non-normal cyclic subgroups

Note Mat. 30 (2010) no. 2, 121–133. doi:10.1285/i15900932v30n2p121

Groups with few

non-normal cyclic subgroups

Davide Oggionni

davide oggionni@alice.it

Gianluca Ponzoni

gianluca.ponzoni@katamail.com

Vittoria Zambelli

Dipartimento di Matematica F. Enriques, Universit` a degli Studi di Milano vittoria.zambelli@unimi.it

Received: 11.10.2009; accepted: 5.7.2010.

Abstract. R. Brandl in [2] and H. Mousavi in [6] classified finite groups which have respec- tively just one or exactly two conjugacy classes of non-normal subgroups. In this paper we determine finite groups which have just one or exactly two conjugacy classes of non-normal cyclic subgropus. In particular, in a nilpotent group if all non-normal cyclic subgroups are con- jugate, then any two non-normal subgroups are conjugate. In general, if a group has exactly two conjugacy classes of non-normal cyclic subgroups, there is no upper bound for the number of conjugacy classes of non-normal subgroups.

Keywords: conjugacy classes, cyclic subgroup MSC 2000 classification: 20D25

1 Preliminaries

Notations

ν(G) will denote the number of conjugacy classes of non-normal subgroups of G.

ν

c

(G) will denote the number of conjugacy classes of non-normal cyclic subgroups of G.

G = [A]B will denote that G is the semidirect product of A and B, with A normal in G.

By A ∼

G

B, with A and B subgroups of G, we mean that A and B are conjugate in G.

Let P be a p-group: then Ω

i

(P ) = hx ∈ P |x

^pi

= 1i, ℧

i

(P ) = ha

^pi

|a ∈ P i.

All groups considered in this paper are finite.

As in [7] Propositions 2.2 and 2.6 one can prove the following Proposition 1. Let G be a group.

(1) If N is a normal subgroup of G, then ν

c G N

≤ ν

c

(G)

(2) If G = A × B, then ν

c

(G) ≥ ν

c

(A)ν

c

(B) + ν

c

(A)µ

c

(B) + µ

c

(A)ν

c

(B), where µ

c

(G) denotes the number of normal cyclic subgroups of G; if (|A|, |B|) = 1, then equality holds.

http://siba-ese.unisalento.it/ c 2010 Universit` a del Salento

Proposition 2. If P is a non-abelian p-group with |P | = p

ⁿ

and exp P = p

ⁿ⁻¹

, then ν

c

(P ) ≤ 2.

It is ν

c

(P ) = 1 if and only if P ≃ M

n

(p) = ha, b|a

^pn−1

= b

^p

= 1, [a, b] = a

^pn−2

i where n ≥ 4 if p = 2. Moreover, ν(P ) = 1.

It is ν

c

(P ) = 2 if and only if p = 2 and P is isomorphic to one of the following groups:

i. D

n

= ha, b|a

²ⁿ⁻¹

= b

²

= 1, [a, b] = a

⁻²

i where n ≥ 3; the non-normal cyclic subgroups of D

n

have order 2. It is ν(D

n

) = 2n − 4.

ii. S

n

= ha, b|a

²ⁿ⁻¹

= b

²

= 1, [a, b] = a

^−2+2n−2

i where n ≥ 4; the non-normal cyclic sub- groups of S

n

have order 2 or 4. It is ν(S

n

) = 2n − 5.

iii. Q

n

= ha, b|a

²ⁿ⁻¹

= 1, a

²ⁿ⁻²

= b

²

, [a, b] = a

⁻²

i where n ≥ 4; the non-normal cyclic subgroups of Q

n

have order 4. It is ν(Q

n

) = 2n − 6.

The proof is a straightforward check on the groups Q

n

, S

n

and D

n

, bearing in mind [2]

and [7], Prop.2.5.

Proposition 3. Let P be a p-group;

(1) if there exists a subgroup N contained in Z(P ) such that ν

_N^P

= 1, then [P : Z(P )] = p

²

and |P

^′

| = p;

(2) if ν 

P Z(P )



= 0, then either

_{Z(P )}^P

is abelian or ν

c

(P ) ≥ 3.

Proof. (1) By [1] it is

_N^P

≃ M

^r

(p), where r ≥ 3 if p odd, r ≥ 4 if p = 2; so

_N^P

has two cyclic maximal subgroups and therefore P has two abelian maximal subgroups. It follows [P : Z(P )] = p

²

. Since P has an abelian maximal subgroup, one has |P

^′

||Z(P )| = p

ⁿ⁻¹

, so |P

^′

| = p.

(2) if

_{Z(P )}^P

is non-abelian, then P

Z(P ) = h¯ x, ¯ y|¯ x

⁴

= 1, ¯ x

²

= ¯ y

²

= [¯ x, ¯ y]i × ¯ E

where ¯ E is an elementary abelian 2-group. The subgroups hxi, hyi and hxyi represent three different conjugacy classes in P ; hence, if ν

c

(P ) ≤ 2, we can assume hxi ⊳ P . Then |x| ≥ 2

³

; since [x, y] ∈ hxi, it is [x, y]

²

= [x

²

, y] = 1, so [x, y] ∈ hx

⁴

i ⊆ Z(P ), a contradiction.

QED

A check on minimal non-abelian groups (see [5]) proves the following Proposition 4. Let G be a minimal non-abelian group; then (1) ν

c

(G) = 1 if and only if ν(G) = 1;

(2) if G is non-nilpotent (so that G = [Q]P ≃ G(q, p, n) with |Q| = q

^m

, |P | = p

ⁿ

), one has ν

c

(G) = 2 if and only if n = 1, m ≥ 2 and p =

^qm_q−1⁻¹

;

(3) if G is a p-group, one has ν

c

(G) = 2 if and only if ν(G) = 2; hence either G = [C

n

]C

4

= [hxi] hyi with [x, y

²

] = 1 or G = [C

4

]C

2

≃ D

3

.

Later we shall use the following Proposition on power-automorphisms. An automorphism φ of a group G is said to be a power-automorphism if φ(H) = H for every subgroup H of G.

Proposition 5. ([4], Hilfsatz 5) Let P be a p-group and α 6= 1 a p

^′

-power automorphism

of P . Then P is abelian and for every a ∈ P it is α(a) = a

^k

where k ∈ Z does not depend on

a.

2 ν _c = 1

Theorem 1. Let G be a non-nilpotent group; one has ν

c

(G) = 1 if and only if G = [N ]P , where N is an abelian group of odd order, P ∈ Syl

p

(G) is cyclic and P induces on N a group of fixed-point-free power-automorphisms of order p.

Proof. Assume ν

c

(G) = 1 and P ∈ Syl

p

(G) with P ⋪ G; let s ∈ P be such that hsi ⋪ G.

For each prime q 6= p and for each q-element a ∈ G we have hai ⊳ G and therefore for Q ∈ Syl

q

(G) it is Q ⊳ G. Furthermore [a, s] 6= 1 and q 6= 2, so s induces a fixed-point-free automorphism on Q; by Proposition 5 the subgroup Q is abelian and s induces a power- automorphism on Q.

Consequently G = [N ]P with N abelian and C

N

(s) = h1i.

For every b ∈ P \ C

^P

(N ) it is hbi ⋪ G so that hbi = hs

^g

i for some g ∈ G; if g = ny with n ∈ N, y ∈ P and b = g

⁻¹

s

^l

g for some l ∈ N, then b = [y, s

^−l

]s

^l

∈ P

^′

hsi. Since P 6= C

P

(N ), it is P = P

^′

hsi = hsi.

Vice versa, since s induces on N a fixed-point-free automorphism of order p, the conjugates

of P are the only non-normal cyclic subgroups of G.

^QED

Lemma 1. Let G be a nilpotent group with ν

c

(G) = 1; then (1) G is a p-group with |Ω

1

(G)| ≥ p

²

;

(2) if |G

^′

| = p, it is |Ω

1

(G)| = p

²

and Φ(G) ⊆ Z(G);

(3) if p = 2 and G

^′

⊆ Z(G), for any a ∈ G\Z(G) with |a| = 2 it is Ω

2

(G) ⊆ C

G

(a) and exp G ≥ 2

³

Proof. (1) It follows from Propositions 1 and 2.

(2) Consider hai ⋪ G.

If |a| = p, then for any b ∈ G with |b| = p we have either hbi ⋪ G or habi ⋪ G, so that respectively either hbi or habi is conjugate to hai; it follows b ∈ haiG

^′

. Therefore Ω

1

(G) ⊆ hai × G

^′

and |Ω

1

(G)| = p

²

.

If |a| = p

^r

with r ≥ 2, then Ω

1

(G) ⊆ Z(G). If it were |Ω

1

(G)| ≥ p

³

, then for any c ∈ G \ haiG

^′

such that |c| = p it would be haci ≁

G

hai and so haci ⊳ G. If g / ∈ N

G

(hai), then 1 6= [a, g] = [ac, g] ∈ haci ∩ G

^′

and therefore G

^′

⊆ Ω

¹

(haci) ⊆ hai, a contradiction.

For any a, b ∈ G it is 1 = [a, b]

^p

= [a

^p

, b] and so Φ(G) = G

^′

℧

1

(G) ⊆ Z(G).

(3) As above we prove Ω

1

(G) ⊆ haiG

^′

. If b ∈ G is such that |b| = 4, then hbi ⊳ G; if it were [a, b] 6= 1, it would be (ab)

²

= 1 and ab ∈ Ω

1

(G) ⊆ C

G

(a), a contradiction. So Ω

2

(G) ⊆ C

G

(a) and exp G ≥ 2

³

.

QED

Theorem 2. Let P be a p-group; it is ν

c

(P ) = 1 (if and) only if it is ν(P ) = 1, that means P ≃ M

n

(p) = ha, b|a

^pn−1

= b

^p

= 1, [a, b] = a

^pn−2

i, where n ≥ 4 if p = 2.

Proof. Suppose ν

c

(P ) = 1 and let P be a minimal counterexample with |P | = p

ⁿ

; by Propo- sition 2 it is exp P ≤ p

ⁿ⁻²

.

Let us consider first the case p 6= 2.

Let N be a minimal normal subgroup of P ; then ν

c P N

≤ ν

c

(P ) = 1.

If ν

c P N

= 0, then

^P_N

is abelian and P

^′

= N has order p. If ν

c P N

= 1, the minimality of P implies ν

^P_N

= 1. By Proposition 3 it is again |P

^′

| = p.

By Lemma 1,2) it is |Ω

¹

(P )| = p

²

; as P is regular, |℧

¹

(P )| = p

ⁿ⁻²

and φ(P ) = ℧

1

(P ) = Z(P ). Then P is a minimal non-abelian group; this contradicts Proposition 4.

Let us suppose now p = 2.

Step 1: P

^′

⊆ Z(P )

By Proposition 3 it is ν 

P Z(P )

 6= 1 and therefore ν

c



P Z(P )

 6= 1; then ν

c



P Z(P )

 =

ν 

P Z(P )

 = 0 and

_{Z(P )}^P

is abelian.

Step 2: Ω

1

(P ) * Z(P )

Suppose Ω

1

(P ) ⊆ Z(P ). Since ν

c

(Q

n

) = 2, P has at least two different minimal normal subgroups N

1

and N

2

and we may suppose

_N^P

1

and

_N^P

2

non-abelian.

If ν

c



P N1



= 1, then ν 

P N1



= 1; by Proposition 3 one has |P

^′

| = 2 and P

N

1

= ha, b|a

²ⁿ⁻²

= b

²

= 1, [a, b] = a

²ⁿ⁻³

i where n ≥ 5.

Then hbi ⋪ P so that |b| = 4 and N

1

⊆ hbi; since exp P ≤ 2

ⁿ⁻²

, it is |a| = 2

ⁿ⁻²

≥ 2

³

and therefore hai ⊳ P . So P = ha, b|a

²ⁿ⁻²

= b

⁴

= 1, [a, b] = a

²ⁿ⁻³

i would be a minimal non-abelian group; this contradicts Proposition 4.

So it must be ν

c



P N1



= ν

c



P N2



= 0 and

_N^P

1

≃

_N^P₂

≃ Q

3

× E, where E is elementary abelian; it follows exp P = 4. Let

P

N

1

= hx, y|x

⁴

= 1, x

²

= y

²

= [x, y]i × ×

ⁿ⁻⁴i=1

he

i

i

Any two of the subgroups hxi, hyi, hxyi are not conjugate; we may suppose hxi ⊳ P , hyi ⊳ P and hx, yi ≃ Q

3

.

Since P is non-hamiltonian, we may suppose |e

¹

| = 4, N ⊆ he

¹

i ⊳ P and [x, e

¹

] = [y, e

1

] = 1; it follows |xe

1

| = 4 and [xe

1

, y] = [x, y] = x

²

∈ hxe /

1

i. Therefore hxe

1

i ⋪ P and hye

1

i ⋪ P ; since ν

c

(P ) = 1, then ye

1

∈ hxe

1

iP

^′

⊆ hx, e

1

i, a contradiction.

Step 3: |Z(P )| ≤ 2

ⁿ⁻³

If [P : Z(P )] = 4, then |P

^′

| = 2 and |Ω

1

(P )| = 4 by Lemma 1,2).

Since Ω

1

(P ) * Z(P ), we have |Ω

¹

(Z(P ))| = 2 and Z(P ) = hzi is cyclic; furthermore there exists a ∈ P such that |a| = 2 and hai ⋪ P . By Proposition 4 there exists a non-abelian maximal subgroup of P and φ(P ) ⊆ hz

²

i. Let x ∈ P \ ha, zi; it is x

²

= z

^2r

so that xz

^−r

∈ Ω

1

(P ) ⊆ ha, zi, a contradiction.

Step 4: Z(P ) is cyclic.

If Z(P ) were not cyclic, there would be at least two minimal normal subgroups N

1

and N

2

with

_N^P

1

and

_N^P

2

non-abelian. By Proposition 3 one would have ν 

P N1

 6= 1 and

ν 

P N2

 6= 1; since P is a minimal counterexample, then ν

c



P N1

 = ν

c



P N2

 = 0. ¿From exp

_N^P

1

= exp

_N^P

2

= 4 it would follow exp P = 4; this contradicts Lemma 1,3).

Step 5: |P

^′

| = 2.

Let N be the only minimal normal subgroup of P ; one has ν

c P N

= 0 by Proposition 3. If it were N 6= P

^′

, it would be

P

N = hx, y|x

⁴

= 1, x

²

= y

²

= [x, y]i × (×

^mi=1

he

i

i) ;

then [x, y] ∈ hx

²

iN = hy

²

iN = P

^′

≃ C

4

and [x, y]

²

= [x

²

, y] = 1, a contradiction.

Conclusion : Since |P

^′

| = 2, for any a, b ∈ P it is [a

²

, b] = [a, b]

²

= 1 and ℧

1

(G) ⊆ Z(P ); from

|Z(P )| ≤ 2

ⁿ⁻³

it follows that every maximal subgroup of P is non-abelian.

Since Ω

1

(P ) * Z(P ), there exists an element a ∈ P such that hai ⋪ P , |a| = 2 and Ω

1

(P ) = hai × P

^′

⊆ C

P

(a).

For every b ∈ P \ C

^P

(a) it is [a, b] = c, where P

^′

= hci; then it follows [P : C

^P

(a)] = 2, P = hbiC

P

(a) and C

P

(a) 6= haiZ(P ).

Let d ∈ C

P

(a) \ haiZ(P ): from b

²

, d

²

∈ Z(P ) it follows either hb

²

i ⊆ hd

²

i or hd

²

i ⊆ hb

²

i.

If b

²

= d

^2r

, then (bd

^−r

)

⁴

= 1 and bd

^−r

∈ Ω

2

(P ) ⊆ C

P

(a) by Lemma 1,3); this contra- dicts b / ∈ C

P

(a).

If d

²

= b

^4s

, then db

^−2s

∈ Ω

1

(P ) ⊆ haiZ(P ), whence d ∈ haiZ(P ), a contradiction.

QED

3 ν _c = 2

3.1 Direct products

Proposition 6. If G is a direct product of proper subgroups and ν

c

(G) = 2, then G = A × B where ν

c

(A) = 1 and |B| = q for some prime q.

Moreover, if q divides |A|, then q = 2.

Proof. Suppose G = A × B with A and B proper subgroups of G. Proposition 1 implies either ν

c

(A) = 0 or ν

c

(B) = 0. Suppose ν

c

(B) = 0; then µ

c

(B) ≥ 2 and either ν

c

(A) = 0 or ν

c

(A) = 1.

If ν

c

(A) = 0, we may suppose A non-abelian, so that A ≃ Q

3

×E, where E is an elementary abelian 2-group. Since ν

c

(Q

3

× Q

³

) > 2, B is abelian and there exists b ∈ B such that |b| = 4.

Let A = hx, y|x

⁴

= 1, x

²

= y

²

= [x, y]i × E: the three subgroups hxbi, hybi and hxybi are non-normal and non-conjugate, a contradiction.

It must be ν

c

(A) = 1 and µ

c

(B) = 2. So B = hbi ≃ C

q

for some prime q.

Suppose q divides |A|. If A is non-nilpotent, then A = [N]P as in Theorem 1 and the subgroups P = hxi and hxbi are non-normal and non-conjugate. If q divides |N|, for any a ∈ N such that |a| = q we would have habi ⋪ G so that ν

c

(G) ≥ 3; therefore q = p.

In any case, if q 6= 2, there exists y ∈ A such that the subgroups hyi, hybi and hy

²

bi are non-normal and pairwise non-conjugate, so that ν

c

(G) ≥ 3.

^QED

Corollary 1. If G is nilpotent and it is not a p-group, then ν

c

(G) = 2 if and only if G ≃ M

ⁿ

(p) × C

^q

, where p, q are distinct primes and n ≥ 4 if p = 2.

3.2 p-groups with ν

c

= 2

Lemma 2. Let P be a non-abelian p-group; if exp P = p, then ν

c

(P ) ≥ 4.

Proof. Let |P | = p

ⁿ

. If P

^′

is the only minimal normal subgroup of P , P has

^p_p−1^n−p

non-normal cyclic subgroups of order p and each of them has p conjugates. Then ν

c

(P ) ≥

_(p−1)p^pn−p

≥ p + 1 ≥ 4.

Otherwise, if N is a minimal normal subgroup of P with N 6= P

^′

, by induction ν

c

(P ) ≥ ν

c P

N

≥ 4.

^QED

Proposition 7. If P is a p-group with ν

c

(P ) = 2, then p = 2.

Proof. Suppose p 6= 2, P a minimal counterexample. If |P | = p

ⁿ

, then n ≥ 4 and p

²

≤ exp P ≤ p

ⁿ⁻²

.

For every minimal normal subgroup N of P we have ν

c P N

≤ 1.

If P

^′

were the only minimal normal subgroup of P , one would have ℧

1

(P ) ⊆ Z(P ).

Therefore every non-normal cyclic subgroup would have order p and would have p conjugates;

moreover P would be regular and exp Ω

1

(P ) = p. Then Ω

1

(P ) would contain exactly 2p + 1 cyclic subgroups; a calculation on the order of Ω

1

(P ) shows that this is impossible.

So there exists a minimal normal subgroup N of P with ν

c P N

= 1; that means P

N = ha, b|a

^pn−2

= b

^p

= 1, [a, b] = a

^pn−3

i,

|P

^′

| = p, |Z(P )| = p

ⁿ⁻²

and Z(P ) = ha

^p

, N i with |a| = p

ⁿ⁻²

= exp P .

If |b| = p

²

, then Z(P ) = ha

^p

, b

^p

i = Φ(P ) and P would be a minimal non-abelian group;

this contradicts Proposition 4. So |b| = p.

If N = hyi, the subgroups hbi, hbyi and hb

²

yi are non-normal and pairwise non-conjugate,

a contradiction.

^QED

If G is not a Dedekind group, let R(G) denote the intersection of all non normal subgroups of G; the groups with R(G) 6= h1i are determined in [1].

Lemma 3. Let P be a non-Dedekind 2-group with R(P ) 6= h1i, |P | = 2

ⁿ

and exp P ≤ 2

ⁿ⁻²

. If ν

c

(P ) = 2, then P = [hai]hbi with |a| = 2

ⁿ⁻²

, |b| = 4.

Proof. It is ν

c

(Q

3

× C

4

) = 3 and ν

c

(Q

3

× Q

3

) = 9. Then by [1], Theorem 1 it is P = hA, x|A is abelian, x

⁴

= 1, 1 6= x

²

∈ A, [x, a] = a

²

for any a ∈ Ai.

For any y ∈ A it is (yx)

²

= x

²

. If hyxi ⊳ P , for any a ∈ A one would have a

²

= [yx, a] = x

²

, so Φ(A) = hx

²

i; it would follow A ≃ C

4

× E with Φ(E) = 1 and P ≃ Q

3

× E hamiltonian. So hyxi ⋪ P for every y ∈ A.

Since P

^′

= Φ(A) = ℧

1

(A), then for any y

1

, y

2

∈ A the subgroups hy

1

xi and hy

2

xi are conjugate if and only if y

₁⁻¹

y

2

∈ hx

²

iΦ(A), so [A : hx

²

iΦ(A)] = 2. If a ∈ A \ hx

²

Φ(A)i it is A = ha, x

²

, Φ(A)i = ha, x

²

i.

Since exp A ≤ exp P ≤ 2

ⁿ⁻²

, we have x

²

6∈ hai, hence P = [hai]hxi with |a| = 2

ⁿ⁻²

,

|x| = 4.

^QED

Remark 1. If P = [hai]hbi, with |a| = 2

ⁿ⁻²

, |b| = 4, [a, b

²

] = 1 6= [a, b] and n ≥ 4, then ν

c

(P ) = 2: if [a, b] = a

²ⁿ⁻³

, the non normal cyclic subgroups not conjugate to hbi are conjugate to ha

²ⁿ⁻⁴

bi, otherwise they are conjugate to habi.

Theorem 3. Let P be a non abelian 2-group, which is not a direct product of proper subgroups, with |P | = p

ⁿ

, exp P ≤ 2

ⁿ⁻²

. Then ν

c

(P ) = 2 if and only if P = [hai]hbi, with

|a| = 2

ⁿ⁻²

, |b| = 4, [a, b

²

] = 1 and n ≥ 4.

Proof. We see that if P = [hai]hbi with |a| = 2

ⁿ⁻²

, |b| = 4 and [a, b

²

] 6= 1, then b

⁻¹

ab = a

¹⁺²ⁿ⁻⁴

or b

⁻¹

ab = a

^−1+2n−4

; the subgroups hbi, hb

²

i and respectively ha

²ⁿ⁻³

bi in the first case and hbai in the second case are non normal and pairwise non-conjugate. So it will suffice to prove that, if ν

c

(P ) = 2, then P = [hai]hbi with |a| = 2

ⁿ⁻²

and |b| = 4.

First of all, note that for any minimal normal subgroup T of P we may suppose ν

c P T

6= 1.

Indeed, if it were

P

T = ha, b|a

²ⁿ⁻²

= b = 1, [a, b] = a

²ⁿ⁻⁵

i

with n ≥ 5, then P = ha, bi × T if |b| = 2 and P = [hai]hbi if |b| = 4.

By Lemma 3, we may suppose R(P ) = h1i. Let P be a minimal counterexample and let H = hhi, K = hki be two non-conjugate non-normal cyclic subgroups of P , with |H| ≤ |K|.

If H 6≤ K, since ν

c

(P ) = 2, then H ∩ K ⊳ P . It follows H ∩ K ≤ H

P

∩ K

P

= R(P ) = h1i.

Then either H ≤ K or H

^g

∩ K = h1i for every g ∈ P . If H ≤ K, then |H| = 2, |K| = 4 and h = k

²

.

Let T be a minimal normal subgroup of P. It is

^KT_T

⋪

^P_T

with

^KT_T

= 4 and ν

c P T

= 2.

If

_T^S

= hsT i ⋪

^PT

with

^S_T

non-conjugate to

^KT_T

, then hsi is conjugate to H and

^HTT

⋪

^P_T

. So

^P_T

has two non-normal cyclic subgroups

^HT_T

and

^KT_T

of order respectively 2 and 4, with

HT

T

≤

^KT_T

. This implies that

^P_T

is not a direct product and exp

^P_T

≤ 2

ⁿ⁻³

.

By minimality of P ,

^P_T

= [haT i]hbT i ≃ [C

2ⁿ⁻³

]C

4

with n ≥ 5. But hbi ⋪ P , so |b| = 4,

|a| = 2

ⁿ⁻³

≥ 4. Since hai is not conjugate to hbi, then hai ⊳ P and P = ha, bi × T , contrary to the hypothesis.

So H

^g

∩ K = h1i for every g ∈ P . We distinguish two cases.

Case 1: |K| ≥ 4.

Every proper subgroup of K is normal in P ; for T ≤ K with |T | = 2 one has ν

c P T

= 2.

If exp

^P_T

= 2

ⁿ⁻²

, then

^P_T

is isomorphic to D

_n−1

or S

_n−1

, because the non-normal cyclic subgroups of Q

n−1

have non-trivial intersection. Then

^P_T

has a non-normal cyclic subgroup

_T^S

= hsT i of order 2, and P = ha, s, T i with |a| = 2

ⁿ⁻²

. Since P is not a direct product, we have P = ha, si = [hai]hsi and |s| = 4, against our hypothesis.

Then exp

^P_T

≤ 2

ⁿ⁻³

.

If

^P_T

were a direct product of proper subgroups, we would have P

T = hx, y|x

²ⁿ⁻³

= y

²

= 1, [x, y] = x

²ⁿ⁻⁴

i × hz|z

²

= 1i

where n ≥ 6; we can suppose

^K_T

= hyi and

^HT_T

= hyzi and so K = hyi, H = hyzi with

|K| = 4, |H| = 2.

The subgroup hzi is conjugate neither to K nor to H, so hzi ⊳ P . Since P is not a direct product, it is |z| = 4. It follows y

²

= z

²

and [y, z] = 1, so P

^′

≤ hxi.

If it were T ≤ hxi, it would be T ≤ P

^′

= hx

²ⁿ⁻⁴

i ≤ hx

⁴

i and 1 = [x, y]

⁴

= [x

⁴

, y]; then 1 = [x, y

²

] = [x, y]

²

, so [x, y] ∈ T , a contradiction.

So T  hxi and [x, z] = 1; it would follow P = hx, yzi × hzi, a contradiction.

By the minimality of P one has

^P_T

= [haT i]hbT i ≃ [C

2ⁿ⁻³

]C

4

and [a, b

²

] ∈ T . By the previous Remark we may suppose K = hbi and |K| = 8, H = habi with |H| = 4. It follows hai ⊳ P , so [a, b

²

] ∈ hai ∩ T .

If T ≤ hai then P = [hai]habi ≃ [C

2ⁿ⁻²

]C

4

.

If T  hai, [a, b

²

] = 1 and P = [hai]hbi. Since hab

²

i ⊳

^P_T

, the subgroup hab

²

i is conjugate neither to K nor to H, hence hab

²

i ⊳ P , a contradiction.

Case 2: |K| = |H| = 2

Let T be a minimal normal subgroup of P . If ν

c P

T

= 2, the non normal cyclic subgroups of

^P_T

are conjugate either to

^HT_T

or to

KT

T

, so their order is 2. By the minimality of P one has

^P_T

≃ D

n−1

or

^P_T

≃ M

n−2

(2)×C

2

.

If

^P_T

= ha, b|a

²ⁿ⁻²

= b

²

= 1, [b, a] = a

²

i, then P = ha, bi × T .

Let

^P_T

= ha, b|a

²ⁿ⁻³

= b

²

= 1, [a, b] = a

²ⁿ⁻⁴

i × hc|c

²

= 1i with n ≥ 6. We can suppose K = hbi, H = hbci. Since hci is conjugate neither to hbi nor to hbci, it is hci ⊳ P If |c| = 2, then P = ha, b, T i × hci.

So |c| = 4 and T = hc

²

i; from 1 = (bc)

²

= c

²

[b, c] it follows [b, c] = c

²

. If T  ha, bi, we have [a, c] = 1 and |a| = 2

ⁿ⁻³

; thus   ca

²ⁿ⁻⁵

 

 = 4 and hca

²ⁿ⁻⁵

i ⊳ P , but [b, ca

²ⁿ⁻⁵

] = [b, a

²ⁿ⁻⁵

][b, c] = c

²

6∈ hca

²ⁿ⁻⁵

i.

If T ≤ ha, bi, then T ≤ hai and |a| = 2

ⁿ⁻²

. ¿From [a, c] ∈ T it follows [a, c

²

] = 1, so (a

²ⁿ⁻⁴

c)

²

= 1. Since [a

⁴

, b] = 1 and n ≥ 6, then [a

²ⁿ⁻⁴

c, b] = c

²

6= 1, so ha

²ⁿ⁻⁴

ci ⋪ P , but ha

²ⁿ⁻⁴

ci is conjugate neither to hai nor to hbci.

So ν

c P T

= 0 for every minimal normal subgroup T of P . Since HT ⊳P , then T = h[h, g]i for any g ∈ P \ N

P

(H); it follows that P has just one minimal normal subgroup T . For

^P_T

= hxT, yT i × E with hxT, yT i ≃ Q

3

and Φ(E) = h1i we may suppose hxi ⊳ P , hence [x, y]

²

= [x

²

, y] = 1, so [x, y] ∈ T , a contradiction.

Then

^P_T

is abelian, P

^′

= T = hti ≤ Z(P ) and Z(P ) is cyclic. Since H and K are not conjugate, it is hk 6= t. If [h, k] = 1, then |hk| = 2, hhki ⋪ P and so ν

c

(P ) ≥ 3.

Therefore (hk)

²

= [h, k] = t.

Let Z(P ) = hzi with |Z(P )| = 2

^s

. If s ≥ 2, it is (z

^2s−2

hk)

²

= 1, thus L = hz

^2s−2

hki ⋪ P with L neither conjugate to H nor to K, a contradiction.

We conclude that Z(P ) = P

^′

= hti; P is an extraspecial group and P is a central product P = S

1

∗ S

2

∗ . . . ∗ S

r

, with S

1

isomorphic either to D

3

or to Q

3

, S

i

≃ D

3

for 2 ≤ i ≤ r and |S

i

∩ S

j

| = 2 for i 6= j (see [8], 5.3).

Let S

2

= ha, b|a

⁴

= b

²

= 1, [a, b] = a

²

i. If S

¹

= hc, d|c

⁴

= d

²

= 1, [c, d] = c

²

i, the sub- groups hbi, hdi and hbdi are non-normal and pairwise non-conjugate. If S

1

= hc, d|c

⁴

= 1, c

²

= d

²

= [c, d]i, the subgroups hbi, haci and hadi are non-normal and pairwise non-conjugate.

QED

Remark 2. If P = [hai]hbi, with |a| = 2

ⁿ⁻²

, |b| = 4, [a, b

²

] = 1 and n ≥ 4 it is ν(P ) = 2 if and only if a

^b

= a

¹⁺²ⁿ⁻³

; this follows from Theorem I in [6].

3.3 Non nilpotent groups with ν

c

= 2

Proposition 8. Let G be a non-nilpotent group such that ν

c

(G) = 2. If H and K are non-normal, non-conjugate cyclic subgroups, then one of the following cases holds:

(1) |H| = p

^α

, |K| = p

^β

(2) |H| = p

^α

, |K| = q

^β

(3) |H| = p

^α

, |K| = p

^α

q where p, q are distinct primes.

The third case holds only if G = A × B, where A and B are proper subgroups of G and (|A|, |B|) = 1.

Proof. Obviously there exists a cyclic subgroup H = hhi such that H ⋪ G and |H| = p

^α

,

where p is a prime. If |K| is not a prime power, then K = R × S, where (|R|, |S|) = 1 and

R ⋪ G, so that R is conjugate to H. Let q be a prime and T a subgroup of S such that |T | = q;

since R × T ⋪ G and ν

^c

(G) = 2, then R × T = K and |K| = p

^α

q. For any prime w 6= p and any w-element y ∈ G one has hyi ⊳ G and any Sylow w-subgroup is normal.

Since [H, T ] = h1i, by Proposition 5 it is [H, Q] = h1i. Suppose H ⊆ P ∈ Syl

p

(G); if it were [P, Q] 6= h1i, for any a ∈ P such that [a, Q] 6= h1i it would be hai ⋪ G, consequently hai ∼

G

H, a contradiction.

So Q ⊆ Z(G) and Q is a direct factor of G.

^QED

Lemma 4. Let G be a non-nilpotent group with ν

c

(G) = 2. If there exists a unique prime p such that the Sylow p-subgroups are non-normal and there are two non-normal cyclic subgroups H and K whose orders are coprime (i.e. |H| = p

^α

, |K| = q

^β

), then G is a minimal non-abelian group.

Proof. By Proposition 6 G is not a direct product of proper subgroups.

Let H = hhi ⊆ P ∈ Syl

p

(G) and K = hki ⊆ Q ∈ Syl

q

(G) and Q ⊳ G. For any prime r different from p and q and for any non trivial r-element a ∈ G one would have hai ⊳ G and hkai = hki × hai ⋪ G, against the hypothesis ν

c

(G) = 2; so G = [Q]P .

In a similar way one proves C

Q

(H) = h1i = C

P

(K) and therefore N

G

(P ) = P and C

P

(Q) = h1i. So for any b ∈ P \ h1i it is hbi ⋪ G and hbi is conjugate to H; this means

|H| = p = exp P . We may suppose H ⊆ Z(P ); then any hbi would be conjugated to H by an element of Q and therefore G = QH with H = P . It will suffice to prove that Q is a minimal normal subgroup of G.

By the Frattini Argument P Φ(Q) ⋪ G, so that ν

c



G Φ(Q)



6= 0. We distinguish two cases:

ν

c



G Φ(Q)

 = 1 and ν

c



G Φ(Q)

 = 2.

Case 1 : ν

c



G Φ(Q)



= 1

Every q-subgroup of

_Φ(Q)^G

is normal.

It is ν

c



G Φ(K)



= 2. If Φ(K) 6= h1i, by induction on the order of the group,

Φ(K)^G

would be a minimal non-abelian group,

_Φ(K)^Q

would be elementary abelian and Φ(Q) = Φ(K), which contradicts ν

c



G Φ(Q)



= 1. Therefore |K| = q.

Since KΦ(Q) ⊳ G, K

^G

6= Q. Let y ∈ Q \ K

^G

, then hyi ⊳ G and hkyi ⊳ G. Hence hy, ki ⊳ G, so that K

^G

⊆ Ω

1

(hy, ki) ⊳ G.

By Theorem 1, q 6= 2 and either hy, ki ≃ M

s

(q) or hy, ki is abelian; this implies

|Ω

1

(hy, ki)| = q

²

and K

^G

= Ω

1

(hy, ki). It follows that K has q conjugates.

We may suppose H = P ⊆ N

^G

(K).

Since ν

c

(Q) = 1, it is Q = ha, k|a

^ql

= k

^q

= 1, [a, k] = a

^ql−1

i and K

^G

= hk, a

^ql−1

i.

Since ν

c



G Φ(Q)



= 1, one has h

⁻¹

ah = a

^r

f

1

and h

⁻¹

kh = k

^r

f

2

where f

1

, f

2

∈ Φ(Q).

Since h ∈ N

^G

(K), it is f

2

= 1; furthermore h

⁻¹

a

^ql−1

h = a

^rql−1

and the automorphism induced by h on Q fixes every non-normal subgroup of Q. From [h, k] 6= 1 it follows that h induces a non-identity q

^′

-automorphism which fixes every subgroup of Q; a contradiction by Proposition 5.

Case 2 : ν

c



G Φ(Q)



= 2 Suppose Φ(Q) 6= h1i.

The subgroups

^HΦ(Q)_Φ(Q)

and

^KΦ(Q)_Φ(Q)

are non-normal in

_Φ(Q)^G

. By induction

_Φ(Q)^G

is a

minimal non-abelian group of order q

^m

p, with m ≥ 2.

There is no normal subgroup of G of order q. Indeed, if hyi ⊳ G with |y| = q, then the subgroup Hhyi cannot be cyclic, so that q ≡ 1(mod p), but m is the least integer such that q

^m

≡ 1(mod p).

It follows |K| = q and exp Q = q.

The proper cyclic subgroups of Q are conjugate in G, then they are contained in Z(Q) and Q is elementary abelian, a contradiction.

So Φ(Q) = h1i.

If Q = K

^G

× T , for any 1 6= t ∈ T it would be h

⁻¹

th = t

^r

and h

⁻¹

(kt)h = k

^s

t

^s

; it would follow h

⁻¹

kh = k

^s

and hki ⊳ G. So Q = K

^G

.

Let A be a proper subgroup of Q normal in G. Since Q = K

^G

, hai ⊳ G for any a ∈ A . By Maschke’s Theorem Q = A × B, where B ⊳ G. Then h

⁻¹

ah = a

^r

for any a ∈ A and h

⁻¹

bh = b

^s

for any b ∈ B, with r 6≡ s(mod q); this means habi ⋪ G for a 6= 1 and b 6= 1.

Let k = ¯ a¯b, ¯ a ∈ A and ¯b ∈ B. Since habi is conjugate to hki, it is hai = h¯ai and hbi = h¯bi, therefore |Q| = q

²

and K has q − 1 conjugates. Then q = 3 and either r ≡ 1(mod q) or s ≡ 1(mod q); it follows respectively A ⊆ C

Q

(H) or B ⊆ C

Q

(H), which contradicts C

Q

(H) = h1i.

So Q is a minimal normal subgroup of G.

QED

Proposition 9. If G is a non-nilpotent group with ν

c

(G) = 2, then there exists just one prime p such that the Sylow p-subgroups of G are not normal.

Proof. Let G be a minimal counterexample.

Let P ∈ Syl

p

(G) and Q ∈ Syl

q

(G) be non-normal in G. There exist two non-normal subgroups H = hhi ⊆ P and K = hki ⊆ Q; since ν

c

(G) = 2, one has [h, k] 6= 1, thus [P, Q] 6= h1i.

For any prime r 6∈ {p, q} and for any r-element x ∈ G we have hxi ⊳ G; therefore if R ∈ Syl

r

(G), then R ⊳ G and R is abelian.

Let N be the product of all the normal Sylow subgroups of G; N is abelian, P and Q induce on N power automorphisms, so that [P, Q] ⊆ C

G

(N ) and [P, Q] ⊳ G.

We can suppose G = [N ](P Q) with C

N P

(Q) = C

N Q

(P ) = h1i. If N 6= h1i then P [P, Q] ⋪ G and Q[P, Q] ⋪ G and ν

c



G [P,Q]



= 2 against the minimality of G.

So N = h1i and G = P Q.

Suppose P

G

6= h1i; by minimality of G it is

^QP_PG^G

⊳

_P^G

G

, so G = [P

G

]N

G

(Q). Then P = [P

G

]N

P

(Q) and without loss of generality we may suppose H ≤ N

P

(Q).

Every subgroup of P is normal in G; since C

PG

(Q) = h1i, it is p 6= 2, P

G

is abelian and Q induces on P

G

a group of power-automorphisms. If a ∈ P

G

, |a| = exp P

G

, then C

Q

(P

G

) = C

Q

(a) and

_C ^Q

Q(PG)

is cyclic; we may suppose Q = KC

Q

(P

G

). Now, hk

^q

i⊳G, so [Q : C

Q

(P

G

)] = q and q divides p − 1. Since N

P

(Q) ≤ N

G

(C

Q

(P

G

)), then C

Q

(P

G

) ⊳ G.

If ν

c



G PG

 = 2, by Lemma 4 it is N

G

(Q) ≃

_P^G_G

≃ G(q, p, 1) and |Q| ≥ q

²

. Q would be a minimal normal subgroup of N

G

(Q). Since [Q, P

G

] 6= h1i, one has C

Q

(P

G

) = h1i, against [Q : C

Q

(P

G

)] = q.

If ν

c



G PG

 = 1, then

_P^G

G

= h

_QP

G PG

i

P

PG

with Q ≃

^QP_PG^G

abelian and

_P^P

G

cyclic. For

_P^P

G

= hxP

G

i we have hxi ⋪ G, hence hx

^p

i ⊳ G; this means   

P^PG

 

 = p = |N

^P

(Q)| and H = N

P

(Q).

We have   

CQ^QH(PG)

 

 = pq; as q < p, one has Q ≤ N

^G

(HC

Q

(P

G

)), so [Q, H] ⊆ HC

^Q

(P

G

) ∩

Q = C

Q

(P

G

). Since [Q]H = N

G

(Q) ≃

_P^G_G

, it is ν

c

(QH) = 1 and for every b ∈ Q one has

h

⁻¹

bh = b

^s

for some s ∈ N; as [h, Q] 6= h1i it is also [h, Ω

¹

(Q)] 6= h1i, so s 6≡ 1(mod q). Then

Q = [h, Q] ≤ C

Q

(P

G

), against C

P

(Q) = h1i.

^QED

Theorem 4. Let G be a non-nilpotent group which is not a direct product of proper subgroups. Then ν

c

(G) = 2 if and only if G is isomorphic to one of the following groups:

1) the minimal non-abelian group [Q]C

p

where Q is elementary abelian of order q

ⁿ

, n ≥ 2 and

|C

p

| = p;

2) hx, A|x

^pn

= 1

G

, a

^x

= a

^r

for any a ∈ Ai, p prime, n ≥ 2, A abelian, (|A|, p) = 1, (r(r

^p

− 1), |A|) = 1, r

^p2

≡ 1 (mod exp A);

3) [A]D

n

= hx, y, A|x

²ⁿ⁻¹

= y

²

= 1

G

, x

^y

= x

⁻¹

, [a, x] = 1

G

, a

^y

= a

⁻¹

for any a ∈ Ai, where A is abelian, |A| is odd and n ≥ 3;

4) [A]S

n

= hx, y, A|x

²ⁿ⁻¹

= y

²

= 1

G

, x

^y

= x

^−1+2n−2

, [a, x] = 1

G

, a

^y

= a

⁻¹

for any a ∈ Ai, where A is abelian, |A| is odd and n ≥ 4;

5) [A]Q

n

= hx, y, A|x

²ⁿ⁻¹

= 1

G

, y

²

= x

²ⁿ⁻²

, x

^y

= x

⁻¹

, [a, x] = 1

G

, a

^y

= a

⁻¹

for any a ∈ Ai, where A is abelian, |A| is odd and n ≥ 4;

6) hx, y, A | x

²ⁿ⁻²

= y

⁴

= 1

G

, x

^y

= x

⁻¹

, [a, x] = 1

G

, a

^y

= a

⁻¹

for any a ∈ Ai, where A is abelian, |A| is odd and n ≥ 4;

7) hx, y, A | x

²ⁿ⁻²

= y

⁴

= 1

G

, x

^y

= x

²ⁿ⁻³⁻¹

, [a, x] = 1

G

, a

^y

= a

⁻¹

for any a ∈ Ai, where A is abelian, |A| is odd and n ≥ 5.

Proof. Let ν

c

(G) = 2 and P ∈ Syl

p

(G) with P ⋪ G.

Suppose G is not the minimal non-abelian group 1); there exist two non-normal non- conjugate cyclic subgroups H = hhi and K = hki contained in P .

By Proposition 9 it is G = [A]P , where A is abelian and P induces on A a group of power- automorphisms; furthermore C

A

(H) = C

A

(K) = h1i, which implies N

A

(H) = N

A

(K) = h1i.

If P is cyclic, then without loss of generality K = P , H = Φ(P ) and G is isomorphic to 2).

Suppose P non-cyclic.

If a ∈ A is such that a

⁻¹

Ha ≤ P or a

⁻¹

Ka ≤ P , then a

⁻¹

Ha ≤ haiH ∩ P = H (respectively a

⁻¹

Ka ≤ haiK ∩ P = K), so a = 1. It follows that every subgroup of P

G

is normal in G and for g ∈ G it is g

⁻¹

Hg ⊆ P (respectively g

⁻¹

Kg ⊆ P ) if and only if g ∈ P ; this implies ν

c

(P ) ≤ 2

If S = hsi ≤ P

G

and S ⋪ G, then S ∼

G

H or S ∼

G

K, so H ≤ P

G

or K ≤ P

G

; since [A, P

G

] = h1i, one would have A = C

A

(H) or A = C

A

(K), a contradiction. So a subgroup T of P

G

is normal in G if and only if T ≤ P

^G

.

Suppose H ⋪ P and K ⋪ P . Then ν

c

(P ) = 2 and by Proposition 7 it is p = 2.

If P = M

_n−1

(2) × C

2

, then G = AM

_n−1

(2) × C

2

and G would be a direct product of proper subgroups, against the hypothesis.

If exp P = 2

ⁿ⁻¹

= |x| with x ∈ P , it is hxi ⊳ G; P is isomorphic to D

n

, S

n

or Q

n

with n ≥ 4 and G is isomorphic to 3), 4) or 5) with n ≥ 4.

Otherwise P = [hxi]hyi ≃ [C

2ⁿ⁻²

]C

4

with x

^y

∈ {x

⁻¹

, x

^−1+2n−3

, x

¹⁺²ⁿ⁻³

}. If x

^y

= x

¹⁺²ⁿ⁻³

with n ≥ 5 it is hxyi ⊳ P , then P = hx, xyi ⊳ G, a contradiction. In the other two cases G is isomorphic to 6) or 7).

Now we prove that

_P^P

G

is cyclic.

It is N

G

(P ) = P × C

A

(P ); since C

A

(P ) ≤ C

A

(H) = h1i, then N

G

(P ) = P .

Let g ∈ G\P and T = g

⁻¹

P g ∩P . If t ∈ T and hti ⋪ G, then hti = a

⁻¹

Ha or hti = a

⁻¹

Ka for some a ∈ P . Since ghtig

⁻¹

≤ P , then ga

⁻¹

Hag

⁻¹

≤ P or ga

⁻¹

Kag

⁻¹

≤ P , so ag

⁻¹

∈ P and g ∈ P , a contradiction. Then for any g ∈ G \ P one has P ∩ g

⁻¹

P g ⊳ G, so P ∩ g

⁻¹

P g = P

G

and

_P^G

G

is a Frobenius group with complement

_P^P

G

(see [8], 10.5). It follows

_P^P

G

cyclic or

P PG

≃ Q

^r

.

If

_P^P

G

= hx, y|x

^2r−1

= 1, x

^2r−2

= y

²

, [y, x] = x

²

i ≃ Q

r

, the subgroups hxi, hx

²

i and hyi would be non-normal in G (because they are not contained in P

G

) and pairwise non-conjugate in G, but then ν

c

(G) ≥ 3.

So

_P^P

G

is cyclic and P

G

 Φ(P ).

Without loss of generality we may distinguish two cases: either H ≤ K, or neither H nor K contains properly a non-normal subgroup of G (equivalently, Φ(H) ⊳ G and Φ(K) ⊳ G).

Case 1 : H ≤ K.

In this case one has H = Φ(K), Φ(H) ⊳ G and Φ(H) = H ∩ P

G

= K ∩ P

G

. Since

_P^P

G

is cyclic, it is P = KP

G

. Since P is not cyclic, Φ(H) 6= P

G

. Let Φ(H) ≤ L ≤ P

G

with [P

G

: L] = p; then

^G_L

=

^AKL_L

×

^P_L^G

. By Proposition 1 it follows ν

c



AK Φ(H)



= ν

c AKL

L

= 1.

If

_Φ(H)^K

⊳

_Φ(H)^AK

, then [A, K] ≤ A ∩ Φ(H) = h1i, a contradiction to C

A

(K) = h1i. So

K

Φ(H)

⋪

_Φ(H)^AK

and analogously

_Φ(H)^H

⋪

_Φ(H)^AK

, so ν

c



AK Φ(H)



≥ 2, a contradiction.

Case 2 : Φ(H) ⊳ G and Φ(K) ⊳ G.

It is Φ(H) = H ∩ P

^G

and Φ(K) = K ∩ P

^G

, so   

^HPP_G^G

   =

  

^KPP_G^G

 

 = p. For

P^P_G

= hsP

^G

i, one has hsi ∼

P

H or hsi ∼

P

K, so P = HP

G

or P = KP

G

. It follows   

P^PG

   = p, so P = HP

G

= KP

G

.

Without loss of generality we may suppose H ⊳ P , so

_Φ(H)^P

=

_Φ(H)^H

×

_Φ(H)^PG

and

_Φ(H)^G

=

AH

Φ(H)

×

_Φ(H)^PG

; it follows Φ(H) 6= h1i.

It is

_Φ(H)^H

⋪

_Φ(H)^AH

, so ν

c



AH Φ(H)

 ≥ 1. It must be µ

c



_P

G Φ(H)

 = 2 and ν

c



_P

G Φ(H)

 = 0, so  



Φ(H)^PG

   = p,

  

Φ(H)^PG

 

 = p

²

and Φ(P ) = Φ(H) ⊳ G.

Let x ∈ P \ Φ(H) be such that hxΦ(H)i 6=

_Φ(H)^PG

and hxΦ(H)i 6=

_Φ(H)^H

; then hxi ⋪ G and hxi ≁

G

H and so hxi ∼

G

K and hxΦ(H)i =

^KΦ(H)_Φ(H)

. Therefore

_Φ(H)^P

has only three proper subgroups and p = 2.

Suppose Φ(K) 6= Φ(H); let Φ(K) ≤ J ≤ Φ(H) with [Φ(H) : J] = 2. Then  

^KJ_J



 = 2 and

^P_J

= 

_H

J



_KJ

J

would be either abelian or isomorphic to D

3

. Since  



^PJ^G

 

 = 4 and every subgroup of

^PJ^G

is normal in

^G_J

,

^P_J

cannot be isomorphic to D

3

. If

_P^P

J

were abelian, then

^G_J

=

^AKJ_J

×

^P_J^G

with µ

c



_P

G J

 ≥ 3, a contradiction by Proposition 1.

We conclude that Φ(H) = Φ(K) = Φ(P ), K ⊳ P and P is a Dedekind group with

|H| = |K| ≥ 4.

If there exists C ≤ P with |C| = 2 and C ∩ H = h1i, then P = H × C with C ⊳ G, so G = AH × C, a contradiction. Then P is isomorphic to Q

3

and G is as in 5) with n = 3.

QED

Remark 3. Theorem I in [6] shows that among the groups presented in Theorem 4 only the alternating group A

4

and the groups of type 2) with |A| = q ( q prime) have just two conjugacy classes of non-normal subgroups.

References

[1] N. Blackburn: Finite groups in which the non-normal subgroups have nontrivial inter- section, J.Algebra, 3 (1996), 30-37

[2] R. Brandl: Groups with few non-normal subgroups, Comm. Alg., 23 (1995), 2091-2098 [3] D. Gorenstein: Finite groups, Harper and Row, New York (1968)

[4] B. Huppert: Zur Sylowstruktur auflosbarer Gruppen, Arch. Math., 12 (1961), 161-169 [5] G. A. Miller, H. Moreno: Non-abelian groups in which every proper subgroup is abelian,

Trans. Am. Math. Soc., 4 (1903), 398-404

[6] H. Mousavi: On finite groups with few non-normal subgroups, Comm. Alg., 27 (7) (1999), 3143-3151

[7] J. Poland, A. Rhemtulla: The number of conjugacy classes of non-normal subgroups in nilpotent groups, Comm.Alg., 24 (1996), 3237-3245

[8] D. J. S. Robinson: A course in the Theory of Groups, Springer Verlag, New York (1982)