Proc. Indian Acad. Sci. (Math. Sci.) Vol. 122, No. 3, August 2012, pp. 329–334.c© Indian Academy of Sciences

Minimal degrees of faithful quasi-permutation representationsfor direct products of p-groups

GHODRAT GHAFFARZADEH andMOHAMMAD HASSAN ABBASPOUR

Department of Mathematics, Islamic Azad University, Khoy Branch, Khoy, IranE-mail: gh.ghaffarzadeh@iaukhoy.ac.ir; m.abbaspour@iaukhoy.ac.ir

MS received 24 October 2009; revised 30 March 2012

Abstract. In [2], the algorithms of c(G), q(G) and p(G), the minimal degrees offaithful quasi-permutation and permutation representations of a finite group G are given.The main purpose of this paper is to consider the relationship between these minimaldegrees of non-trivial p-groups H and K with the group H × K .

Keywords. Quasi-permutation representations; p-groups; character theory.

1. Introduction

By quasi-permutation matrix we mean a square matrix over the complex field C withnon-negative integral trace. Thus every permutation matrix over C is a quasi-permutationmatrix. For a given finite group G, let p(G) denote the minimal degree of a faithful permu-tation representation of G (or of a faithful representation of G by permutation matrices),let q(G) denote the minimal degree of a faithful representation of G by quasi-permutationmatrices over the rational field Q, and let c(G) denote the minimal degree of a faithfulrepresentation of G by complex quasi-permutation matrices. Then it is easy to see that

c(G) ≤ q(G) ≤ p(G),

where G is a finite group.In Theorem 2 and Corollary 2 of [7], it is shown that if H and K are non-trivial nilpotent

groups, then p(H × K ) = p(H) + p(K ). The purpose of this article is to get similarresults on q(G) and c(G), for all non-trivial p-groups H and K , that is

q(H × K ) = q(H) + q(K ) and c(H × K ) = c(H) + c(K ).

By Theorem 3.2 of [3], p(G) = q(G) for each p-group and if p �= 2, then c(G) =q(G) = p(G). Now as p(H × K ) = p(H)+ p(K ), q(H × K ) = q(H)+ q(K ). Also inthe case p odd, c(H × K ) = c(H) + c(K ). Hence it remains to prove the last equation,in the case p = 2. For this purpose, we will show that the number of Galois conjugacyclasses of complex irreducible characters of a p-group G, needed in the algorithm of c(G)

given in Theorem 2.1 is equal to the minimal number of generators of its center.We mention that the last two equations do not remain true, in the case of arbitrary

nilpotent groups H and K . For example, if G = C2 × C3, then c(G) = q(G) = 4,

329

330 Ghodrat Ghaffarzadeh and Mohammad Hassan Abbaspour

however c(C2) = q(C2) = 2 and c(C3) = q(C3) = 3. In general, the above equationsare not true for abelian groups with a direct summand C6. In fact it is a well-known resultthat for a non-trivial abelian group G ∼= ∏r

i=1 Cmi , where each mi is a prime power, wehave c(G) = q(G) = T (G) − n, where T (G) = ∑r

i=1 mi and n is maximal such that Ghas a direct summand Cn

6 (see [1]).

2. The main results

We begin by recalling the formula of c(G), which is valid for all finite groups.Let G be a finite group. Let Ci for 0 ≤ i ≤ r be the Galois conjugacy classes over Q

of irreducible complex characters of the group G. For 0 ≤ i ≤ r , suppose that ψi is arepresentative of the class Ci with ψ0 = 1G . Write �i = ∑

Ci and Ki = ker ψi . ClearlyKi = ker �i . For I ⊆ {0, 1, . . . , r}, put K I = ⋂

i∈I Ki . Also, if the ni ’s are non-negativeintegers and I ⊆ {1, . . . , r}, then we will use the notation m(χ) for χ = ∑

i∈I ni�i todenote m(χ) = −min{∑i∈I ni�i (g) : g ∈ G}.Theorem 2.1. Let G be a finite group. Then in the above notation

c(G) = min{ξ(1) + m(ξ) : ξ =

∑

i∈I

�i , K I = 1 for I ⊆ {1, . . . , r}

and K J �= 1 i f J ⊂ I}.

Proof. See Lemma 2.2 of [3]. �

We express a lemma on p-groups before the principal results.

Lemma 2.2. Let A = {A1, . . . , An} be a collection of subgroups of an abelian p-group Asuch that

(a)⋂n

i=1 Ai = 1,(b) for all 1 ≤ j ≤ n,

⋂ni=1i �= j

Ai �= 1,

(c) for all 1 ≤ j ≤ n, A/A j is cyclic;

then n = d(A), the minimal number of generators of A.

Proof. See Lemma 2 of [6]. �

Theorem 2.3. Let G be a p-group whose center Z(G) is minimally generated by d ele-ments. Let c(G) = ξ(1) + m(ξ) and ξ = ∑

i∈I �i . Let �i ’s satisfy the conditions of thealgorithm c(G). Then

(a) m(ξ) = 1p−1

∑i∈I �i (1),

(b) |I | = d.

Proof.

(a) For every k ∈ I , let Ik = I − {k}. By Theorem 2.1, Ck = ⋂i∈Ik

ker ψi �= 1, sochoose zk ∈ Ck

⋂Z(G) of order p. Since K I = 1, we have zk �∈ ker ψk . Let z =

�i∈I zi . Clearly, o(z) = p and z �∈ ⋃i∈I ker ψi . Thus, ψi (z) = εiψi (1) for all i ∈ I ,

where εi is a complex p-th root of 1 and εi �= 1. Hence, εi is the primitive p-th rootof unity. Set Hi = GalQ(Q(εi )). Since ε

p−1i + · · · + εi = −1 and |Hi | = p − 1,

Minimal degrees of faithful quasi-permutation representations 331

it follows that∑

σ∈Hiψi (z)σ = −ψi (1). As Q(εi ) ⊆ Q(ψi ), we set i = GalQ(ψi )

and ′i = GalQ(εi )(Q(ψi )). Note that Q(ψi ) is a finite degree Galois extension of Q

and the Galois group i is abelian. So the restriction map φ : i → Hi defined byφ(σ) = σ |Q(εi ) induces an isomorphism φ : i/

′i

∼= Hi . By definition of φ and φ, it

follows that �i (z) = ∑σ∈i

ψi (z)σ = −diψi (1), where di = |′i | = |i |

p−1 . Therefore,∑

i∈I �i (z) = −∑i∈I diψi (1) = − 1

p−1

∑i∈I |i |ψi (1) = − 1

p−1

∑i∈I �i (1). Since

m(ξ) = −min{∑i∈I �i (g) : g ∈ G}, we have m(ξ) ≥ 1p−1

∑i∈I �i (1). In the case

p = 2, since 1p−1 = 1 and

∑i∈I �i (z) = −∑

i∈I �i (1), by definition of m(ξ) and thefact that �i (g) ≤ �i (1) for each i ∈ I and g ∈ G, it follows that m(ξ) = ∑

i∈I �i (1).Therefore, let p > 2 and m(ξ) > 1

p−1

∑i∈I �i (1). Then by Theorem 3 of [4], there exist

subgroups G1i < G2i in G such that |G2i : G1i | = p and �i = λ1iG − λ2i

G , where λhi

is the principal character of Ghi . Hence,

ξ =∑

i∈I

�i =∑

i∈I

(λ1iG − λ2i

G) =∑

i∈I

λ1iG −

∑

i∈I

λ2iG .

Since ξ is faithful, so is ξ + ∑i∈I λ2i

G and thus∑

i∈I λ1iG . However, by Lemma 5.11

of [5], ker λ1iG = (G1i )G , where (G1i )G = ⋂

x∈G G1ix is the core of G1i in G. Hence,

1 = ker(∑

i∈I λ1iG) = ⋂

i∈I (G1i )G . Therefore,

c(G)=ξ(1) + m(ξ)>1

p−1

∑

i∈I

�i (1)+∑

i∈I

�i (1)= p

p − 1

∑

i∈I

�i (1)

= p

p−1

∑

i∈I

(|G : G1i |−|G : G2i |)= p

p−1

∑

i∈I

p−1

p|G : G1i |

=∑

i∈I

|G : G1i |. (1)

Thus, c(G) >∑

i∈I |G : G1i |. On the other hand, let �i = {G1i x : x ∈ G} and� = ⋃

i∈I �i . So |�i | = |G : G1i |. Let ρ denote the permutation representationof G corresponding to the action of G on � by right multiplication. We know thatker ρ = ⋂

ω∈� StabG(ω). But

⋂

ω∈�

StabG(ω) =⋂

i∈I

(⋂

x∈G

G1ix

)

=⋂

i∈I

(G1i )G = 1.

So ρ is a faithful permutation (hence quasi-permutation) representation of G with degreeρ(1) = ∑

i∈I |�i | = ∑i∈I |G : G1i |. Therefore, by definition of c(G), it follows that

c(G) ≤ ∑i∈I |G : G1i |, which is a contradiction. Thus m(ξ) = 1

p−1

∑i∈I �i (1).

(b) Put Ki = ker �i = ker ψi , i ∈ I . We show that the set {Ki⋂

Z(G) : i ∈ I }satisfies the hypothesis of Lemma 2.2. Then it follows that |I | = d. Clearly, the subgroupsKi

⋂Z(G) have the properties (a) and (b) in Lemma 2.2. Also

Z(G)/(Ki

⋂Z(G)

) ∼= Z(G)Ki/Ki ≤ Z (G/Ki ) .

But by Lemma 2.27 of [5], Z (G/Ki ) is a cyclic group. So Z(G)/(Ki⋂

Z(G)) is cyclic.Therefore Lemma 2.2 can be applied. Hence the proof is complete. �

332 Ghodrat Ghaffarzadeh and Mohammad Hassan Abbaspour

By combining Theorems 2.3 and the next one, we have a programme for finding theminimal degree of a faithful representation of direct products of p-groups by complexquasi-permutation matrices.

Theorem 2.4. Let G = H × K , where H and K are non-trivial p-groups. Let m =d(Z(H)) and n = d(Z(K )) and suppose that G = {G1, . . . , Gm+n} is a collection ofnormal subgroups of G such that

(a)⋂m+n

i=1 Gi = 1,

(b) for all 1 ≤ j ≤ m + n,⋂m+n

i=1i �= j

Gi �= 1;

then the subgroups Gi can be reordered such that

(m⋂

i=1

Gi

)⋂

H = 1 and

(n⋂

i=1

Gm+i

)⋂

K = 1.

Proof. It is obvious that

d(Z(G)) = d(Z(H)) + d(Z(K )) = m + n.

For all 1 ≤ i ≤ m + n, put Gi =(

⋂m+nj=1j �=i

G j

)⋂

Z(G). Then by hypothesis, Gi �= 1.

Also

Gi⋂ m+n∏

j=1j �=i

G j ≤ Gi⋂

Gi =(

m+n⋂

j=1G j

)⋂

Z(G) = 1.

So G1 × · · · × Gm+n is a direct product of non-trivial subgroups of Z(G). Sinced(Z(G)) = m + n, it follows that each Gi is a non-trivial cyclic group. ThusGi

⋂�1(Z(G)) = 〈zi 〉, say, a cyclic group and in the elementary abelian group

�1(Z(G)), one also deduces that

(i) Gi⋂

�1(Z(G)) = 〈z j : 1 ≤ j ≤ m + n, j �= i〉, 1 ≤ i ≤ m + n

(ii) �1(Z(H)) × �1(Z(K )) = �1(Z(G)) = 〈zi : 1 ≤ i ≤ m + n〉 .

By Note 1 after Theorem 1 of [7], we conclude that the Gi can be reordered so that

(iii) 〈zi : 1 ≤ i ≤ m〉⋂�1(Z(K )) = 1 and 〈zm+i : 1 ≤ i ≤ n〉 ⋂

�1(Z(H)) = 1.So from (i), it follows that

(m⋂

i=1

Gi

)⋂

�1(Z(G)) = 〈zm+i : 1 ≤ i ≤ n〉

and(

n⋂

i=1

Gm+i

)⋂

�1(Z(G)) = 〈zi : 1 ≤ i ≤ m〉 .

Minimal degrees of faithful quasi-permutation representations 333

Hence by (iii), we deduce(

m⋂

i=1

Gi

)⋂

Z(H) = 1 and

(n⋂

i=1

Gm+i

)⋂

Z(K ) = 1.

The result is now apparent. �

Theorem 2.5. Let G = H × K with H and K non-trivial 2-groups. Then

c(H × K ) = c(H) + c(K ).

Proof. Suppose that m = d(Z(H)) and n = d(Z(K )). By Theorem 2.1, c(G) = m(ξ) +ξ(1), where ξ = ∑

i∈I �i ,⋂m+n

i=1 ker �i = 1 and⋂m+n

i=1i �= j

ker �i �= 1 for all 1 ≤ j ≤m + n. Also by Theorem 2.3, |I | = d(Z(G)) = m + n.

Put Gi = ker �i , 1 ≤ i ≤ m + n. By Theorem 2.4, we conclude that Gi can bereordered so that

m⋂

i=1

(H⋂

Gi ) = 1 andn⋂

i=1

(K⋂

Gm+i ) = 1.

For each 1 ≤ i ≤ m + n, let ψi be a representative of the Galois conjugacy class of theirreducible characters of G with the sum �i . Then ψi = λi × φi with λi ∈ Irr(H) andφi ∈ Irr(K ). Hence

ker λi = H⋂

Gi and ker φi = K⋂

Gi .

For each 1 ≤ i ≤ m + n, put

�i =∑

σ∈GalQ(λi )

λσi , �i =

∑

σ∈GalQ(φi )

φσi

and

ξ1 =m∑

i=1

�i , ξ2 =n∑

i=1

�i+m .

By Theorem 2.1, it follows that

c(H) ≤ m(ξ1) +m∑

i=1

�i (1) and c(K ) ≤ m(ξ2) +n∑

i=1

�i+m(1).

As Q(λi ) ⊆ Q(ψi ), we get |GalQ(λi )| ≤ |GalQ(ψi )|. Hence

�i (1) = |GalQ(λi )|λi (1) ≤ |GalQ(ψi )|ψi (1) = �i (1).

Similarly �i (1) ≤ �i (1). Since p = 2, Theorem 2.3(a) specifically says that m(ξ) =∑m+ni=1 �i (1), and so ξ(1) + m(ξ) = 2

∑m+ni=1 �i (1). Thus

c(G) = 2m+n∑

i=1

�i (1) = 2m∑

i=1

�i (1) + 2m+n∑

i=m+1

�i (1)

≥ 2m∑

i=1

�i (1) + 2m+n∑

i=m+1

�i (1) ≥ c(H) + c(K ).

334 Ghodrat Ghaffarzadeh and Mohammad Hassan Abbaspour

The reverse inequality holds obviously for all finite groups. Hence the proof iscomplete. �

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