2 2  l=1 kr α := r 2 k ! r  r  kr if r is odd and kr is even  kr  2 2 ( )!  2  0 if r and kr are odd

Then |FixP2(I)(g)| = α1α2 ··· α2m

Proof. Let g ∈ G and suppose kr1 , kr2 , . . . , krt 6= 0. We have:

g = g × · · · × g × · · · × g × · · · × g r1,1 r1,kr1 rt,1 rt,krt

99 100

where gri,j is a cycle of length ri. We assume

gri,j = (ai,j,1ai,j,2 . . . ai,j,ri ).

kri Let Iri := ∪j=1{ai,j,1, ai,j,2, . . . , ai,j,ri }. We will show

Yt Fix (g) ' Fix (g × · · · × g ). P2(I) P2(Iri ) ri,1 ri,kri j=1

Let S ∈ FixP2(I)(g) we will construct Sr1 ,Sr2 ,...,Srt such that Sri ∈ Fix (g × · · · × g ) and S = ∪t S . P2(Iri ) ri,1 ri,kri i=1 ri t For all 1 ≤ i ≤ t let Sri := S ∩ Iri . Then, since I = ∪i=1Iri we have

t ∪i=1 Sri ⊆ S.

Let {ai,j,k, au,v,w} ∈ S with i 6= u and assume ri < ru. Since g.S = S we have

ri ri ri g .S = S and g .{ai,j,k, au,v,w} ∈ S. We have g .{ai,j,k, au,v,w} = {ai,j,k, b} with b 6= au,v,w since ri < ru. Thus, if {a, b} ∈ S then there exists a unique 1 ≤ i ≤ t such

t t that a, b ∈ Iri . Hence, S ⊆ ∪i=1(S ∩ Iri ). Thus, S = ∪i=1Sri . It remains to show that S ∈ Fix (g × · · · × g ) for all 1 ≤ i ≤ t. ri P2(Iri ) ri,1 ri,kri

Let {a, b} ∈ Sri , then {a, b} ∈ S and g.{a, b} ∈ S. Since a, b ∈ Iri we have g.{a, b} = (g × · · · × g ).{a, b} hence, g.a, g.b ∈ I . Thus, {g.a, g.b} ∈ S and ri,1 ri,kri ri ri S ∈ Fix (g × · · · × g ). ri P2(Iri ) ri,1 ri,kri Conversely, given S ∈ Fix (g × · · · × g ) for all 1 ≤ i ≤ t we have ri P2(Iri ) ri,1 ri,kri t that S := ∪i=1Sri ∈ FixP2(I)(g) . Thus,

Yt |Fix (g)| = |Fix (g × · · · × g )|. P2(I) P2(Iri ) ri,1 ri,kri i=1 We will show that |Fix (g × · · · × g )| = α for all 1 ≤ i ≤ t. P2(Iri ) ri,1 ri,kri ri Let i ∈ {1, 2, . . . , t} and letg ¯ := g × · · · × g . Let S ∈ Fix (¯g ) and i ri,1 ri,kri ri P2(Iri ) i suppose ri is odd. Suppose that {ai,j,k, ai,j,l} ∈ Sri with k < l. Then 101

g¯i.{ai,j,k, ai,j,l} = gri,j.{ai,j,k, ai,j,l} = {g .a , g .a } ri,j i,j,k ri,j i,j,l  {ai,j,k+1, ai,j,l+1} if l < ri =   {ai,j,k+1, ai,j,1} if l = ri

Hence, if {ai,j,k, ai,j,l} ∈ Sri then Sri must contain elements of the form {a, b} with both a and b in {ai,j,1, ai,j,2, . . . , ai,j,ri }. Since ri is odd there will be an element c ∈ {ai,j,1, ai,j,2, . . . , ai,j,ri } such that {c, d} ∈ Sri and d 6∈ {ai,j,1, ai,j,2, . . . , ai,j,ri }.

In this caseg ¯i.{c, d} 6∈ Sri which is a contradiction. Hence, if {a, b} ∈ Sri then there exist 1 ≤ j, k ≤ ri, j 6= k such that a ∈ {ai,j,1, ai,j,2, . . . ai,j,ri } and b ∈ {ai,k,1, ai,k,2, . . . , ai,k,ri }. Hence, the letters of an odd cycle must be paired with letters of another odd cycle of the same length. Therefore, if we have an odd number of cycles of the same odd length no elements of P2(I) are ﬁxed and |Fix (¯g )| = 0 = α . If k is even then all k -cycles must be paired and for P2(Iri ) i ri ri ri each pairing we have ri-possibilities to pair the letters. Hence,

kri µ ¶µ ¶ µ ¶ r 2 k k − 2 2 |Fix (¯g )| = ³ i ´ ri ri ··· = α . P2(Iri ) i ri kri 2 2 2 2 !

Suppose ri is even. In this case we can have {a, b} ∈ Sri with either a, b ∈

{ai,j,1, ai,j,2, . . . , ai,j,ri } for some 1 ≤ j ≤ kri or a ∈ {ai,j,1, ai,j,2, . . . , ai,j,ri } and b ∈

{ai,k,1, ai,k,2, . . . , ai,k,ri } with 1 ≤ j 6= k ≤ kri . If both a, b ∈ {ai,j,1, ai,j,2, . . . , ai,j,ri } for some 1 ≤ j ≤ kri we have only one possibility to pair the letters and if a and b belong to diﬀerent sets of letters we have ri-possibilities to pair the letters from the two sets. Hence, in the case of an even cycle we have

k b ri c µ ¶µ ¶ µ ¶ X2 l ri kri kri − 2 kri − 2l + 2 |FixP (I )(¯gi)| = 1 + ··· = αr . 2 ri l! 2 2 2 i l=1 102

Corollary 5.1.2. Let m be a positive integer and let I = {1, 2,..., 2m}. Then

(2m)! |P (I)| = . 2 2mm!

Proof. Let G = {id}. Then G acts trivially on P2(I) hence |P2(I)| = |FixP2(I)(id)|. 1m (2m)! (2m)! By Theorem 5.1.1, we have |Fix (id)| = α = · = . Thus, P2(I) 1 m! 2m 2mm! (2m)! |P (I)| = . 2 2mm!

Corollary 5.1.3. Let m ≥ 2 be an integer and let I = {1, 2,..., 2m}. Let

G = Sn1 × · · · × Snt with n1 + ··· + nt = 2m acting on P2(I). Assume that at least one of the n1, . . . , nt is at least 2. Then there exists g ∈ G, g 6= id such that

|FixP2(I)(g)| < |P2(I)|.

Proof. Without loss of generality we can assume that n1 ≥ 2. Then g = (12) × id ∈ G and g 6= id. By Theorem 5.1.1 we have

|FixP2(I)(g)| = α1 · α2

(2m − 2)! (2m)! where α = and α = 1. By Corollary 5.1.2 we have |P (I)| = . 1 2m−1(m − 1)! 2 2 2mm! (2m − 2)! (2m)! It is easy to see that < , hence, 2m−1(m − 1)! 2mm!

|FixP2(I)(g)| < |P2(I)|.

Lemma 5.1.4. Let I = {1, 2,..., 2m} and let M = {m1, m2, . . . , mt} with t ≤ 2m − 1 and m ≥ 2. Let f : I → M be a surjective map. Then,

(2m)! |P (I)/ ∼ | < . 2 f 2mm! 103

−1 Proof. For all 1 ≤ i ≤ t let ni := |f (mi)| and let G = Sn1 × · · · × Snt . Since t ≤ 2m − 1 we have that G acts non-trivially on P2(I). Hence, the number of orbits is less than the number of elements of P2(I). Thus,

(2m)! |P (I)/ ∼ | < |P (I)| = . 2 f 2 2mm!

Theorem 5.1.5. Let m, n be positive integers such that n − 2m − 1 ≥ 0.

Let (µ, e, λ, φ) ∈ CCT n,m. Recall then from Deﬁnition 4.8.1 that we also have

µ = (µ1 ≥ ... ≥ µi0 = e > µi0+1 ≥ ... ≥ µk) and there exist 1 ≤ i1 < . . . < i2m such that λi =µ ¯i + 1 if i ∈ {i1, . . . , i2m} and λi =µ ¯i otherwise, where µ¯ = (µ1 ≥ . . . >

µi0 − 1 = e − 1 ≥ µi0+1 ≥ ... ≥ µk, 0,..., 0).

a) Suppose e 6= 1 or {1 ≤ t ≤ 2m : λit = e} = ∅. Then

(2m)! |cct−1 ((µ, e, λ, φ))| ≤ . n,m 2mm!

The upper bound is attained if and only if one of the following conditions is satisﬁed: i) m = 1 or

ii)µ ¯i1 ,..., µ¯i2m are all distinct or

iii) there exists 0 ≤ s ≤ 2m − 2 such that µ¯is+1 =µ ¯is+2 = e − 1 and

µ¯i1 ,..., µ¯is+1 , µ¯is+3 ,..., µ¯i2m are all distinct.

b) Suppose e = 1 and {1 ≤ t ≤ 2m : λit = e} 6= ∅. Then

(2m)! |cct−1 ((µ, e, λ, φ))| ≤ 2 . n,m 2mm!

The upper bound is attained if and only if one of the following conditions is satisﬁed i) m = 1 or

ii)µ ¯i1 ,..., µ¯i2m are all distinct. 104

Proof. a) Suppose e 6= 1 or {1 ≤ t ≤ 2m : λit = e} = ∅. Then, by Theorem 4.10.14, we have:

−1 cctn,m((µ, e, λ, φ)) = {(µ, e, α((µ, e, λ, φ),S)) : [S] ∈ P2(I)/ ∼F }

where I = {1, 2,..., 2m}, F : I → {fi1 , . . . , fi2m }, F(j) = fij .

From Lemma 4.10.10 we have that α((µ, e, λ, φ),S1) = α((µ, e, λ, φ),S2) if and only if S1 ∼F S2. Hence,

(2m)! |cct−1 ((µ, e, λ, φ))| = |P (I)/ ∼ | ≤ |P (I)| = . nm 2 F 2 2mm!

If m = 1 then by the above formula we always have equality.

Assume that m ≥ 2. Then by Lemma 5.1.4 the upper bound is attained if and

only if all fi1 , . . . , fi2m .

If φ = 2 or {1 ≤ t ≤ 2m : λit = e} = ∅ then   µ¯ij + 1 ifµ ¯ij ≤ e − 1 fij =  µ¯ij + 2 ifµ ¯ij ≥ e

Hence the fi1 , . . . , fi2m are all distinct if and only if allµ ¯i1 ,..., µ¯i2m are distinct.

If φ = 1 and {1 ≤ t ≤ 2m : λit = e} = {s + 1, . . . , s + r} then    µ¯i + 1 ifµ ¯i ≤ e − 1, ij 6= is+1  j j fi = e + 1 if i = i j  j s+1   µ¯ij + 2 ifµ ¯ij ≥ e

and fi1 , . . . , fi2m are distinct if and only if |{1 ≤ t ≤ 2m : λit = e}| ≤ 2

and for all 1 ≤ j, k ≤ 2m such thatµ ¯ij 6= e − 1 6=µ ¯ik we haveµ ¯ij 6=µ ¯ik .

If |{1 ≤ t ≤ 2m : λit = e}| = 1 then allµ ¯i1 ,..., µ¯i2m are distinct and if

|{1 ≤ t ≤ 2m : λit = e}| = 2 then there exists 0 ≤ s ≤ 2m − 2 such that

µ¯is+1 =µ ¯is+2 = e − 1 andµ ¯i1 ,..., µ¯is+1 , µ¯is+3 ,..., µ¯i2m are distinct. 105

b) Suppose that e = 1 and {1 ≤ t ≤ 2m : λit = e} = {s + 1, . . . , s + r} for some 0 ≤ s ≤ 2m − r. Since e = 1 we have φ = 1. Then by Theorem 4.10.15 we have

−1 cctn,m((µ, e, λ, φ)) = {(µ, e, α((µ, e, λ, φ),S):[S] ∈ P2(I)/ ∼F }

0 0 0 ∪ {(µ, e, α ((µ, e, λ, φ),S ):[S ] ∈ P2(I)/ ∼F 0 }.

By Corollary 4.10.16 we have that the above union is disjoint hence

−1 (2m)! |cct ((µ, e, λ, φ))| = |P (I)/ ∼ | + |P (I)/ ∼ 0 | ≤ 2 . n,m 2 F 2 F 2mm!

The upper bound is attained if and only if either m = 1 or fi1 , . . . , fi2m are all

0 0 distinct and fi1 , . . . , fi2m are all distinct. Since    1 ifµ ¯i = 0, i 6= is+1  j fi = 2 if i = i j  s+1   µ¯ij + 2 ifµ ¯ij ≥ 1 and    1 ifµ ¯ = 0 0 ij fi = j  µ¯ij + 2 ifµ ¯ij ≥ 1

we have that the upper bound is attained if and onlyµ ¯i1 ,..., µ¯i2m are all distinct.

Corollary 5.1.6. Let m, n be positive integers such that n − 2m − 1 > 0. Let ρ¯ ∈ P(n − 2m − 1) and let e¯ be a part of ρ¯ possibly 0. Let ρ be the partition of n − 2m obtained from ρ¯ by increasing by 1 at a part of size e¯. Let λ be a partition of n − 1 obtained from ρ¯ by increasing by 1 at 2m-distinct parts. Then

(2m)! |cct−1 ((ρ, e¯ + 1, λ, φ))| ≥ . n,m 2mm!

Proof. Letρ ¯i1 ,..., ρ¯i2m be the 2m-distinct parts where we increase by 1. Let

E = {1 ≤ t ≤ 2m : λit = e} and let e =e ¯ + 1. By Theorem 5.1.5 we have that 106

5.2 The Number of λ’s

Deﬁnition 5.2.1. Let m, n be positive integers such that n − 2m − 1 > 0. Let

ρ¯ ∈ P(n − 2m − 1) and lete ¯ be a part ofρ ¯ possibly 0. We denote by Pρ¯ the set of distinct partitions of n − 1 that are obtained fromρ ¯ by increasing by 1 at 2m parts. We deﬁne

Γ: Pρ¯ → P(2m)

Γ(λ) = η where η is the following partition of 2m: suppose λ is obtained fromρ ¯ by increasing

by 1 atρ ¯i1 , ρ¯i2 ,..., ρ¯i2m and

ρ¯ = ... =ρ ¯ > ρ¯ = ... =ρ ¯ > . . . > ρ¯ = ... =ρ ¯ i1 iη1 iη1+1 iη1+η2 iη1+...+ηr−1+1 iη1+...+ηr

Then η is the partition of 2m whose parts are η1, . . . , ηr deﬁned above. Lemma 5.2.2. Let m, n be positive integers such that n − 2m − 1 > 0. Let

ρ¯ ∈ P(n − 2m − 1) and let e¯ be a part of ρ¯ possibly 0. Let Pρ¯ and Γ: Pρ¯ → P(2m) be as in Deﬁnition 5.2.1. Then µ ¶µ ¶ µ ¶ d d − m(η ) d − (m(η ) + m(η ) + ··· + m(η )) |Γ−1(η)| = ρ,η¯ 1 ρ,η¯ 2 1 ··· ρ,η¯ r 1 2 r−1 m(η1) m(η2) m(ηr) where η has distinct parts η1 > . . . > ηk with multiplicities m(η1), . . . , m(ηk), and

dρ,η¯ i denotes the number of distinct parts of ρ¯ with multiplicity at least ηi. Here we always count 0 as a part with large multiplicity. 107

m(η1) m(ηk) Proof. Let η = (η1 , . . . , ηk ) ∈ P(2m). For all 1 ≤ i ≤ k we will increase by 1 at m(ηi)-parts ofρ ¯ with multiplicity at least ηi. Since η1 > . . . > ηk we have ¡ ¢ dρ,η¯ 1 dρ,η¯ ≥ ... ≥ dρ,η¯ and we have -possibilities to increase by 1 at m(η1)-parts 1 k m(η1) ¡ ¢ dρ,η¯ 2 −m(η1) ofρ ¯ with multiplicity at least η1. Then we have -possibilities to increase m(η2) at m(η2)-parts ofρ ¯ with multiplicity at least η2. Hence, the number of possibilities is µ ¶µ ¶ µ ¶ d d − m(η ) d − (m(η ) + m(η ) + ··· + m(η )) ρ,η¯ 1 ρ,η¯ 2 1 ··· ρ,η¯ r 1 2 r−1 . m(η1) m(η2) m(ηr)

Theorem 5.2.3. Let m,n be integers with n−2m−1 ≥ 0 and let ρ¯ ∈ P(n−2m−1).

Let e¯ be a part of ρ¯ possibly 0. For all 1 ≤ i ≤ 2m let dρ,i¯ denote the number of distinct parts of ρ¯ with multiplicity at least i. Let Pρ¯ denote the set of distinct partitions of n − 1 that are obtained from ρ¯ by increasing by 1 at 2m-parts. Then

X µ ¶ µ ¶ dρ,η¯ 1 dρ,η¯ r − (m(η1) + m(η2) + ··· + m(ηr−1)) |Pρ¯| = ··· m(η1) m(ηr) η∈P(2m) where for η ∈ P(2m) we denote by η1 > η2 > . . . > ηk the distinct parts of η and by m(ηi) we denote the multiplicity of ηi.

Proof. Let Γ : Pρ¯ → P(2m) be the map deﬁned in Deﬁnition 5.2.1. We have X −1 |Pρ¯| = |Γ (η)| η∈P(2m) and using Lemma 5.2.2 we obtain

X µ ¶ µ ¶ dρ,η¯ 1 dρ,η¯ r − (m(η1) + m(η2) + ··· + m(ηr−1)) |Pρ¯| = ··· . m(η1) m(ηr) η∈P(2m) 108

m(η1) m(ηr) Proposition 5.2.4. Let m be a positive integer and let η = (η1 , . . . , ηr ) be a partition of 2m. Let µ ¶µ ¶ µ ¶ X X − m(η1) X − (m(η1) + m(η2) + ··· + m(ηr−1)) Rη(X) = ··· . m(η1) m(η2) m(ηr)

Then we have:

2m a) deg Rη ≤ 2m with equality if and only if η = (1 ).

2m 2m−2 b) If η 6= (1 ), (2, 1 ) then deg Rη < 2m − 1.

Proof. We have degRη = m(η1) + ··· + m(ηr) and 2m = η1m(η1) + ··· + ηrm(ηr). ¡ ¢ 2m X a) If η = (1 ) then η1 = 1, m(η1) = 2m and Rη = 2m , hence degRη = 2m.

Conversely, suppose η1 ≥ 2. Then we have:

2m = η1m(η1) + ··· ηrm(ηr)

≥ 2m(η1) + m(η2) + ··· + m(ηr)

= m(η1) + 2m.

Hence, m(η1) = 0 which implies that all parts must be equal to 1. Thus, η = (12m).

2m 2m−2 b) Suppose there exists η ∈ P(2m) \{(1 ), (2, 1 )} such that deg Rη ≥

2m − 1. If η1 ≥ 3 then we have

2m = m(η1)η1 + ··· + m(ηr)ηr

≥ 3m(η1) + m(η2) + ··· + m(ηr)

= 2m(η1) + m(η1) + m(η2) + ··· + m(ηr)

≥ 2m(η1) + 2m − 1.

We obtain 1 ≥ 2m(η1) which is impossible as m(η1) ≥ 1. Thus, η1 = 1, 2. If η1 = 1

2m 2m−2 then η = (1 ) which is excluded, hence η1 = 2 and m(η1) ≥ 2 as η 6= (2, 1 ). We have 109

2m = 2m(η1) + m(η2)

= m(η1) + m(η1) + m(η2)

≥ m(η1) + 2m − 1.

2m 2m−2 Hence, m(η1) ≤ 1 which is a contradiction. Thus, if η 6= (1 ), (2, 1 ) we have deg Rη ≤ 2m − 2.

Corollary 5.2.5. Let m, n be positive integers such that n − 2m − 1 > 0. Then, there exists a polynomial R(X) ∈ Q[X] of degree at most 2m − 1 which depends on m but not on n and such that for all ρ¯ ∈ P(n − 2m − 1) we have

d2m |P | ≤ ρ¯ + R(d ) ρ¯ (2m)! ρ¯ where Pρ¯ denotes the set of distinct partitions of n−1 obtained from ρ¯ by increasing by 1 at 2m parts and dρ¯ denotes the number of distinct parts of ρ¯, including 0 as a part.

Proof. By Theorem 5.2.3 we have

X µ ¶ µ ¶ dρ,η¯ 1 dρ,η¯ r − (m(η1) + m(η2) + ··· + m(ηr−1)) |Pρ¯| = ··· m(η1) m(ηr) η∈P(2m) µ ¶ µ ¶ X d d − (m(η ) + m(η ) + ··· + m(η )) ≤ ρ¯ ··· ρ¯ 1 2 r−1 m(η1) m(ηr) η∈P(2m) µ ¶ d = ρ¯ 2m µ ¶ µ ¶ X d d − (m(η ) + m(η ) + ··· + m(η )) + ρ¯ ··· ρ¯ 1 2 r−1 . m(η1) m(ηr) η ∈ P(2m) η 6= (12m)

Let µ ¶µ ¶ µ ¶ X X − m(η1) X − (m(η1) + m(η2) + ··· + m(ηr−1)) Rη(X) := ··· m(η1) m(η2) m(ηr) 110

Then by Proposition 5.2.4 we have deg Rη ≤ 2m with equality if and only if η = (12m). Hence, µ ¶ d X |P | ≤ ρ¯ + R (d ) ρ¯ 2m η ρ¯ η ∈ P(2m) η 6= (12m) d2m X = ρ¯ + R¯(d ) + R (d ) (2m)! ρ¯ η ρ¯ η ∈ P(2m) η 6= (12m) d2m = ρ¯ + R(d ) (2m)! ρ¯ X ¯ ¯ where R = R + Rη and since deg Rη, deg R < 2m for all η ∈ P(2m), η 6=

η ∈ P(2m) η 6= (12m) (12m) we have deg R < 2m.

Corollary 5.2.6. Let m, n be positive integers such that n − 2m − 1 > 0. Let ρ¯ ∈ P(n − 2m − 1) and let e¯ be a part of ρ¯ possibly zero. Let ρ be the partition of n − 2m obtained from ρ¯ by increasing by 1 at a part of size e¯. Let Pρ¯ denote the set of partitions of n − 1 obtained from ρ¯ by increasing by 1 at 2m parts. Let

(2m)! M := {λ ∈ P : |cct−1 ((ρ, e¯ + 1, λ, φ))| < }. ρ¯ ρ¯ n,m 2mm!

Then there exists a polynomial R(X) ∈ Q[X] of degree at most 2m − 2 depending only on m such that d2m−1 |M | ≤ ρ¯ + R(d ) ρ¯ (2m − 2)! ρ¯ where dρ¯ denotes the number of distinct parts of ρ¯ including 0 as a part.

Proof. Note that if m = 1 then Mρ¯ = ∅, hence, the conclusion holds trivially.

Assume that m ≥ 2. We have Mρ¯ ⊆ Pρ¯ and by Theorem 5.2.3 we have X µ ¶ µ ¶ dρ,η¯ 1 dρ,η¯ r − (m(η1) + m(η2) + ··· + m(ηr−1)) |Pρ¯| = ··· . m(η1) m(ηr) η∈P(2m) 111

(2m)! Since for all λ ∈ M we have |cct−1 ((ρ, e¯ + 1, λ, φ))| < by Corollary ρ¯ n,m 2mm! 5.1.6 we have that λ is obtained fromρ ¯ by increasing at 2m parts not all distinct. Thus,

X µ ¶ µ ¶ dρ,η¯ 1 dρ,η¯ r − (m(η1) + m(η2) + ··· + m(ηr−1)) |Mρ¯| ≤ ··· m(η1) m(ηr) η ∈ P(2m) η 6= (12m) µ ¶ µ ¶ X d d − (m(η ) + m(η ) + ··· + m(η )) ≤ ρ¯ ··· ρ¯ 1 2 r−1 m(η1) m(ηr) η ∈ P(2m) η 6= (12m) µ ¶ d − 1 = d ρ¯ ρ¯ 2m − 2 µ ¶ µ ¶ X d d − (m(η ) + m(η ) + ··· + m(η )) + ρ¯ ··· ρ¯ 1 2 r−1 m(η1) m(ηr) η ∈ P(2m) η 6= (12m), (2, 12m−2) µ ¶ d − 1 X = d ρ¯ + R (d ) ρ¯ 2m − 2 η ρ¯ η ∈ P(2m) η 6= (12m), (2, 12m−2) d2m−1 X = ρ¯ + R¯(d ) + R (d ) (2m − 2)! ρ¯ η ρ¯ η ∈ P(2m) η 6= (12m), (2, 12m−2) where µ ¶µ ¶ µ ¶ X X − m(η1) X − (m(η1) + m(η2) + ··· + m(ηr−1)) Rη(X) = ··· m(η1) m(η2) m(ηr) for all η ∈ P(2m), η 6= (12m), (2, 12m−2). By Proposition 5.2.4 b) we have ¯ deg Rη < 2m − 1 and since degR < 2m − 1 we have deg R < 2m − 1 where X ¯ R(X) := R(X) + Rη(X).

η ∈ P(2m) η 6= (12m), (2, 12m−2)

Thus, d2m−1 |M | ≤ ρ¯ + R(d ) ρ¯ (2m − 2)! ρ¯ with deg R < 2m − 1. 112

5.3 The Average Number of Distinct Parts of a Partition of an Integer

Proposition 5.3.1. Let a1, a2, . . . , an be non-negative real numbers. Assume that there exists M > 0 such that ai ≤ M for all 1 ≤ i ≤ n. Then for all t ≥ 1 we have

Ã !t Ã ! Ã ! 1 Xn 1 Xn 1 Xn a ≤ at ≤ M t−1 a n i n i n i i=1 i=1 i=1 Proof. If t = 1 then the inequalities hold trivially. Assume that t > 1. We have

1 Xn 1 Xn at ≤ M t−1a n i n i i=1 i=1 1 Xn = M t−1 a n i i=1 For the other inequality we will use Minkowski’s Inequality. Let q be a positive

1 1 real number such that t + q = 1. We have   Ã !t Ã ! 1 Ã µ ¶ ! 1 t 1 Xn Xn t Xn 1 q q a ≤  at  n i i n i=1 Ã i=1 ! i=1 n ³ ´ X 1 t t 1 q = ai (n nq ) Ã i=1 ! n X ¡ ¢ 1 t 1 t(1− q ) = ai n Ã i=1 ! n X ¡ ¢ 1 t 1 t t = ai n Ã i=1 ! 1 Xn = at . n i i=1

The following theorem is Theorem B in Wilf [12].

Theorem 5.3.2. The average number of diﬀerent sizes of parts that a partition of the integer n has is √ 6 d¯ ∼ n1/2. n π 113

Remark 5.3.3. The result of Theorem 5.3.2 holds true if we allow our partitions to have parts equal to 0 and we count 0 as a part.

5.4 lim NG(n, m)/NCT (n, m) n→∞ m Proposition 5.4.1. Let m be a positive integer and let R(X) = amX +

m−1 am−1X + ··· + a1X + a0 ∈ R[X] be such that am > 0. Then there exists

R1(X) ∈ R[X] with deg R1 ≤ m − 2 such that for all positive integers n we have X R(dρ) √ a ρ∈P(n) √ √ a a (d¯ )m−R (2 n)d¯ + 0 ≤ ≤ a (2 n)m−1d¯ +R (2 n)d¯ + 0 m n 1 n |P(n)| |P(n)| m n 1 n |P(n)| where dρ denotes the number of distinct parts of ρ for all ρ ∈ P(n) counting 0 as a P ¯ 1 distinct part and dn = |P(n)| ρ∈P(n) dρ.

m−2 m−3 Proof. Let R1(X) := |am−1|X + |am−2|X + ··· + |a1|. Then, R1(X) ∈ R[X] P ¯t 1 t with deg R1 ≤ m − 2. For all t ≥ 1 let d n := |P(n)| ρ∈P(n) dρ. It is known, see for √ instance [6] that dρ ≤ 2 n for all ρ ∈ P(n). Hence, we can use Proposition 5.3.1 and we have P P ρ∈P(n) R(dρ) 1 m m−1 |P(n)| = |P(n)| ρ∈P(n) amdρ + am−1dρ + ··· + a1dρ + a0

¯m m−1 ¯ a0 = amd n + am−1d n + ··· + a1dn + |P(n)|

¯m m−1 ¯ a0 ≤ amd n + |am−1|d n + ··· + |a1|dn + |P(n)| √ √ m−1 ¯ m−2 ¯ ¯ a0 ≤ am(2 n) dn + |am−1|(2 n) dn + ··· + |a1|dn + |P(n)| √ √ m−1 ¯ ¯ a0 = am(2 n) dn + R1(2 n)dn + |P(n)| .

For the other inequality we have P P ρ∈P(n) R(dρ) 1 m m−1 |P(n)| = |P(n)| ρ∈P(n) amdρ + am−1dρ + ··· + a1dρ + a0

¯m m−1 ¯ a0 = amd n + am−1d n + ··· + a1dn + |P(n)|

¯m m−1 ¯ a0 ≥ amd n − |am−1|d n − · · · − |a1|dn + |P(n)| √ ¯ m m−2 ¯ ¯ a0 ≥ am(dn) − |am−1|(2 n) dn − · · · − |a1|dn + |P(n)| √ ¯ m ¯ a0 = am(dn) − R1(2 n)dn + |P(n)| . 114

Corollary 5.4.2. Let m be a positive integer and let R(X) ∈ R[X] be a polynomial of degree m with positive leading coeﬃcient. Then, P R(dρ) lim ρ∈P(n) = ∞ n→∞ |P(n)| where dρ denotes the number of distinct parts of ρ counting 0 as a part. Further- P more, for n large enough we have ρ∈P(n) R(dρ) > 0

m m−1 Proof. Suppose that R(X) = amX + am−1X + ··· + a1X + a0. Then, from

Proposition 5.4.1 we have that there exists R1(X) a polynomial of degree at most m − 2 such that P R(dρ) √ a ρ∈P(n) ≥ a (d¯ )m − R (2 n)d¯ + 0 |P(n)| m n 1 n |P(n)|

Hence, P µ ¶ R(d ) √ ρ∈P(n) ρ ¯ m ¯ a0 lim ≥ lim am(dn) − R1(2 n)dn + n→∞ |P(n)| n→∞ |P(n)| Ã √ √ ! 6√ m √ 6√ a0 = lim am( n) − R1(2 n) n + n→∞ π π |P(n)| = ∞

as deg R1 ≤ m − 2. Furthermore, for n >> 0 we obtain P R(dρ) ρ∈P(n) > 0 |P(n)| P hence, ρ∈P(n) R(dρ) > 0.

Theorem 5.4.3. Let R(X),Q(X) ∈ R[X] with positive leading coeﬃcient. Suppose that deg R < deg Q. Then X R(dρ) ρ∈P(n) lim X = 0 n→∞ Q(dρ) ρ∈P(n) 115

where dρ denotes the number of distinct parts of ρ counting 0 as a part.

Proof. By Corollary 5.4.2 we have that there exists N ∈ N such that for all n ≥ N P P we have ρ∈P(n) R(dρ), ρ∈P(n) Q(dρ) > 0. For all n ≥ N let P ρ∈P(n) R(dρ) An := P . ρ∈P(n) Q(dρ)

Then we have An > 0.

r r−1 Suppose that R(X) = arX + ar−1X + ··· + a1X + a0 and that Q(X) =

s s−1 bsX + bs−1X + ··· + b1X + b0 with r < s and ar, bs > 0. By Proposition 5.4.1 there exist R1(X),Q1(X) ∈ R[X] with deg R1 ≤ r − 2 and deg Q1 ≤ s − 2 such that P √ a R(dρ) √ √ a a (d¯ )r−R (2 n)d¯ + 0 ≤ ρ∈P(n) ≤ a (2 n)r−1d¯ +R (2 n)d¯ + 0 , r n 1 n |P(n)| |P(n)| r n 1 n |P(n)| P √ b Q(dρ) √ √ b b (d¯ )s−Q (2 n)d¯ + 0 ≤ ρ∈P(n) ≤ b (2 n)s−1d¯ +Q (2 n)d¯ + 0 . s n 1 n |P(n)| |P(n)| s n 1 n |P(n)| We obtain 1 P |P(n)| ρ∈P(n) R(dρ) An = 1 P |P(n)| ρ∈P(n) Q(dρ) √ √ r−1 ¯ ¯ a0 ar(2 n) dn + R1(2 n)dn + ≤ √ |P(n)| . ¯ s ¯ b0 bs(dn) − Q1(2 n)dn + |P(n)| Hence, √ √ r−1 ¯ ¯ a0 ar(2 n) dn + R1(2 n)dn + |P(n)| lim An ≤ lim √ ¯ s ¯ b0 n→∞ n→∞ bs(dn) − Q1(2 n)dn + √ √ √ √ |P(√n)|√ r−1 6 6 a0 ar(2 n) π n + R1(2 n) π n + |P(n)| = lim √ √ √ √ √ n→∞ 6 s s 6 b0 bs( π ) ( n) − Q1(2 n) π n + |P(n)| = 0

√ √ ¯ 6 as by Theorem 5.3.2 we have dn ∼ π n and r < s. Since for all n ≥ N we have

An > 0 we obtain that lim An = 0. n→∞ Theorem 5.4.4. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. Let NCT (n, m) denote the number of non-isomorphic character tables of 116 p-groups P with |P | = pn, |P 0| = p and [P : Z(P )] = p2m. Then

1 X N (n, m) ≥ (d2m+1 + Q¯(d )) CT (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) where dρ¯ denotes the number of distinct parts of ρ¯ counting 0 as a part and Q¯(X) ∈ R[X] depends only on m and has degree less than or equal to 2m.

Proof. By Corollary 4.11.5 we have X X X ¯ NCT (n, m) = f(¯ρ, e,¯ λ)

ρ¯∈P(n−2m−1) e∈D(¯ρ)∪{0} λ∈Pρ¯ X X X ≥ 1

ρ¯∈P(n−2m−1) e∈D(¯ρ)∪{0} λ∈Pρ¯ X = dρ¯|Pρ¯| (|Pρ¯| does not depend one ¯) ρ¯∈P(n−2m−1) X = dρ¯ ρ¯∈P(n−2mµ−1) ¶ µ ¶ X d d − (m(η ) + ··· + m(η )) ρ,η¯ 1 ··· ρ,η¯ r 1 r−1 m(η1) m(ηr) η∈P(2m) ·µ ¶ X d = d ρ¯ ρ¯ 2m ρ¯∈P(n−2m−1)  µ ¶ µ ¶ X d d − (m(η ) + ··· m(η ))  ρ,η¯ 1 ρ,η¯ r 1 r−1  + ···  m(η1) m(ηr)  η ∈ P(2m), η 6= (12m) µ ¶ X d ≥ d ρ¯ ρ¯ 2m ρ¯∈P(n−2m−1) X d (d − 1) ··· (d − 2m + 1) = d ρ¯ ρ¯ ρ¯ ρ¯ (2m)! ρ¯∈P(n−2m−1) 1 X ¡ ¢ = d2m+1 + Q¯(d ) (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) where deg Q¯ ≤ 2m.

Theorem 5.4.5. Let m, n be positive integers such that n − 2m − 1 ≥ 0. Let

(2m)! C := {(ρ, e, λ, φ) ∈ CCT : |cct−1 ((ρ, e, λ, φ))| < }. < n,m n,m 2mm! 117

Then there exist A > 0 and R(X) ∈ R[X] with deg R ≤ 2m − 1 such that

X ¡ 2m ¢ |C<| ≤ A dρ¯ + R(dρ¯) ρ¯∈P(n−2m−1) where dρ¯ denotes the number of distinct parts of ρ¯ and we consider 0 as a distinct part.

Proof. We remark ﬁrst that if m = 1 the set C< is empty so the inequality holds trivially.

Assume that m ≥ 2. Then using Proposition 4.11.4 we have      ρ¯ ∈ P(n − 2m − 1), e¯ 6= 0, λ ∈ Pρ¯    |C<| ≤ 2| (¯ρ, e,¯ λ): and there exists some φ such that |    (2m)!   |cct−1 ((ψ(¯ρ, e¯), λ, φ))| <   n,m 2mm!    ρ¯ ∈ P(n − 2m − 1), λ ∈ Pρ¯ and  + | (¯ρ, λ): (2m)! |  |cct−1 (((¯ρ, 1), 1, λ, 1))| <  n,m 2mm! where Pρ¯ is the set of partitions of n − 1 obtained fromρ ¯ by increasing by 1 at 2m parts, and ψ is the map from Proposition 4.11.4. Note that (ρ, e, λ, φ) ∈ C< if and only if λ ∈ Mρ¯, where Mρ¯ was deﬁned in Corollary 5.2.6. Hence, X X X |C<| ≤ 2

ρ¯∈P(n−2m−1) e¯∈D(¯ρ)∪{0} λ∈Mρ¯ where D(¯ρ) denotes the set of distinct parts ofρ ¯. By Corollary 5.2.6 we have

d2m−1 |M | ≤ ρ¯ + R¯(d ) ρ¯ (2m − 2)! ρ¯ with deg R¯ < 2m − 1. Thus, µ ¶ X X d2m−1 |C | ≤ 2 ρ¯ + R¯(d ) < (2m − 2)! ρ¯ ρ¯∈P(n−2m−1) e¯∈D(¯ρ)∪{0} 118

µ ¶ X d2m−1 = 2 d ρ¯ + R¯(d ) ρ¯ (2m − 2)! ρ¯ ρ¯∈P(n−2m−1)µ ¶ X d2m−1 = 2 ρ¯ + 2d R¯(d ) . (2m − 2)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) d2m−1 Let A = 2 ρ¯ and R(d ) := 2d R¯(d ). We have A > 0, deg R ≤ 2m − 1 and (2m − 2)! ρ¯ ρ¯ ρ¯ X ¡ 2m ¢ |C<| ≤ A dρ¯ + R(dρ¯) . ρ¯∈P(n−2m−1)

Theorem 5.4.6. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. Let NG(n, m) denote the number of non-isomorphic p-groups P with

n 2m |P | = p , derived subgroup of order p and [P : Z(P )] = p and let NCT (n, m) denote the number of non-isomorphic character tables of these groups. Then

(2m)! N (n, m) ≥ (N (n, m) − |C |) G 2mm! CT < (2m)! where C = {(ρ, e, λ, φ) ∈ CCT : |cct−1 ((ρ, e, λ, φ))| < }. < n,m n,m 2mm!

Proof. From Corollary 4.11.2 we have that there exists a bijection between CT m,n,p and CCT n,m such that the following diagram is commutative

Θm,n,p Pm,n,p / Sn,m

ctm,n,p cctn,m CT m,n,p / CCT n,m

Hence, (2m)! |C | = |{[CT ] ∈ CT : |ct−1 ([CT ])| < }|. < m,n,p m,n,p 2mm! Thus,

(2m)! N (n, m) − |C | = |{[CT ] ∈ CT : |ct−1 ([CT ])| ≥ }|. CT < m,n,p m,n,p 2mm! 119

(2m)! Let C¯ := {[CT ] ∈ CT : |ct−1 ([CT ])| ≥ }. Then, since for each m,n,p m,n,p 2mm! (2m)! [CT ] ∈ C¯ we have at least non-isomorphic groups having their irreducible 2mm! character table in the isomorphism class [CT ] we have

(2m)! (2m)! N (n, m) ≥ |C|¯ = (N (n, m) − |C |). G 2mm! 2mm! CT <

Corollary 5.4.7. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. Let NCT (n, m) denote the number of non-isomorphic irreducible character tables of p-groups P with |P | = pn, derived subgroup of order p and (2m)! [P : Z(P )] = p2m. Let C = {(ρ, e, λ, φ) ∈ CCT : |cct−1 ((ρ, e, λ, φ))| < }. < n,m n,m 2mm! Then |C | lim < = 0. n→∞ NCT (n, m) Proof. By Theorem 5.4.5 we have

X ¡ 2m ¢ |C<| ≤ Adρ¯ + R(dρ¯) ρ¯∈P(n−2m−1) where A > 0, dρ¯ denotes the number of distinct parts ofρ ¯ and R is a polynomial of degree at most 2m − 1. By Theorem 5.4.4 we have

1 X ¡ ¢ N (n, m) ≥ d2m+1 + Q(d ) CT (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) where Q is a polynomial of degree at most 2m. We obtain X ¡ 2m ¢ Adρ¯ + R(dρ¯) |C | ρ¯∈P(n−2m−1) < ≤ N (n, m) 1 X ¡ ¢ CT d2m+1 + Q(d ) (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) 120 and X ¡ 2m ¢ Adρ¯ + R(dρ¯) |C<| ρ¯∈P(n−2m−1) lim ≤ lim X = 0 n→∞ N (n, m) n→∞ 1 ¡ ¢ CT d2m+1 + Q(d ) (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) |C | by Theorem 5.4.3. Since < ≥ 0 we obtain NCT (n, m)

|C | lim < = 0. n→∞ NCT (n, m)

Theorem 5.4.8. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. Let NG(n, m) denote the number of non-isomorphic p-groups P with

n 2m |P | = p , derived subgroup of order p and [P : Z(P )] = p and let NCT (n, m) denote the number of non-isomorphic character tables of such groups. Then

(2m)! N (n, m) ≤ (N (n, m) + |C |) G 2mm! CT 1 where C1 = {(ρ, 1, λ, 1) ∈ CCT n,m}.

Proof. Let C2 = CCT n,m \C1. Then, by Theorem 5.1.5 a) we have that

(2m)! |cct−1 ((ρ, e, λ, φ))| ≤ n,m 2mm! for all (ρ, e, λ, φ) ∈ C2 and by Theorem 5.1.5 b) we have

(2m)! |cct−1 ((ρ, 1, λ, 1))| ≤ 2 n,m 2mm! for all (ρ, 1, λ, 1) ∈ C1. Using Theorem 8 in Blackburn [2] and Theorem 4.10.17 we have −1 NG(n, m) = |cctn,m(CCT n,m)| (2m)! (2m)! ≤ 2 |C | + |C | 2mm! 1 2mm! 2 121

(2m)! = (|C | + |C | + |C |) 2mm! 1 2 1 (2m)! = (N (n, m) + |C |) 2mm! CT 1 since by Corollary 4.11.2 we have NCT (n, m) = |CCT n,m| = |C1| + |C2|. Thus,

(2m)! N (n, m) ≤ (N (n, m) + |C |). G 2mm! CT 1

Proposition 5.4.9. Let m, n be positive integers such that n − 2m − 1 ≥ 0. Let

C1 = {(ρ, 1, λ, 1) ∈ CCT n,m}. Then there exists a bijection

C1 '{(¯ρ, λ) :ρ ¯ ∈ P(n − 2m − 1), λ ∈ Pρ¯}

where Pρ¯ is the set of distinct partitions of n − 1 obtained from ρ¯ by increasing by 1 at 2m parts.

Proof. We deﬁne

f : C1 → {(¯ρ, λ) :ρ ¯ ∈ P(n − 2m − 1), λ ∈ Pρ¯}

f((ρ, 1, λ, 1)) = (¯ρ, λ) whereρ ¯ is obtained from ρ by decreasing by 1 at its last part of size 1 and adding 2m − 1 parts of size 0. Thus,ρ ¯ ∈ P(n − 2m − 1) and since λ is obtained from ρ by ﬁrst decreasing by 1 at a part of size 1 and then increasing by 1 at 2m parts we have that f is well deﬁned.

Moreover ¯ f : {(¯ρ, λ) :ρ ¯ ∈ P(n − 2m − 1), λ ∈ Pρ¯} → C1

f¯((¯ρ, λ)) = ((¯ρ, 1), 1, λ, 1) is well deﬁned and an inverse of f. Thus, f is a bijection. 122

Corollary 5.4.10. Let m, n be positive integers such that n − 2m − 1 ≥ 0 and let p be a prime. Let NCT (n, m) denote the number of non-isomorphic irreducible character tables of p-groups P with |P | = pn, derived subgroup of order p and

2m [P : Z(P )] = p . Let C1 = {(ρ, 1, λ, 1) ∈ CCT n,m}. Then

|C | lim 1 = 0. n→∞ NCT (n, m)

Proof. By Proposition 5.4.9 we have

C1 '{(¯ρ, λ) :ρ ¯ ∈ P(n − 2m − 1), λ ∈ Pρ¯}. X Hence, |C1| = |Pρ¯|. Using Corollary 5.2.6 we obtain ρ¯∈P(n−2m−1) µ ¶ X d2m |C | ≤ ρ¯ + R(d ) 1 (2m)! ρ¯ ρ¯∈P(n−2m−1) where dρ¯ denotes the number of distinct parts ofρ ¯ counting 0 as a part and R is a polynomial of degree at most 2m − 1. By Theorem 5.4.4 we have

1 X ¡ ¢ N (n, m) ≥ d2m+1 + Q(d ) CT (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) where Q is a polynomial of degree at most 2m. Hence, µ ¶ X d2m ρ¯ + R(d ) (2m)! ρ¯ |C | ρ¯∈P(n−2m−1) 1 ≤ . N (n, m) 1 X ¡ ¢ CT d2m+1 + Q(d ) (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1)

We obtain µ ¶ X d2m ρ¯ + R(d ) (2m)! ρ¯ |C1| ρ¯∈P(n−2m−1) lim ≤ lim X = 0 n→∞ N (n, m) n→∞ 1 ¡ ¢ CT d2m+1 + Q(d ) (2m)! ρ¯ ρ¯ ρ¯∈P(n−2m−1) 123

|C | by Theorem 5.4.3. Since 1 ≥ 0 we obtain NCT (n, m)

|C | lim 1 = 0. n→∞ NCT (n, m)

Theorem 5.4.11. Let m, n be integers with m ≥ 2 and n − 2m − 1 ≥ 0. Let

n NG(n, m) denote the number of non-isomorphic p-groups P with |P | = p , derived

2m subgroup of order p and [P : Z(P )] = p and let NCT (n, m) denote the number of non-isomorphic character tables of such groups. Then

NG(n, m) (2m)! lim = m . n→∞ NCT (n, m) 2 m!

Proof. By Theorem 5.4.8 we have

(2m)! N (n, m) ≤ (N (n, m) + |C |) G 2mm! CT 1 where C1 = {(ρ, 1, λ, 1) ∈ CCT n,m}. Hence, µ ¶ NG(n, m) (2m)! |C1| ≤ m 1 + . NCT (n, m) 2 m! NCT (n, m)

Thus, µ ¶ NG(n, m) (2m)! |C1| (2m)! lim ≤ m 1 + lim = m n→∞ NCT (n, m) 2 m! n→∞ NCT (n, m) 2 m! |C | as by Corollary 5.4.10 we have lim 1 = 0. n→∞ NCT (n, m) By Theorem 5.4.6 we have

(2m)! N (n, m) ≥ (N (n, m) − |C |) G 2mm! CT < (2m)! where C = {(ρ, e, λ, φ) ∈ CCT : |cct ((ρ, e, λ, φ))| < }. Hence, < n,m n,m 2mm! µ ¶ NG(n, m) (2m)! |C<| ≥ m 1 − . NCT (n, m) 2 m! NCT (n, m) 124

We obtain µ ¶ NG(n, m) (2m)! |C<| (2m)! lim ≥ m 1 − lim = m n→∞ NCT (n, m) 2 m! n→∞ NCT (n, m) 2 m!

|C | since by Corollary 5.4.7 we have lim < = 0. n→∞ NCT (n, m) We have proved that

(2m)! NG(n, m) (2m)! m ≤ lim ≤ m 2 m! n→∞ NCT (n, m) 2 m! hence, by the Squeeze Theorem we have

NG(n, m) (2m)! lim = m . n→∞ NCT (n, m) 2 m!

Proposition 5.4.12. Let n ≥ 3 be an integer. Then

NG(n, 1) − NCT (n, 1) ≥ |P(n − 3)|.

Proof. From Corollary 4.11.2 we have the following commutative diagram

Θn,1 P1,n,p / Sn,1

ct1,n,p cctn,1 CT 1,n,p / CCT n,1.

In Theorem 5.1.5 we have proved that for all (ρ, e, λ, φ) ∈ CCT n,1 we have

−1 |cctn,1((ρ, e, λ, φ))| ∈ {1, 2}.

−1 Moreover, |cctn,1((ρ, e, λ, φ))| = 2 if and only if e = 1 and λ is obtained fromρ ¯ by increasing by 1 at one part of size 0. Hence,

−1 NG(n, 1) = |{(ρ, e, λ, φ) ∈ CCT n,1 : |cctn,1((ρ, e, λ, φ))| = 1}|

−1 + |{(ρ, 1, λ, 1) ∈ CCT n,1 : |cctn,1((ρ, 1, λ, 1))| = 2}|. 125

We obtain that

−1 NG(n, 1) − NCT (n, 1) = |{(ρ, 1, λ, 1) ∈ CCT n,1 : |cctn,1((ρ, 1, λ, 1))| = 2}|.

It is easy to see that there is a bijection between

{ρ ∈ P(n − 2) : ρ has a part of size 1} → P(n − 3).

Hence,

−1 |{(ρ, 1, λ, 1) ∈ CCT n,1 : |cctn,1((ρ, 1, λ, 1))| = 2}| ≥ P(n − 3).

Theorem 5.4.13. For all ε > 0 we have

NG(n) − NCT (n) lim √ √ = ∞. n→∞ eπ 2/3 n n1+ε Proof. We have

b n−1 c b n−1 c X2 X2 NG(n) − NCT (n) = NG(n, m) − NCT (n, m) m=1 m=1 ≥ NG(n, 1) − NCT (n, 1)

≥ |P(n − 3)| by Proposition 5.4.12.

It is well known that the number of partitions of a positive integer n is asymptotically 1 √ √ |P(n)| ∼ √ eπ 2/3 n. 4n 3 Hence, the conclusion follows now immediately. CHAPTER 6 NUMERICAL RESULTS We wrote a computer programm in GAP that allows us to compute for any integer n ≥ 3 and any prime p the number of non-isomorphic p-groups of order pn and derived subgroup of order p and the number of non-isomorphic irreducible character tables of these p-groups. The results are presented in the next table

Table 6–1. The number of groups and character tables

n NG(n) NCT (n, m) 3 2 1 4 6 4 5 17 12 ...... 10 584 397 ...... 15 8838 4849 ...... 20 91824 38603 ...... 25 763121 238282 ...... 30 5421603 1233858

126 REFERENCES

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127 BIOGRAPHICAL SKETCH

I was born in Bucharest, Romania. I have a Bachelor of Science and a Master of Science from the University of Bucharest. Since August 1998 I have been a researcher at the Institute of Mathematics of the Romanian Academy. In August 2001 I started graduate studies in mathematics at the University of Florida. My research interests are in algebra.

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