CUBO A Mathematical Journal

Vol.10, N
o
¯ 03, (195–210). October 2008

On the Structure of Primitive n-Sum Groups

Eloisa Detomi and Andrea Lucchini

Dipartimento di Matematica Pura ed Applicata,

Via Trieste 63, 35121 Padova, Italy

email: detomi@math.unipd.it

email: lucchini@math.unipd.it

ABSTRACT

For a finite group G, let σ(G) be least cardinality of a collection of proper subgroups

whose set-theoretical union is all of G. We study the structure of groups G containing

no normal non-trivial subgroup N such that σ(G/N ) = σ(G).

RESUMEN

Para un grupo G, sea σ(G) la menor cardinalidad de la colección de subgrupos propios

cujas union (de conjuntos) es todo G. Nosotros estudiamos la estructura de grupos G

contiendo no trivial no normales subgrupos N tal que σ(G/N ) = σ(G).

Key words and phrases: n-sum groups; minimal coverings; monolithic groups.

Math. Subj. Class.: 20D60.

1 Introduction

If G is a non cyclic finite group, then there exists a finite collection of proper subgroups whose

set-theoretical union is all of G; such a collection is called a cover for G. A minimal cover is one



196 Eloisa Detomi and Andrea Lucchini CUBO
10, 3 (2008)

of least cardinality and the size of a minimal cover of G is denoted by σ(G) (and for convenience

we shall write σ(G) = ∞ if G is cyclic). The study of minimal covers was introduced by J.H.E.

Cohn [8]; following his notation, we say that a finite group G is an n-sum group if σ(G) = n and

that a group G is a primitive n-sum group if σ(G) = n and G has no normal non-trivial subgroup

N such that σ(G/N ) = n. We will say that G is σ-primitive if it is a primitive n-sum group for

some integer n. Notice that if N is a normal subgroup of G, then σ(G) ≤ σ(G/N ); indeed a cover

of G/N can be lifted to a cover of G.

It is clear that if G is a non cyclic monolithic primitive group (i.e. G has a unique minimal

normal subgroup and the Frattini subgroup of G is trivial) and G/ soc(G) is cyclic, then G is a

σ-primitive group.

Moreover Cohn proved that an abelian σ-primitive group is the direct product of two cyclic

groups of order p, a prime number.

Tomkinson [14] showed that in a finite solvable group G, σ(G) = |V | + 1, where V is a chief

factor of G with least order among chief factors of G with multiple complements. This allows to

prove (see for example [5]) that a σ-primitive solvable group G is as described above, i.e. either G

is abelian or G is monolithic and G/ soc(G) is cyclic.

However there exist examples of σ-primitive groups with G/ soc(G) non cyclic: actually with

G/ soc(G) ∼= Alt(p) for some prime p (see Corollary 9 and Corollary 12).

The aim of this paper is to collect information on the structure of the σ-primitive groups. In

particular we prove that if G is σ-primitive, then G contains at most one abelian minimal normal

subgroup; moreover two non-abelian minimal normal subgroups of G are not G-equivalent (we

refer to an equivalence relation among the chief factors of a finite group introduced in [10] and

[9], whose main properties are summarized at the beginning of Section 2). Furthermore if G is

non-abelian, then all the solvable factor groups of G/ soc(G) are cyclic.

No example is known of a non-abelian σ-primitive group containing two distinct minimal

normal subgroups. This leads to conjecture that a non-abelian σ-primitive group is monolithic.

We prove a partial result supporting this conjecture.

Theorem 1. Let G be a σ-primitive group with no abelian minimal normal subgroups. Then either

G is a primitive monolithic group and G/ soc(G) is cyclic, or G/ soc(G) is non-solvable and all the

non-abelian composition factors of G/ soc(G) are alternating groups of odd degree.

A better knowledge of the σ-primitive groups is useful in dealing with several questions about

the minimal covers. For example, confirming a conjecture of Tomkinson, we prove:

Theorem 2. There is no finite group G with σ(G) = 11.

Another application concerns the study of σ(G) when G = H ×K is a direct product of two fi-

nite groups. Cohn proved that if H and K have coprime order, then σ(H ×K) = min{σ(H), σ(K)}.

We prove the following more general result:



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On the Structure of Primitive n-Sum Groups 197

Theorem 3. Let G = H ×K be the direct product of two subgroups. If no alternating group Alt(n)

with n odd is a homomorphic image of both H and K, then either σ(G) = min{σ(H), σ(K)} or

σ(G) = p + 1 and the cyclic group of order p is a homomorphic image of both H and K.

2 Preliminary results and easy remarks

For the reader’s convenience, we recall the definition of an equivalence relation among the elements

of the set CF(G) of the chief factors of G, that was introduced in [10] and studied in details in

[9]. A group G is said to be primitive if it has a maximal subgroup with trivial core. The

socle soc(G) of a primitive group G can be either an abelian minimal normal subgroup (I), or

a non-abelian minimal normal subgroup (II), or the product of two non-abelian minimal normal

subgroups (III); we say respectively that G is primitive of type I, II, III and in the first two cases

we say that G is monolithic. Two chief factors of a finite group G are said to be G-equivalent if

either they are G-isomorphic between them or to the two minimal normal subgroups of a primitive

epimorphic image of type III of G. This means that two G-equivalent chief factors of G are either

G-isomorphic between them or to two chief factors of G having a common complement (which

is a maximal subgroup of G). A chief factor H/K is called Frattini if H/K ≤ Φ(G). For any

A ∈ CF(G) we denote by IG(A) the set of those elements of G which induces by conjugation an

inner automorphism in A. Moreover we denote by RG(A) the intersection of the normal subgroups

N of G contained in IG(A) and with the property that IG(A)/N is non-Frattini and G-equivalent

to A. We collect here a sequence of basic properties of the subgroups IG(A) and RG(A), proved

and discussed in [9]:

Proposition 4. Let A ∈ CF(G) and let I/R = IG(A)/RG(A). Then:

1. either R = I, in which case we set δG(A) = 0, or I/R = soc(G/R) and it is a direct product

of δG(A) minimal normal subgroups G-equivalent to A;

2. each chief series of G contains exactly δG(A) non-Frattini chief factors G-equivalent to A;

3. if A is abelian, then I/R has a complement in G/R;

4. if δG(A) ≥ 2, then any two different minimal normal subgroups of I/R have a common

complement, which is a maximal subgroup;

5. a chief factor H/K of G is non-Frattini and G-equivalent to A if and only if RH/RK 6= 1

and RH ≤ I.

Note that if δG(A) = 1, then G/RG(A) is a monolithic primitive group (the monolithic prim-

itive group associated to A).

In the rest of the section we will discuss some basis results on the relation between σ(G) and

σ(G/N ) when N is a minimal normal subgroup of G. We start summarizing some known properties

of σ.



198 Eloisa Detomi and Andrea Lucchini CUBO
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Lemma 5. Let N be a minimal normal subgroup of a group G. If σ(G) < σ(G/N ), then

1. if N has c complements which are maximal subgroups, then c + 1 ≤ σ(G);

2. if N = Sr where S is a non-abelian simple group and l(S) is the minimal index of a maximal

subgroup of S, then l(S)r + 1 ≤ σ(G).

Proof. (1) (See e.g. [14, Proof of Theorem 2.2]) Let M = {Mi | i = 1, . . . , σ(G)} be a set of maximal

subgroups whose union covers G and let M be a complement of N . Clearly M =
⋃

1≤i≤σ(G)
M ∩Mi,

however σ(M ) = σ(G/N ) > σ(G), hence M = M ∩ Mi for some i; in particular if M is a maximal

subgroup of G, then M = Mi ∈ M. So M contains all the c complements of N which are maximal;

since the union of these complements does not cover N , we need at least c + 1 subgroups in M.

(2) Let lG(N ) be the smallest index of a proper subgroup of G supplementing N . By Lemma

3.2 in [14] a minimal cover M of G contains at least lG(N ) subgroups which supplement N . On

the other hand, if all the subgroups in M are supplements of N , then by [8, Lemma 1] we have

lG(N ) ≤ σ(G) − 1. In any case we conclude σ(G) ≥ lG(N ) + 1 ≥ l(S)
r + 1.

Corollary 6. Let N be a minimal normal subgroup of a group G. If σ(G) < σ(G/N ), then

1. if N is abelian, complemented and non-central, then |N| + 1 ≤ σ(G);

2. if N = Sr where S is a non-abelian simple group, then 5r + 1 ≤ σ(G).

Proposition 7. Let N be a non-solvable normal subgroup of a finite group G. Then σ(G) ≤ |N|−1.

Proof. Consider the centralizers in G of the nontrivial elements of N : if there exists an element

g ∈ G which does not belong to
⋂

16=n∈N
CG(n) then the subgroup 〈g〉 acts fixed point freely on

N . By the classification of finite simple groups (see e.g. [15]), it follows that N is solvable, a

contradiction. Hence σ(G) ≤ |N| − 1.

Corollary 8. If N is a non-abelian minimal normal subgroup of G and δG(N ) > 1, then σ(G) =

σ(G/N ).

Proof. Assume by contradiction that σ(G) < σ(G/N ). Since δG(N ) > 1, there exists a maximal

subgroup M of G, such that G/MG is a primitive group of type III and M/MG is a common

complement of the two minimal normal subgroups of the socle H/MG × N MG/MG of G/MG. In

particular M is a non-normal complement of N and it has |N| conjugates, hence |N| + 1 ≤ σ(G)

by Lemma 5. This contradicts Proposition 7.

Corollary 9. Let p a large prime not of the form (qk − 1)/(q − 1) where q is a prime power and

k an integer; then σ(Alt(5) ≀ Alt(p)) < σ(Alt(p)).

Proof. By Proposition 7, σ(Alt(5) ≀ Alt(p)) < | Alt(5)|p. On the other hand, by Theorem [12, 4.4],

σ(Alt(p)) ≥ (p − 2)! > 60p for a large enough prime not of the form (qk − 1)/(q − 1).



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On the Structure of Primitive n-Sum Groups 199

Proposition 10. Let G be a finite group. If V is a complemented normal abelian subgroup of G

and V ∩ Z(G) = 1, then σ(G) < 2|V |. In particular, if V is a minimal normal subgroup, then

σ(G) ≤ 1 + q + · · · + qn where q = | EndG(V )| and |V | = q
n.

Proof. Let H be a complement of V in G; we shall prove that G is covered by the family of

subgroups A = {Hv | v ∈ V } ∪ {CH (v)V | 1 6= v ∈ V }. Let g = hw ∈ G, where h ∈ H, w ∈ V .

If h /∈ CH (v) for every v ∈ V \ {1}, then CV (h) = 1 and the cardinality of the set {h
v | v ∈ V }

is |V : CV (h)| = |V |. Therefore {h
v | v ∈ V } = {hv | v ∈ V } and g = hw ∈ Hv for some v ∈ V .

Thus σ(G) ≤ |A| ≤ |V | + (|V | − 1) < 2|V |. In particular, if V is H-irreducible, then EndG(V ) =

EndH (V ) = F is a finite field. We may identify H with a subgroup of GL(n, q), where |F| = q and

dimF V = n. In this case G is covered by A = {H
v | v ∈ V } ∪ {CH (W )V | W ≤ V, dimF W = 1},

so σ(G) ≤ qn + (1 + · · · + qn−1).

Corollary 11. Let H be a finite group, V an H-module, G = V ⋊ H the semidirect product of V

by H and assume that CV (H) = 0. Then

1. if H1(H, V ) 6= 0, then σ(G) = σ(H);

2. if σ(H) ≥ 2|V |, then H1(H, V ) = 0.

Proof. Assume by contradiction that σ(G) < σ(H). By Lemma 5, c + 1 ≤ σ(G) where c is the

number of complements of V in G. If H1(H, V ) 6= 0, then there are at least two conjugacy classes

of complements for V in G and, since CV (H) = 0, any conjugacy class consists of |V | complements,

hence c ≥ 2|V | and σ(G) > 2|V | against Proposition 10.

Corollary 12. Let V the fully deleted module for Alt(n) over F2 and let G be the semidirect

product of V by Alt(n).

1. If n = p is a large odd prime not of the form (qk − 1)/(q − 1) where q is a prime power and

k an integer, then σ(G) < σ(Alt(n)).

2. If n is even, then σ(G) = σ(Alt(n))

Proof. 1) Since |V | = 2p−1 (see e.g. [11, Prop. 5.3.5]), Proposition 10 gives that σ(G) < 2|V | < 2p.

On the other hand, by Theorem [12, 4.4], σ(Alt(p)) ≥ (p − 2)! > 2p for a large enough prime not

of the form (qk − 1)/(q − 1).

2) This follows from Corollary 11 and the fact that H1(Alt(n), V ) 6= 0 whenever n is even (see e.g.

[2, p. 74]).

Corollary 13. Let V 6= W be non-Frattini non-central abelian minimal normal subgroups of G.

Then

1. if δG(V ) > 1, then σ(G) = σ(G/V );



200 Eloisa Detomi and Andrea Lucchini CUBO
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2. σ(G) = min{σ(G/V ), σ(G/W )}.

Proof. 1) By a result in [3], the number c of complements of V in G is

c = | Der(G/V , V )| = | EndG/V (V )|
δG(V )−1| Der(G/CG(V ), V )|

hence c ≥ 2|V | whenever δG(V ) > 1. If σ(G) < σ(G/V ), then by Lemma 5 and Proposition 10,

2|V | < c + 1 ≤ σ(G) < 2|V |, a contradiction.

2) If V and W are G-equivalent, then by (1) σ(G) = σ(G/V ) = σ(G/W ). So assume that V and W

are not G-equivalent and, by contradiction, that σ(G) < min{σ(G/V ), σ(G/W )}. A complement

of V in G has at least |V | conjugates and it is a maximal subgroup of G, so we can find at least

|V | complements of V . In the same way there are at least |W | distinct complements of |W | in G.

Moreover, since V and W are not G-equivalent, V and W cannot have a common complement.

Arguing as in Lemma 5 we see that all the complements of V and W have to appear in a minimal

cover of G. Therefore σ(G) ≥ |V | + |W | ≥ min{2|V |, 2|W |}, against Proposition 10.

3 The structure of σ-primitive groups

We collect some known properties of σ-primitive groups and some consequences of the previous

section.

Corollary 14. Let G be a non-abelian σ-primitive group. Then:

1. Z(G) = 1;

2. the Frattini subgroup of G is trivial;

3. if N is a minimal normal subgroup of G, then δG(N ) = 1;

4. there is at most one abelian minimal normal subgroup of G;

5. the socle soc(G) = G1 × · · · × Gn is a direct product of non-G-equivalent minimal normal

subgroups and at most one of them is abelian.

6. G is a subdirect product of the monolithic primitive groups Xi = G/RG(Gi) associated to the

minimal normal subgroups Gi, 1 ≤ i ≤ n.

Proof. Part (1) is Theorem 4 in [8]. If Φ(G) is the Frattini subgroup of G and H is a proper

subgroup of G, then also H Φ(G) is a proper subgroup of G. Hence we can assume that Φ(G) is

contained in every subgroup of a minimal cover of G so that σ(G) = σ(G/ Φ(G)) and therefore (2)

holds. Parts (3) and (4) follows from Corollaries 8 and 13. Then (3) and (4) implies (5). To prove

(6) we consider the intersection R =
⋂n

i=1
RG(Gi). If R 6= 1, then R contains a minimal normal



CUBO
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On the Structure of Primitive n-Sum Groups 201

subgroup N of G. By (2) and (5), N is non-Frattini and G-equivalent to Gi for some 1 ≤ i ≤ n.

Hence by Proposition 4 (5), RG(Gi)N 6= RG(Gi), in contradiction with N ≤ R ≤ RG(Gi).

Definition 15. Let X be a primitive monolithic group and let N be its socle. For any non-empty

union Ω =
⋃

i
ωiN of cosets of N in X with the property that 〈Ω〉 = X, define σΩ(X) to be the

minimum number of supplements of N in G needed to cover Ω. Then we define

σ
∗(X) = min

{

σΩ(X) | Ω =
⋃

i

ωiN, 〈Ω〉 = X

}

.

Proposition 16. Let G be a non-abelian σ-primitive group, G1, . . . , Gn the minimal normal

subgroups, and X1, . . . Xn the monolithic primitive groups associated to Gi, i = 1, . . . n. Then

σ(G) ≥
∑n

i=1
σ∗(Xi).

Proof. Let M be a set of σ = σ(G) maximal subgroups whose union is G. Define M¬Gi = {M ∈

M | M 6≥ Gi}; note that

• M¬Gi 6= ∅ for each 1 ≤ i ≤ n; otherwise every maximal subgroup of M would contain Gi

and the set {M/Gi | M ∈ M} would cover G/Gi with σ(G) < σ(G/Gi) subgroups.

• M¬Gi ∩M¬Gj = ∅ for i 6= j; indeed if there exists M ∈ M¬Gi ∩M¬Gj , then GiMG/MG and

Gj MG/MG are minimal normal subgroups of the primitive group G/MG, hence δG(Gi) ≥ 2,

contrary to Corollary 14.

Therefore M contains the disjoint union of the non-empty sets M¬Gi , 1 ≤ i ≤ n, and we are

reduced to prove that |M¬Gi| ≥ σ
∗(Xi), for every i. Let us fix an index i and let π : G 7→ X be

the projection of G over X = Xi. We set N = soc X ∼= Gi, Mi = {M ∈ M | M ≥ Gi} = M\M¬Gi
and

Ω =

{

π(g) | g ∈ G \
⋃

M∈Mi

M

}

.

By minimality of the cover M, G 6=
⋃

M∈Mi
M hence Ω 6= ∅. Moreover, as Gi ≤ M ∈ Mi and

π(Gi) = soc X = N , we get that for every x ∈ Ω the coset xN is contained in Ω. If 〈Ω〉 = H 6= X,

then G is covered by the set Mi ∪ {π
−1(H)} and this actually is a minimal cover of G, since

|Mi| + 1 ≤ σ. But then, as π
−1(H) ≥ Gi, we would have σ(G/Gi) ≤ |Mi| + 1 = σ(G), a

contradiction. Hence 〈Ω〉 = X.

Now we shall prove that |M¬Gi| ≥ σΩ(X) ≥ σ
∗(X). By [9, Proposition 11] the kernel

R = RG(Gi) of the projection πi of G over X has the property that if H is a proper subgroup of

G such that HGi = G then HR 6= G. Therefore every maximal subgroup M ∈ M¬Gi contains

R, M = π−1(π(M )) and π(M ) is a a maximal subgroup of X supplementing N. Clearly, as
⋃

M∈M¬G
i

M covers G \
⋃

M∈Mi
M , we have that

⋃

M∈M¬G
i

π(M ) covers Ω. Therefore |{π(M ) |

M ∈ M¬Gi}| = |M¬Gi| ≥ σΩ(X) ≥ σ
∗(X).



202 Eloisa Detomi and Andrea Lucchini CUBO
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Remark 17. For every primitive monolithic group Xi, σ
∗(Xi) ≤ lXi (soc(Xi)), where lXi (soc(Xi))

is the smallest index of a proper subgroup of Xi supplementing soc(Xi). Indeed, if a supplement

of Ni = soc(Xi) in Xi has non trivial intersection with a coset gNi, then |gNi ∩ M| = |Ni ∩

M| = |gNi|/|G : M|, and therefore we need at least lXi (soc(Xi)) supplements to cover gNi. So in

particular the previous proposition implies that σ(G) ≥
∑n

i=1
lXi (Ni).

Lemma 18. Let N be a normal subgroup of a group X. If a set of subgroups covers a coset yN

of N in X, then it also covers every coset yαN with α prime to |y|.

Proof. Let s be an integer such that sα ≡ 1 mod |y|. As s is prime to |y|, by a celebrated result

of Dirichlet, there exists infinitely many primes in the arithmetic progression {s + |y|n | n ∈ N};

we choose a prime p > |X| in {s + |y|n | n ∈ N}. Clearly, yp = ys. As p is prime to |X|, there

exists an integer r such that pr ≡ 1 mod |X|. Hence, if yN ⊆ ∪i∈I Mi, for every g ∈ y
αN we have

that gp ∈ (yα)pN = (yα)sN = yN ⊆ ∪i∈I Mi and therefore also g = (g
p)r belongs to ∪i∈I Mi.

Corollary 19. Let G be a non-abelian σ-primitive group, N a minimal normal subgroup and X

the monolithic primitive groups associated to N . Then:

1. if X = N , then G = N ;

2. if |X/N| is a prime, then G = X.

Proof. Note that if X = N , then there is only one coset of N in X hence Ω = N , σ∗(N ) =

σN (N ) = σ(N ). By Proposition 16, σ
∗(N ) = σ(N ) ≤ σ(G). As N = X is a homomorphic image

of G, we get G = N .

Now let |X/N| be a prime. Let Ω be a non-empty union of cosets of N in X with the property

that 〈Ω〉 = X; then Ω contains a coset yN which is a generator for X/N . By Lemma 18 we have

that if
⋃

i
Mi covers Ω, then

⋃

i
Mi covers every coset of N with the exception, at most, of the

subgroup N itself. Hence, σ(X) ≤ σΩ(X) + 1 that is σ
∗(X) ≥ σ(X) − 1. By Proposition 16,

σ(G) ≥
∑n

i=1
σ∗(Xi). Moreover, by Remark 16, σ

∗(Xi) ≥ 2. Therefore, as σ(G) ≤ σ(X), there is

no room for another minimal normal subgroup in G.

Corollary 20. If N = Alt(n), n 6= 6, is a normal subgroup of G, then either σ(G) = σ(G/N ) or

G ∈ {Sym(n), Alt(n)}.

Proof. It is sufficient to consider a σ-primitive image of G and then apply Corollary 19.

Actually, the corollary holds also for n = 6, thanks to the following proposition.

Proposition 21. Let G be a σ-primitive group and let O∞(G) be the smallest normal subgroup of

G such that G/ O∞(G) is solvable. If G is non solvable, then G/ O∞(G) is a cyclic group.



CUBO
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On the Structure of Primitive n-Sum Groups 203

Proof. By Corollary 14, G is a subdirect product of the monolithic primitive groups Xi associated

to the minimal normal subgroups Gi, 1 ≤ i ≤ n; call Ni = soc(Xi) ∼= Gi. Let M be a set of

σ = σ(G) maximal subgroups whose union is G and define M¬Gi = {M ∈ M | M 6≥ Gi}. Let mi

be the minimal index of a supplement of Ni in Xi: by Remark 17, σ(G) ≥
∑n

i=1
mi.

Let R = O∞(G) and assume by contradiction that G/R is not cyclic. Then, by Tomkinson’s

result [14], σ(G/R) = q + 1 where q is the order of the smallest chief factor A = H/K of G/R

having more than a complement in G/R. As G is not solvable, then σ(G) < σ(G/R) = q + 1.

Since G is the subdirect product of the Xi’s, without loss of generality we can assume that A is a

chief factor of X = X1.

If N = soc(X) is an elementary abelian p-group, then, by Corollary 6 and Corollary 14 (1),

|N| + 1 ≤ σ(G) < q + 1. Therefore |N| < q and A is a chief factor, say U/V , of an irreducible

linear group X/N ≤ GL(N ) acting on N . By Clifford Theorem, U is a completely reducible linear

group hence Op(U ) = 1. Then, by Theorem 3 in [4], |U/U
′| < |N| < q, against |A| = |U/V | = q.

Assume now that N = S is a simple non-abelian group. Then A is isomorphic to a chief

factor of a subgroup of Out(S) hence q = |A| ≤ | Out(S)| < m1 (see e.g. Lemma 2.7 [4]). But

σ(G) ≥
∑n

i=1
mi ≥ m1 > q, against σ(G) < q + 1.

We are left with the case N = Sr where S is a simple non-abelian group. Then X/N is

isomorphic to a subgroup of Out(S) ≀ Sym(r). If A is isomorphic to a chief factor of a transitive

subgroup of Sym(r), then Theorem 2 in [4] gives that q = |A| ≤ 2r < (n1)
r ≤ m1, where n1 is

the minimal index of a subgroup of S. But this contradicts m1 ≤ σ(G) ≤ q. Therefore A has to

be a chief factor of a subgroup of Out(S)r. Then q = |A| ≤ | Out(S)|r ≤ nr
1
≤ m1 gives the final

contradiction.

Lemma 22. Let G be a non-solvable transitive permutation group of degree n. Then either σ(G) ≤

4n or every non-abelian composition factor of G is isomorphic to an alternating group of odd degree.

Proof. Let G be a non-solvable transitive permutation group of degree n. We can embed G into a

wreath product of its primitive components, let say G ≤ K1 ≀ K2 ≀ · · · ≀ Kt where Ki is a primitive

permutation group of degree ni and n1n2 · · · nt = n (see for example [7]). Let Kj be a non-solvable

component and assume that Kj is not an alternating or symmetric group of odd degree; then G

has an homomorphic image G which is embedded in a wreath product K ≀ H where K = Kj is a

permutation group of degree a = nj and H has degree b with ab ≤ n. If K does not contain Alt(a)

then |K| ≤ 4a [13] and G has a non-solvable normal subgroup of order at most 4ab. By Proposition

7 this implies that σ(G) ≤ σ(G) ≤ 4ab ≤ 4n. So assume that K contains Alt(a) where a is even.

We identify G with its image in K ≀ H: G is a transitive group of degree ab, with a system of

imprimitivity B with blocks of size a and K is the permutation group induced on a block by its

stabilizer. Let M1 be the set of subgroups G ∩ M where M is a maximal intransitive subgroup of

Sym(ab) and let M2 be the set of subgroups G∩(M ≀H) where M ∼= Sym(a/2)≀Sym(2) is a maximal

imprimitive subgroup of Sym(a); if T ∈ M2 and B ∈ B, then the permutation group induced on

B by the stabilizer TB is isomorphic to the imprimitive proper subgroup Sym(a/2) ≀ Sym(2) of K,



204 Eloisa Detomi and Andrea Lucchini CUBO
10, 3 (2008)

hence T is a proper subgroup of G. Now let x ∈ G: if x is not a cycle of length ab then there exists

T ∈ M1 containing x; otherwise there exists T ∈ M2 containing x. Hence the set M1 ∪M2 covers

G with
ab/2
∑

i=1

(

ab

i

)

+
1

2

(

a

a/2

)

≤ 2ab ≤ 2n

proper subgroups. Therefore σ(G) ≤ 2n.

Proposition 23. Let G be a σ-primitive group with a non-abelian minimal normal subgroup N .

If G/N CG(N ) is not cyclic, then all the non-abelian composition factors of G/N CG(N ) are alter-

nating groups of odd degree.

Proof. Let N = Sr, where S is a non-abelian simple group. By Corollary 6, 5r + 1 ≤ σ(G).

Denote by X the monolithic primitive group associated to the G-group N ; then X is a subgroup

of Aut(S) ≀ Sym(r). Let K be the image of X in Sym(r). If K is solvable, then, by Schreier

Conjecture, X/ soc(X) ∼= G/N CG(N ) is solvable. By Proposition 21 it follows that G/N CG(N ) is

cyclic.

Thus, if G/N CG(N ) is not cyclic, then K is non-solvable. Since 5
r + 1 ≤ σ(G) ≤ σ(K), the

previous lemma implies that every non-abelian composition factor of K is an alternating group of

odd degree. Then, by Schreier Conjecture, the same holds for G/N CG(N ).

Theorem 24. Let G be a σ-primitive group with no abelian minimal normal subgroups. Then

either G is a primitive monolithic group and G/ soc(G) is cyclic, or G/ soc(G) is non-solvable and

all the non-abelian composition factors of G/ soc(G) are alternating groups of odd degree.

Proof. By Corollary 14, G is a subdirect product of the monolithic primitive groups Xi associated

to the minimal normal subgroups Gi, 1 ≤ i ≤ n. By Proposition 23 and Proposition 21, for every i,

G/GiCG(Gi) ∼= Xi/ soc(Xi) is either cyclic or non-solvable and all of its non-abelian composition

factors are alternating groups of odd degree. Therefore either G/ soc(G) is solvable (hence cyclic

by Proposition 21) or non-solvable and all of its non-abelian composition factors are alternating

groups of odd degree.

We are left to prove that if G/ soc(G) is cyclic then n = 1. Assume by contradiction that

n ≥ 2.

Let ui be the number of distinct prime divisors of the order of the cyclic groups Xi/ soc(Xi)

and assume that u1 ≤ · · · ≤ un.

Step 1. Let mi be the minimal index of a supplement of soc(Xi) in Xi; then mi ≥ ui

If soc(Xi) = S is a simple group, then Xi/S is isomorphic to a subgroup of Out(S), and thus

ui ≤ 2
ui ≤ | Out(S)| ≤ mi (see e.g. Lemma 2.7 [4]).

If soc(Xi) = S
r where r 6= 1, then Xi/ soc(Xi) is isomorphic to a subgroup Y of Out(S) ≀

Sym(r). Let K be the intersection of Y with the base subgroup (Out(S))r of the wreath product



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On the Structure of Primitive n-Sum Groups 205

Out(S) ≀ Sym(r) and let a be the number of distinct prime divisors of |K|; since |K| divides

| Out(S)|r, we get that 2a ≤ | Out(S)| ≤ ni where ni is the minimal index of a subgroup of

S. Now b = ui − a is smaller or equal than the number of distinct prime divisors of the order

of Y /K which is isomorphic to a non trivial subgroup of Sym(r), hence 1 ≤ b < r and thus

ui = a + b ≤ (2
a)b ≤ (2a)r ≤ (ni)

r ≤ mi whenever a > 0. If a = 0, then Xi/ soc(Xi) is isomorphic

to a subgroup of Sym(r) and thus ui < r ≤ (ni)
r ≤ mi. This proves the first step.

Let π be the projection of G over X = X1 and call N = soc X. Note that there exist precisely

u1 maximal subgroups of the cyclic group X/N ; let H1, . . . , Hu1 be the maximal subgroups of G

such that their images in X/N give all the maximal subgroups of X/N .

Let M be a set of σ = σ(G) maximal subgroups whose union is G and define A to be the set

of maximal subgroups of M containing G1, B = M \ A and

Ω =

{

π1(g) | g ∈ G \
⋃

M∈A

M

}

.

Step 2. Assume that Ω contains a coset yN such that 〈yN〉 = X/N .

By Lemma 18, if Ω is covered by σΩ(X) maximal subgroups, then the same subgroups cover

every coset yαN with α prime to |y|. All the other elements of X are covered by the u1 maximal

subgroups π(H1), . . . , π(Hu1 ), since the images of these elements are not generators of X/N . Then

σ(X) ≤ σΩ(X)+u1. On the other hand, by Proposition 16, σΩ(X)+
∑

i6=1
σ∗(Xi) ≤ σ(G) < σ(X),

hence
∑

i6=1
σ∗(Xi) < u1. Remark 17 and Step 1 give that

∑

i6=1
ui ≤

∑

i6=1
mi < u1, and this

contradicts the minimality of u1.

Step 3. Assume that Ω does not contain a coset yN such that 〈yN〉 = X/N .

Then Ω is covered by the images in X of the subgroups H1, . . . , Hu1 and thus, by definition of

Ω, G is covered by the subgroups in A and H1, . . . , Hu1 . It follows that |B|+|A| = σ(G) ≤ u1 +|A|,

hence, by Step 1, |B| ≤ u1 ≤ m1, against Lemma 3.2 in [14]. This final contradiction implies that

G has to be a primitive monolithic group and proves the proposition.

4 There is no group for which σ(G) = 11

In this section we will show that σ(G) can never be equal to 11. The first trivial observation is

that σ(G) 6= 11 whenever G is solvable, since in this case by Tomkinson’s result σ(G) = q + 1, for

a prime power q.

Assume by contradiction that there exists a primitive 11-sum group G. By Corollary 14,

soc(G) is the direct product of n non G-equivalent minimal normal subgroups G1, . . . , Gn, where

at most one of them is abelian.



206 Eloisa Detomi and Andrea Lucchini CUBO
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Lemma 25. Suppose that G is a primitive 11-sum group. Then G has no abelian minimal normal

subgroups.

Proof. Assume by contradiction that G1 is abelian. By Corollary 14, G1 is a complemented non-

central factor of G, hence, by Corollary 6, |G1| + 1 ≤ σ(G) = 11. Moreover, by Proposition 10,

11 = σ(G) < 2|G1|. Hence |G1| can only be 7, 2
3 or 32. Actually, if |G1| = 7, then the bound in

Proposition 10 gives σ(G) ≤ 1 + 7, against σ(G) = 11.

Note that, by Proposition 16, σ(G) = 11 ≥
∑n

i=1
σ∗(Xi) where Xi are the monolithic groups

associated to the Gi’s; since G1 is the only abelian subgroup and σ
∗(Xi) ≥ 5 if Gi is non-abelian,

then G1 is the unique minimal normal subgroup of G and G ≤ G1 ⋊ Aut(G1).

If |G1| = 9, then G ≤ F
2

3
⋊ GL(2, 3); hence G is solvable, a contradiction.

Thus |G1| = 8 and G = F
3

2
⋊ GL(3, 2), since every proper subgroup of GL(3, 2) is solvable.

Let M = {M1, · · · , M11} be a set of 11 maximal subgroups covering G. In [6] it is proved that

σ(GL(3, 2)) = 15 and, in particular, that one needs at least 7 subgroups to cover the seven point

stabilizers of GL(3, 2). It follows that all the 8 complement of G1 in G occur in M, let say they

are M1, . . . , M8. As in the proof of Proposition 10, for every point stabilizer g ∈ GL(3, 2) there

exists an element vg ∈ G1 such that gvg does not belong to any complement of G1 in G. Hence

the remaining subgroups M9, M10, M11 of M have to cover all the elements gvv where g is a point

stabilizer. Since M9, M10 and M11 contain G1, this would imply that we can cover the seven point

stabilizers of GL(3, 2) with only three subgroups, a contradiction.

Theorem 26. There is no group G with σ(G) = 11.

Proof. Suppose that G is a primitive 11-sum group and let G1, . . . , Gn be its minimal normal

subgroups. By the previous lemma every Gi is non-abelian. If Gi = Alt(5) for some i, then,

by Corollary 20, G = Alt(5) or Sym(5). Otherwise, σ∗(Xi) ≥ lXi (Gi) > 5 for every i and

Proposition 16 implies that there is at most one minimal normal subgroup in G. By the same

argument, if G1 = S
r, where S is a simple non-abelian group, since lX1 (G1) ≥ 5

r and, by Lemma

5, 5r +1 ≤ σ(G) = 11, we have that G1 = S and lX1 (G1)+1 ≤ 11. Therefore G is an almost-simple

group with socle S and lG(S) ≤ 10, in particular

S ∈ {Alt(n) | 5 ≤ n ≤ 10} ∪ {Sym(n) | 5 ≤ n ≤ 10} ∪ {PSL(2, q) | 7 ≤ q ≤ 8}.

Thanks to the works of Maroti [12] and Bryce et al. [6], we can exclude most of these cases: indeed

σ(Alt(n)) ≥ 2n−2 if n 6= 7, 9, σ(Alt(5)) = 10, σ(Alt(9)) ≥ 80, σ(Sym(n) = 2n−1 if n is odd and

n 6= 9, σ(Sym(9)) ≥ 172, σ(PSL(2, 7)) = 15, σ(PGL(2, 7)) = 29, σ(PSL(2, 8)) = 36. Moreover,

σ(Aut(Alt(6))) ≤ σ(C2 × C2) = 3 and σ(Sym(6)) = 13 (see e.g. [1]). The remaining cases are

G = Alt(7), Sym(8), Sym(10), M10, PGL(2, 9) and Aut(PSL(2, 8)).

• G 6= Alt(7). Assume by contradiction σ(Alt(7)) = 11. There are seven maximal subgroups of

Alt(7) isomorphic to Alt(6); since σ(Alt(6)) = 16 > 11, each of them has to appear in a minimal

cover of G. Moreover, there are two conjugacy classes with 15 maximal subgroups isomorphic to



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On the Structure of Primitive n-Sum Groups 207

PSL(3, 2) and since σ(PSL(3, 2)) = σ(PSL(2, 7)) = 15 > 11 we have that σ(Alt(7)) is at least

7 + 15 + 15.

• G 6= Sym(8). If σ(Sym(8)) ≤ 11 then, since σ(Sym(7)) = 26 and σ(Alt(8)) ≥ 26, arguing as

in the previous case we get that a minimal cover M of Sym(8) contains the 8 point stabilizers

and Alt(8). Let g1 = (1, 2, 3, 4, 5, 6, 7, 8), g2 = (1, 2, 3, 7, 4, 5, 6, 8) and g3 = (1, 2, 3, 5, 4, 6, 7, 8);

any couple of them generate Sym(8) so that we need at least 3 more subgroups in M, and thus

σ(Sym(8)) > 11.

• G 6= Sym(10). If σ(Sym(10)) ≤ 11, then, as σ(Sym(9)) = 28 and σ(Alt(10)) ≥ 28, a minimal

cover M of Sym(10) contains 10 point stabilizers and Alt(10). But these subgroups do not cover

the 10-cycles. Thus σ(Sym(10)) > 11.

• G 6= M10. In M10 there are 180 elements of order 8. The only maximal subgroups containing

elements of order 8 are the Sylow 2-subgroups and each of them contains 4 of these elements; thus

we need at least 180/4 = 45 subgroups to cover the elements of order 8.

• G 6= PGL(2, 9). In PGL(2, 9) there are 144 elements of order 10. The only maximal subgroups

containing elements of order 10 are the normalizers of the Sylow 5-subgroups and each of them

contains 4 of these elements; thus we need at least 144/4 = 36 subgroups to cover the elements of

order 10.

• G 6= Aut(PSL(2, 8)). In Aut(PSL(2, 8)) \ PSL(2, 8) there are 336 elements of order 9. The only

maximal subgroups containing elements of this kind are the normalizers of the Sylow 3-subgroups;

each of them contains 12 of these elements thus we need at least 336/12 = 28 subgroups to cover

Aut(PSL(2, 8)).

5 Direct products

Proposition 27. Let G = H1 × H2 be the direct product of two subgroups. Let Ni be the smallest

normal subgroup of Hi such that Hi/Ni is a direct product of simple groups. If H1/N1 and H2/N2

have at most one non-abelian simple group S in common and the multiplicity of S in H1/N1 is at

most one, then either σ(G) = min{σ(H1), σ(H2)}, or the cyclic group Cp is an epimorphic image

of both H1 and H2 and σ(G) = p + 1.

Proof. Let G be a counterexample with minimal order. We first prove that G is a σ-primitive group.

As Φ(G) = Φ(H1) × Φ(H2), we have Φ(G) = 1. Let N be a minimal normal subgroup of G and

assume by contradiction that σ(G) = σ(G/N ). If N ≤ H1, then, by minimality of |G|, we have that

either σ(G/N ) = σ(H1/N × H2) = min{σ(H1/N ), σ(H2)} ≥ min{σ(H1), σ(H2)} ≥ σ(G), and so

σ(G) = min{σ(H1), σ(H2)}, or Cp is a common factor of H1/N N1 and H2/N2, and σ(G/N ) = p+1;

in this case σ(G) = σ(G/N ) = p + 1. Now assume that N is not contained in H1 or H2. Then N

is a central minimal normal subgroup of G, N = Cp ∼= N1N/N1 ∼= N2N/N2 and G has a factor

group isomorphic to Cp × Cp; therefore σ(G) ≤ p + 1. On the other hand, N = N H2 ∩ H1 ∼= N is



208 Eloisa Detomi and Andrea Lucchini CUBO
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a central minimal normal subgroup of G contained in H1; by the previous case, σ(G) < σ(G/N ).

Since δG(N ) ≥ 2, N has at least |N| = p complements; hence, by Lemma 5, σ(G) ≥ p + 1 and

therefore σ(G) = p + 1. Thus a counterexample G with minimal order is a σ-primitive group.

If G is solvable, then either G ∼= C2p and σ(G) = p+1 or G is monolithic: the second possibility

cannot occur as G is the direct product of two non trivial normal subgroups. So from now on we

may assume that G is non solvable, and in particular, by Proposition 21, that H1/N1 and H2/N2

have no common abelian factor.

Now observe that if M is a maximal subgroup of G and M does not contain H1 and H2,

then G/MG is a primitive group with nontrivial normal subgroups H1MG/MG and H2MG/MG. If

H1MG/MG = H2MG/MG, then G/MG = H1MG/MG = H2MG/MG is a central factor of G/MG

and H1/N1 and H2/N2 have a common abelian factor, a contradiction. Thus H1MG/MG 6=

H2MG/MG, and since G/MG = H1MG/MG × H2MG/MG is a primitive group, H1MG/MG and

H2MG/MG are isomorphic simple groups. Therefore, if H1/N1 and H2/N2 have no simple groups

in common, then every maximal subgroup M of G contains either H1 or H2, and we obtain the

result arguing as in Lemma 4 of [8].

So, we assume that H1/N1 and H2/N2 have precisely one non-abelian simple group S in

common and the multiplicity of S in H1/N1 is one: let Ki ≥ Ni be the normal subgroups of

Hi such that H1/K1 = S and H2/K2 = S
n, being n the multiplicity of S in H2/N2 , and set

K = K1 × K2.

Let M be a minimal cover of G given by σ(G) maximal subgroups of G. We set:

M1 = {L ∈ M | L ≥ H1} = {H1 × M | M a maximal subgroup of H2},

M2 = {L ∈ M | L ≥ H2} = {M × H2 | M a maximal subgroup of H1},

M3 = M \ (M1 ∪ M2).

Then we define the two sets

Ω1 = H1 \
⋃

M×H2∈M2

M, Ω2 = H2 \
⋃

H1×M∈M1

M,

and their images under the projection πKi of Hi over Hi/Ki

Ωi = {πKi (w) | w ∈ Ωi}.

As H1/K1 is not cyclic, we can cover Ω1 with |Ω1| subgroups. Hence we can cover H1 =

{
⋃

M×H2∈M2
M} ∪ Ω1 with the images of the maximal subgroups in M2 plus |Ω1| maximal sub-

groups, and thus σ(H1) ≤ |M2| + |Ω1|. On the other hand, |M2| + |M3| ≤ σ(G) < σ(H1), and we

obtain that

|Ω1| > |M3|.

Now observe that the elements of the set Ω1 × Ω2 can not belong to any of the subgroup

of M1 or M2, thus the set Ω1 × Ω2 has to be covered by the subgroups of M3. If M ∈ M3,



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On the Structure of Primitive n-Sum Groups 209

then G/MG is a primitive group and G/MG = H1MG/MG × H2MG/MG = S × S; in particular

M ≥ K and M/K is a maximal subgroup of diagonal type of G/K. This means that there exists

an automorphism α of S and an index i ∈ {1, . . . n}, such that the set (M/K) ∩ (Ω1 × Ω2) is given

by elements of the type (x, y1, y2, . . . , yn) where x ∈ Ω1, (y1, y2, . . . , yn) ∈ Ω2 and yi = x
α. For

every y ∈ S we denote by sy the number of vectors (y1, y2, . . . , yn) such that (y1, y2, . . . , yn) ∈ Ω2

and yi = y: note that
∑

y∈S

sy = |Ω2| = |Ω1 × Ω2|/|Ω1|.

On the other hand

|(M/K) ∩ (Ω1 × Ω2)| ≤
∑

y∈S

sy = |Ω1 × Ω2|/|Ω1| < |Ω1 × Ω2|/|M3|,

since |Ω1| > |M3|. This implies that we can not cover Ω1 × Ω2 with the |M3| subgroups of M3, a

contradiction.

Theorem 28. Let G = H1 × H2 be the direct product of two subgroups. If no alternating group

Alt(n) with n odd is a homomorphic image of both H1 and H2, then either σ(G) = min{σ(H1), σ(H2)}

or σ(G) = p + 1 and S = Cp is a homomorphic image of both H1 and H2.

Proof. Let G be a counterexample with minimal order. Let Ni be the minimal normal subgroup

of Hi such that Hi/Ni is a direct product of simple groups. As in the proof of Proposition 27, it

is easy to see that G is a σ-primitive group, H1/N1 and H2/N2 have at least one simple group S

in common and S is non-abelian.

By Corollary 14, G has at most one abelian minimal normal subgroup, so we can assume that

every minimal normal subgroup of H1 is non-abelian.

Let K be a normal subgroup of G with G/K ∼= S. Note that δG(G/K) ≥ 2, indeed δG(G/K)

coincides with the multiplicity of S in G/(N1 ×N2). Hence, by Corollary 14 (3), no minimal normal

subgroup of G is G-equivalent to G/K. This implies in particular that S is an epimorphic image

of H1/ soc(H1), and consequently S is an homomorphic image of X/N where X is a monolithic

primitive group associated to a minimal normal subgroup N of H1. By the remark above N is non-

abelian, so N = T r with T a non-abelian simple group. Since X is a subgroup of Aut(T ) ≀ Sym(r)

and S is non-abelian, S is an homomorphic image of a transitive group Y of degree r. Then Y

satisfies the assumption of Lemma 22 and, since S is not an alternating group of odd degree, we

get σ(Y ) ≤ 4r. Since, by Corollary 6, 5r + 1 ≤ σ(G) ≤ σ(Y ), we get a contradiction.

Received: July 2008. Revised: August 2008.



210 Eloisa Detomi and Andrea Lucchini CUBO
10, 3 (2008)

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