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ANNALES MATHÉMATIQUES
BLAISE PASCAL
Yasuhiro Kishi
On D5-polynomials with integer coefficients
Volume 16, no 1 (2009), p. 113-125.
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Annales mathématiques Blaise Pascal 16, 113-125 (2009)
On D5-polynomials with integer coefficients
Yasuhiro Kishi
Abstract
We give a family of D5-polynomials with integer coefficients whose splitting
fields over Q are unramified cyclic quintic extensions of quadratic fields. Our poly-
nomials are constructed by using Fibonacci, Lucas numbers and units of certain
cyclic quartic fields.
1. Introduction
The following is a fundamental problem in the theory of quadratic fields:
For a given positive integer N , find quadratic fields whose class num-
ber is divisible by N . Several authors (for example, T. Nagell [6], N. C.
Ankeny and S. Chowla [1], Y. Yamamoto [12], P. J. Weinberger [11] and H.
Ichimura [3]) gave an infinite family of quadratic fields whose class number
is divisible by arbitrary given integer N . If limited to the case N = 5, C.
J. Parry [8], J.-F. Mestre [5], M. Sase [10] and D. Byeon [2] gave a family
of quadratic fields whose class number is divisible by 5. In particular, Sase
[10] gave a family of polynomials whose splitting field is a D5-extension
of Q and an unramified C5-extension of containing the quadratic field. In
the present paper, we will give other such polynomials by the use of the
result of our previous result [4]. Then we get a new family of quadratic
fields whose class number is divisible by 5. As a consequence, the following
conjecture arises:
Conjecture 1.1. Let Fn denote the n-th number in the Fibonacci se-
quence:
1, 1, 2, 3, 5, 8, 13, . . .
Then the class number of the quadratic field Q(
√
−F50s+25) (s ≥ 0) is
divisible by 5.
Keywords: class number, Fibonacci number, polynomial.
Math. classification: 11R29.
113
Y. Kishi
The organization of this paper is as follows. In Section 1, we state the
main theorem that the above conjecture is true under some conditions. In
Section 2, we give a parametric D5-polynomial with integer coefficients.
It is a review of [4]. In Section 3, we study the Fibonacci and the Luca
sequences. In Section 4, we give a proof of the main theorem. We give a
numerical example in the last Section 5.
We list here those symbols which will be used throughout this article.
Let Q denote the field of rational numbers and Z denote the ring of
rational integers.
For an integer n, let Cn and Dn denote the cyclic group of order n and
the dihedral group of order 2n, respectively.
For an extension L/K, denote the norm map and the trace map of L/K
by NL/K and by TrL/K , respectively. For simplicity, we denote NL and
TrL if the base field K = Q. For a Galois extension L/K, we denote the
Galois group of L/K by Gal(L/K).
For a polynomial f(X) and a field K, we denote the minimal splitting
field of f(X) over K by SplK(f).
2. The main Theorem
To state the main theorem, we prepare some notations.
Let (Fn) and (Ln) be the Fibonacci and the Luca sequences, respec-
tively, defined as follows:
F1 = 1, F2 = 1, Fn+2 = Fn+1 + Fn (n ≥ 1),
L1 = 1, L2 = 3, Ln+2 = Ln+1 + Ln (n ≥ 1).
Let ζ := e2πi/5 be a primitive fifth root of unity. For a non-negative
integer m, we set km := Q(
√
−F2m+1). Moreover we define a cyclic quartic
field Mm as follows: Mm is the proper subextension of km(ζ)/Q(
√
5) other
than km(
√
5) and Q(ζ). Then we can express Mm = Q(
√
−F2m+1(ζ −
ζ−1)). We define an element δm of Mm by
δm := 1 + εm
√
−F2m+1(ζ − ζ−1),
where ε := (1 +
√
5)/2 is a fundamental unit of Q(
√
5). As we will see in
Section 4, δm is a unit in Mm. Let τ be a generator of Gal(km(ζ)/km) (∼=
C4) with ζτ = ζ2.
114
On D5-polynomials
For a non-negative integer m, we define a polynomial gm(X) of degree
5 with integer coefficients by
gm(X) := X5 − 10X3 − 20X2 + 5(20F 22m+1 − 3)X
+ 40F 22m+1((−1)mL2m+1 + 1)− 4.
Under the above notations and assumptions, we have the following.
Theorem 2.1. Let m be a non-negative integer. If δ2+3τ+τ2m is not a fifth
power in Mm, then SplQ(gm) is a D5-extension of Q containing km. If,
moreover, m ≡ 12 (mod 52), then SplQ(gm) is an unramified cyclic quintic
extension of km. Hence, by putting m = 25s + 12, the class number of the
quadratic field Q(
√
−F50s+25) is divisible by 5.
Remarks 2.2. The author think the assumption “δ2+3τ+τ2m is not a fifth
power in Mm” can be excluded (cf. Remark 6.2).
3. Construction of D5-polynomials
This section is a review of [4] in the case p = 5 and the base field Q. Let ζ
be a primitive fifth root of unity, and let k = Q(
√
D) be a quadratic field
which does not coincide with Q(
√
5). Then there exists a unique proper
subextension of the bicyclic biquadratic extension k(ζ)/Q(
√
5) other than
k(
√
5) and Q(ζ). We denote it by M . Then M is a cyclic quartic field. Let
us call M the associated field with k. Fix the generator τ of Gal(k(ζ)/k)
with ζτ = ζ2, and define a subset M(k) of k(ζ)× as follows:
M(k) := {γ ∈ k(ζ)× | γ3+4τ+2τ2+τ3 6∈ k(ζ)5}.
For an element γ ∈ M , we define a polynomial fγ(X) by
fγ(X) := X5 − 10NM (γ)X3 − 5NM (γ)NT (γ)X2
+ 5NM (γ){NM (γ)−NT (γ1+τ )}X −NM (γ)NT (γ2+τ ),
where NT = NQ(√5)TrM/Q(√5).
Applying [4, Theorem 2.1, Corollary 2.6] to the case p = 5, we get the
following proposition.
Proposition 3.1. Let the notation be as above. Then for δ ∈M(k)∩M ,
SplQ(fδ) is a D5-extension of Q containing k. Putting E := SplQ(fδ),
115
Y. Kishi
moreover, we have
E = k
(
TrE(ζ)/E
( 5√
δ3+4τ+2τ2+τ3
))
.
Remarks 3.2. In [4], we can see that every D5-extension E of Q containing
k is given as E = SplQ(fδ) for some δ ∈M(k) ∩M .
4. Properties of Fibonacci and Lucas numbers
There are many relations between Fibonacci and Lucas numbers. (See for
example [9] and [7].) In this section, we show the following four properties
which we need in the next section.
(A) The power of ε = (1 +
√
5)/2 is expressed by
εm =
Lm + Fm
√
5
2
.
(B) For positive integer m, we have
Fm+1 =
Lm + Fm
2
, (4.1)
Lm+1 =
Lm + 5Fm
2
. (4.2)
(C) Let m be a positive integer. If m is divisible by 52, then so is Fm.
(D) Let n and m be positive integers. If d = gcd(n, m), then we have
gcd(Fn, Fm) = Fd.
We easily get the property (A) by mathematical induction on m, using
εm+1 =
Lm + Fm
√
5
2
· 1 +
√
5
2
=
(Lm + 5Fm)/2 + (Lm + Fm)/2 ·
√
5
2
,
and the property (B).
We will prove the property (B) by mathematical induction on m. The
equations (4.1) and (4.2) hold clearly for m = 1. Assume that (4.1) and
116
On D5-polynomials
(4.2) hold for m. Then we have
Lm+1 + Fm+1
2
=
(Lm + 5Fm)/2 + Fm+1
2
=
Lm + 5Fm + 2Fm+1
4
=
(2Fm+1 − Fm) + 5Fm + 2Fm+1
4
= Fm+1 + Fm = Fm+2
and
Lm+1 + 5Fm+1
2
=
Lm+1 + 5(Lm + Fm)/2
2
=
2Lm+1 + 5Lm + 5Fm
4
=
2Lm+1 + 5Lm + (2Lm+1 − Lm)
4
= Lm+1 + Lm = Lm+2.
Hence (4.1) and (4.2) hold for m + 1.
Before proving the property (C), we show that the relation
Fn+m = FmFn+1 + Fm−1Fn (4.3)
holds for positive integers n, m. For any n, we have
Fn+1 = 1 · Fn+1 + 0 · Fn = F1Fn+1 + F0Fn,
Fn+2 = 1 · Fn+1 + 1 · Fn = F2Fn+1 + F1Fn.
(For convenience we define F0 = 0.) Then (4.3) holds for m = 1, 2. Assume
that (4.3) holds for m = k, k − 1:
Fn+k = FkFn+1 + Fk−1Fn,
Fn+(k−1) = Fk−1Fn+1 + Fk−2Fn.
Then we have
Fn+(k+1) = Fn+k + Fn+(k−1)
= (FkFn+1 + Fk−1Fn) + (Fk−1Fn+1 + Fk−2Fn)
= (Fk + Fk−1)Fn+1 + (Fk−1 + Fk−2)Fn = Fk+1Fn+1 + FkFn.
Hence (4.3) holds for m = k + 1.
Next we prove
m ≡ 0 (mod n) =⇒ Fm ≡ 0 (mod Fn), (4.4)
namely,
Fnk ≡ 0 (mod Fn)
117
Y. Kishi
for each k ∈ Z, k ≥ 1. It holds clearly for k = 1. Assume that Fnk ≡
0 (mod Fn). Then by (4.3), we have
Fn(k+1) = Fnk+n = FnFnk+1 + Fn−1Fnk ≡ 0 (mod Fn).
Hence (4.4) is proved.
The property (C) follows from (4.4) and F25 = 75025 ≡ 0 (mod 52).
For positive integers n and m, express n = qm + r, 0 ≤ r ≤ m. Then
we claim
gcd(Fn, Fm) = gcd(Fr, Fm). (4.5)
Using (4.3) and (4.4), we have
gcd(Fn, Fm) = gcd(Fqm+r, Fm)
= gcd(FqmFr+1 + Fqm−1Fr, Fm)
= gcd(Fqm−1Fr, Fm).
Here we have
gcd(Fqm−1, Fqm) = gcd(Fqm−1, Fqm−1 + Fqm−2) = gcd(Fqm−1, Fqm−2)
= · · · · · · = gcd(F2, F1) = gcd(1, 1) = 1.
From this together with Fm | Fqm, we have gcd(Fqm−1, Fm) = 1. Then
we get (4.5). Hence by using the Euclidean algorithm, the property (D)
follows.
Remarks 4.1. The inverse of the property (C) also holds true.
5. Proof of the main theorem
The goal of this section is to give a proof of our main theorem.
By the definition, Mm is the associated field with km. Hence we can
apply Proposition 3.1. Now let us calculate fδm(X);
fδm(X) = X
5 − 10NM (δm)X3 − 5NM (δm)NT (δm)X2
+ 5NM (δm){NM (δm)−NT (δ1+τm )}X −NM (δm)NT (δ2+τm ).
We note that τ satisfies the following:
ζτ = ζ2, (
√
5)τ = −
√
5, (
√
−F2m+1)τ =
√
−F2m+1.
118
On D5-polynomials
Put ε := ετ ; then we have
NM (δm) = (1 + εm
√
−F2m+1(ζ − ζ−1))(1 + εm
√
−F2m+1(ζ2 − ζ−2))
× (1− εm
√
−F2m+1(ζ − ζ−1))(1− εm
√
−F2m+1(ζ2 − ζ−2))
= (1 + ε2mF2m+1(ζ − ζ−1)2)(1 + ε2mF2m+1(ζ2 − ζ−2)2)
=
(
1− 5 +
√
5
2
ε2mF2m+1
)(
1− 5−
√
5
2
ε2mF2m+1
)
= 1− 5 +
√
5
2
ε2mF2m+1 −
5−
√
5
2
ε2mF2m+1
+ 5NQ(√5)(ε)
2mF 22m+1
= 1− F2m+1
{
TrQ(√5)
(
5 +
√
5
2
ε2m
)
− 5F2m+1
}
,
by using
(ζ − ζ−1)2 =
(
2i sin
2π
5
)2
= −5 +
√
5
2
and
(ζ2 − ζ−2)2 =
(
2i sin
4π
5
)2
= −5−
√
5
2
.
Here, by (4.1) we have
TrQ(√5)
(
5 +
√
5
2
ε2m
)
= TrQ(√5)
(
5 +
√
5
2
· L2m + F2m
√
5
2
)
=
5L2m + 5F2m
2
= 5F2m+1.
Therefore we get NM (δm) = 1. By similar calculations, we have
NT (δm) = 4,
NT (δ1+τm ) = 4− 20F 22m+1,
NT (δ2+τm ) = 4− 40F 22m+1(1 + (−1)nL2m+1).
119
Y. Kishi
Substituting them into fδm(X), we have
fδm(X) = X
5 − 10X3 − 20X2 + 5(20F 22m+1 − 3)X
+ 40F 22m+1((−1)nL2m+1 + 1)− 4
= gm(X).
Assume that δ2+3τ+τ2m is not a fifth power in Mm. Then we have
δ3+4τ+2τ
2+τ3
m = NM (δm)δ
2+3τ+τ2
m = δ
2+3τ+τ2
m 6∈ km(ζ)5,
and hence δ3+4τ+2τ2+τ3m ∈M(km). Then by Proposition 3.1, SplQ(gm) (=
SplQ(fδm)) is a D5-extension of Q containing km.
Assume in addition that m ≡ 12 (mod 52). We will show that the
cyclic quintic extension SplQ(gm)/km is unramified. Let θ be a root of
gm(X). Let q be a prime number in general. A prime divisor of q in km is
ramified in SplQ(gm) if and only if q is totally ramified in Q(θ) because
[SplQ(gm) : km] and [km : Q] are relatively prime. Hence we have only to
verify that no primes are totally ramified in Q(θ). This can be proved by
using the following Sase’s result. For a prime number p and for an integer
m, we denote the greatest exponent µ of p such that pµ | m by vp(m).
Proposition 5.1. [10, Proposition 2] Let p (6= 2) and q be prime numbers.
Suppose that the polynomial
ϕ(X) = Xp +
p−2∑
j=0
ajX
j , aj ∈ Z
is irreducible over Q and satisfies the condition
vq(aj) < p− j for some j, 0 ≤ j ≤ p− 2. (5.1)
Let θ be a root of ϕ(X).
(1) If q is different from p, then q is totally ramified in Q(θ)/Q if and
only if
0 <
vq(a0)
p
≤ vq(aj)
p− j
for every j, 1 ≤ j ≤ p− 2.
120
On D5-polynomials
(2) The prime p is totally ramified in Q(θ)/Q if and only if one of the
following conditions (S-i), (S-ii) holds:
(S-i) 0 <
vp(a0)
p
≤ vp(aj)
p− j
for every j, 1 ≤ j ≤ p− 2;
(S-ii) (S-ii-1) vp(a0) = 0,
(S-ii-2) vp(aj) > 0 for every j, 1 ≤ j ≤ p− 2,
(S-ii-3)
vp(ϕ(−a0))
p
≤ vp(ϕ
(j)(−a0))
p− j
for every j, 1 ≤ j ≤ p− 2,
and
(S-ii-4) vp(ϕ(j)(−a0)) < p− j for some j, 0 ≤ j ≤ p− 1,
where ϕ(j)(X) is the j-th differential of ϕ(X).
Now let us apply Proposition 4.1 to our polynomial gm(X). First, we
easily verify that gm(X) satisfies (4.1) for each prime. Next, we see from
Proposition 4.1 (1) that no primes except for 5 are totally ramified in
Q(θ)/Q because the greatest common divisor of the coefficient of X3 and
that of X is equal to 5. We will show, therefore, that 5 is not totally
ramified. Denote the constant term of gm(X) by c0;
c0 := 40F 22m+1((−1)nL2m+1 + 1)− 4.
Since c0 is not divisible by 5, the condition (S-i) does not hold. By the
assumption m ≡ 12 (mod 52), we have 2m + 1 ≡ 0 (mod 52). Then by
the property (C), in Section 3, we have F2m+1 ≡ 0 (mod 52), and hence
−c0 ≡ 4 (mod 55). Therefore we have
gm(−c0) ≡ 45 − 10 · 43 − 20 · 42 − 5(20d2 − 3) · 4− 4 ≡ 0 (mod 55),
g(1)m (−c0) ≡ 5 · 44 − 30 · 42 − 40 · 4− 15 ≡ 0 (mod 54),
g(2)m (−c0) ≡ 20 · 43 − 60 · 42 − 40 ≡ 0 (mod 53),
g(3)m (−c0) ≡ 60 · 42 − 60 ≡ 0 (mod 52).
Then the condition (S-ii-4) does not hold. (We can easily check that
(S-ii-1), (S-ii-2) and (S-ii-3) hold.) Hence 5 is not totally ramified in Q(θ).
This completes the proof of the main theorem.
Next, we consider when δ2+3τ+τ2m is not a fifth power in Mm.
121
Y. Kishi
Suppose that δ2+3τ+τ2m is a fifth power in Mm; δ2+3τ+τ
2
m = α
5, α ∈ Mm.
Since
δ3+4τ+2τ
2+τ3
m = NM (δm)δ
2+3τ+τ2
m = δ
2+3τ+τ2
m = α
5
and Mm is normal over Q, we have
TrE(ζ)/E
(
5
√
δ3+4τ+2τ
2+τ3
m
)
= TrE(ζ)/E(α) ∈ Mm,
where E = SplQ(gm). By the last half of Proposition 3.1, therefore, we
have
SplQ(gm) = km(TrE(ζ)/E(α)) ⊂ Mm.
Hence the degree [SplQ(gm) : km] is less than 5. Then gm(X) must be
reducible over Q. Therefore, we have
Proposition 5.2. The element δ2+3τ+τ2m is not a fifth power in Mm if
and only if gm(X) is irreducible over Q.
Finally in this section, we prove the following.
Proposition 5.3. The set{
Q(
√
−F50s+25)
∣∣∣ s ≥ 0}
is infinite.
Proof. On the contrary, we assume #{Q(
√
−F50s+25)
∣∣ s ≥ 0} < ∞. For
an integer m, we denote the square free part of m by sf(m). By the as-
sumption, the set
P :=
⋃
s≥0
{prime factors of sf(F50s+25)}
is finite. Then there exists a positive integer t so that we have
P =
⋃
0≤s≤t
{prime factors of sf(F25(2s+1))}.
Take a prime q with q > 2t + 1. Then for each prime factor r of sf(F25q),
we have
sf(F25(2s+1)) ≡ 0 (mod r) for some s, 0 ≤ s ≤ t. (5.2)
On the other hand, because q is prime, for each s, 0 ≤ s ≤ t, we have
gcd(25(2s + 1), 25q) = 25,
122
On D5-polynomials
and hence by the property (D),
gcd(F25(2s+1), F25q) = F25 = 3001 · 52.
From this together with (5.2), we can express
F25q = 3001A2q
for some Aq ∈ Z. Then we have
−1 = NQ(√5)(ε
25q) =
L225q − 5F 225q
4
=
L225q − 5 · 30012A4q
4
This implies that (Aq, L25q) is an integer solution of the equation
Y 2 = 5 · 30012X4 − 4. (5.3)
The values of L25q (q is prime), of course, are different from each other.
However by Siegel’s theorem, there are only finitely many integer solutions
(X, Y ) of Eq. (5.3), This is a contradiction. 
6. Numerical examples
Example 6.1. Let m = 12. By F12 = 144, L12 = 322 and F25 = 3001 · 52,
we have
δ12 = 1 +
322 + 144
√
5
2
√
−3001 · 52(ζ − ζ−1)
and g12(X) = fδ12(X) is given by
X5 − 10X3 − 20X2 + 562875062485X + 37771618494049996.
Since g12(X) is irreducible over Q, it follows from Proposition 5.2 that δ12
is not a fifth power in Mm. Then by the main theorem, the splitting field
of g12(X) is an unramified cyclic quintic extension of Q(
√
−F25).
In the following table, we list the prime decompositions of −F2m+1 and
the structure of the ideal class groups of km = Q(
√
−F2m+1) for m ≤ 87
with m ≡ 12 (mod 52).
Remarks 6.2. For this table we use GP/PARI (Version 2.1.5). By using
the same calculator, the author verified that gm(X) is irreducible over Q
for every m, 0 ≤ m ≤ 2000. (The disit of the constant term of g2000(X) is
2510.)
123
Y. Kishi
m −F2m+1
Structure of the ideal class
group of km = Q(
√
−F2m+1)
12 −3001 · 52 C40
37 −2 · 61 · 3001 · 230686501 · 52 C2461460 × C2 × C2
62 −5 · 3001 · 158414167964045700001 · 52 C79285156360 × C8 × C2
87 −13 · 701 · 3001 · 141961 C1737032019043290
×17231203730201189308301 · 52 ×C6 × C2 × C2 × C2
Acknowledgments
The author is grateful to Abdelmalek Azizi, Mohammed Ayadi, Moulay C.
Ismaili and their colleagues for their kind hospitality during the conference
Congrès International Algèbre, Théorie des Nombres et leurs Applications.
Together, they made it possible to spend a wonderful week in Morocco.
Added in proof
After this paper had been written, the author proved our conjecture (Con-
jecture 1.1) himself. For the details, see in “A new family of imaginary qua-
dratic fields whose class number is divisible by five”, J. Number Theory
128 (2008), 2450–2458.
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Yasuhiro Kishi
Department of Mathematics
Fukuoka University of Education
1-1 Bunkyoumachi Akama, Munakata-shi
Fukuoka, 811-4192
Japan
ykishi@fukuoka-edu.ac.jp
125


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Lemnouar Noui

ANNALES MATHÉMATIQUES
BLAISE PASCAL
Lemnouar Noui
Properties of subgroups not containing their centralizers
Volume 16, no 2 (2009), p. 267-275.
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Annales mathématiques Blaise Pascal 16, 267-275 (2009)
Properties of subgroups not containing their
centralizers
Lemnouar Noui
Abstract
In this paper, we give a generalization of Baer Theorem on the injective prop-
erty of divisible abelian groups. As consequences of the obtained result we find a
sufficient condition for a group G to express as semi-direct product of a divisible
subgroup D and some subgroup H. We also apply the main Theorem to the p-
groups with center of index p2, for some prime p. For these groups we compute
Nc(G) the number of conjugacy classes and Na the number of abelian maximal
subgroups and Nna the number of nonabelian maximal subgroups.
1. Introduction
We shall recall some definitions:
If H is a subgroup of a group G, a subgroupK is called a complement of
H in G if G = HK and H ∩K = {1}. Therefore if HCG and KCG, then
G is said to be the direct product of H and K, in symbols, G = H K.
If H CG, then G is said to be the semi-direct product of H and K, in
symbols, G = H oK.
An abelian group D is called divisible if for every x ∈ D and every
positive integer n there is a y ∈ D so that x = ny, i.e., each element of
D is divisible by every positive integer. The main property of divisible
groups is that they satisfy the following “injectivity" condition:
Theorem 1.1 (Baer Theorem [3]). If D is a divisible group, then any
homomorphism f : A→ D = from any abelian group A into D extends to
any abelian group G which contains A i.e., there exists a homomorphism
f : G→ D so that f |A = f .
The purpose of this paper is to generalize this result to the nonabelian
groups. To this end, we introduce the property “N” in subgroups: Let H
be a subgroup of an arbitrary group G.
Keywords: Maximal subgroup, divisible groups, p-groups, center, conjugacy classes.
Math. classification: 14L05, 20D25, 20K27, 20E28.
267
L. Noui
H satisfies the condition “N" if and only if ∃g ∈ G−H [g, H] = 1.
This is equivalent to saying that CG(H) * H where CG(H) is the cen-
tralizer of H in G which is defined to be the set of all g in G such that
hg = gh for all h in H, it is clearly a subgroup of G.
By the definition of the condition “N” we deduce that
1) If G is abelian, then every proper subgroup H satisfies the condition
"N".
2) If G is a nonabelian nilpotent group, then every maximal normal
abelian subgroup H of G does not satisfy the condition “N” because
CG(H) = H, [3].
3) There exist a nonabelian groups G whose a subgroup H satisfies
the condition“N”, for example let G = Q8 × Z/2Z where H = Q8 is the
quaternion group of order 8, [2].
2. Main results and proofs
Theorem 2.1. Let G be a group and let H be a subgroup of G such that
each proper subgroup H ′ of G which contains H, satisfies the condition
“N”. Then any homomorphism f : H → D from H into divisible group D
extends to the group G.
Proof. Let us consider the set S of all pairs (Hi, fi ) whereHi is a subgroup
of G containing H and fi : Hi → D is an extension of f , i.e., f|Hi = fi. Let
(Hi, fi) ≤ (Hj , fj) ifHi ⊂ Hj and fj|Hi = fi. The set S is partially ordered
by the relation ≤. We aim to apply Zorn’s Lemma to S and to this end we
consider a chain (Hi, fi)i∈I . It has an upper bound
(
∪i ∈IHi, f
′
)
where
f
′ : ∪i ∈IHi → D is defined by f ′(xi) = fi(xi) for every xi ∈ Hi. This
is unambiguous since xi ∈ Hi ⊂ Hm ⇒ fm (xi) = fi (xi). Consequently,
by Zorn’s Lemma, S has a maximal element, say (H, f ). We claim that
H = G and f is the desired extension of f to G. To see this suppose
H 6= G. By hypotheses H satisfies the condition “N”, consequently, there
is an g ∈ G−H such that
[
g, H
]
= 1, therefore
〈
H, g
〉
= H · 〈 g 〉. There
are two cases:
Case 1.
H ∩ 〈 g 〉 6= {0}. Let n the smallest positive integer so that gn = u ∈ H.
Since D is divisible there is a x ∈ D so that f(u) = nx. Let f∗ : H ·〈 g 〉 →
D be defined by f ∗(agt) = f(a) + tx where a ∈ H and 0 ≤ t ≺ n, if
268
Subgroups not containing their centralizers
agt = a′gt′ then a′−1a = gt′−t ∈ H, so t′ − t = 0 and a = a′. Hence f ∗ is
well-defined mapping. Let z1 = a1gt1 , z2 = a2gt2 be two elements of the
group H ·〈 g 〉, then f∗(z1 )+f
∗(z2 ) = f (a1a2)+( t1+t2 )x and t1+t2 =
kn+ t0 where 0 ≤ t0 ≺ n. In the other hand f
∗(z1z2 ) = f
∗(a1gt1a2gt2),
since
[
g, H
]
= 1, f ∗(z1z2 ) = f
∗(a1a2gt1+t2 ) = f
∗(a1a2ukgt0 ), so
f
∗(z1z2 ) = f (a1a2) + ( kn + t0 )x. Finally f
∗(z1z2 ) = f
∗(z1 ) +
f
∗(z2 ), so f
∗ is a homomorphism so that f ∗|H = f . This contradicts the
maximality of (H, f).
Case 2.
H ∩ 〈 g 〉 = {0}. Then H · 〈 g 〉 = H  〈 g 〉. Let x0 be an element of
D, in this case we can define f ∗ : H  〈 g 〉 → D by writing f ∗(agt) =
f(a)+kx0, it is easy to verify that f
∗ is a homomorphism so that f ∗|H = f
contradicting the maximality of (H, f). Thus H = G and f is the desired
extension of f to G. 
Since every subgroup of an abelian group satisfies the condition“N”,
we can apply Theorem 2.1 to abelian groups, also we deduce the known
result, [3]:
Corollary 2.2. Any divisible subgroup D of an abelian group G splits,
i.e., D has a complement H so that G = H ⊕D.
If G is a group (not necessarily abelian), we write:
Corollary 2.3. Let G be a group and let D be a divisible subgroup of G
such that every subgroup H of G which contains D, satisfies the condition
“N”. Then D has a complement H so that G = H oD.
Proof. We consider the identity map: idD : D → D, by Theorem 2.1, idD
extends to the group G, i.e., there exists a homomorphism f : G → D
so that f |D = idD. Let i : D → G be the inclusion map, then f ◦ i =
idD implies that f(G) = D. Let H = Kerf , if x ∈ G, then x = xf
(x−1)f (x). Since f
[
xf (x−1)
]
= f (x).f ◦ f (x−1) and f |D = idD,
we have f
[
xf (x−1)
]
= 1, so G = H.D. If x belongs to H ∩ D, then
x = f (x′) and f (x) = 1, thus f (x) = f ◦ f (x′) = f (x′), that is,
H ∩D = {1}. Hence G = H oD. 
269
L. Noui
Corollary 2.4. Let G be a finite p-group with center of index p2. If H is
a nonabelian maximal subgroup of G then any homomorphism f : H → D
from H into divisible group D extends to the group G.
To prove Corollary 2.4, we need the following.
Lemma 2.5. Let G be a finite p-group such that its center Z(G) has
index p2. If H is a maximal subgroup of G, then the following properties
are equivalent.
i)H is abelian
ii) Z(G) ⊂ H
iii) H does not satisfy the condition “N”.
Proof of Lemma 2.5. “i) ⇒ ii)”. Let us assume that H is abelian. If
Z(G) * H, there exists g ∈ Z(G) − H and G = H · 〈 g 〉. Then G is
abelian, this contradicts |G : Z(G)| = p2.
“ii) ⇒ iii)”. Assume that Z(G) ⊂ H. Then Z(G) ⊂ Z(H) ⊂ H ⊂ G.
By hypothesis |G : Z(G)| = p2. Since the center does not have a prime
index and |G : H| = p, Z(H) = H, consequently H is abelian. Hence
H is a maximal normal abelian subgroup of the nilpotent group G, so
CG(H) = H, [3], and H does not satisfy the condition “N”.
“iii)⇒ i)”. If CG(H) ⊂ H, then Z(G) ⊂ Z(H) ⊂ H. By the same way
we deduce that H is abelian. 
Proof of Corollary 2.4. Let H be a subgroup of G so that H ⊂ H ′, since
H is nonabelian, by Lemma 2.5, there is g ∈ G −H so that [g, H] = 1.
Then [g, H ′] = 1 by maximality of H. Thus the conditions of Theorem 2.1
are satisfied, so we obtain Corollary 2.4. 
3. Subgroups satisfying the condition “N”
If A is finitely generated abelian group, the rank of A is defined by rk(A)
the minimum number of generators of A.
We denote us by xG the conjugacy class of x in an arbitrary groupG and
CG(x) the centralizer of x in G and Nc(G) the number of the conjugacy
classes.
If G is a finite p-group of class c, then from [4], we know that
Nc(G) ≥ c |G |1/c − c+ 1.
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Subgroups not containing their centralizers
Let G be a finite p-group of order pn such that its center has index
p2. In this section, we compute the number Nc(G) and N0 the number of
maximal subgroups in G satisfying the condition “N”.
Theorem 3.1. Let G be a finite p-group of order pn such that its center
has index p2, then
1) G has precisely p+ 1 abelian maximal subgroups.
2) The number of maximal subgroups satisfying the condition “N” equals
N0 = (pr−p2)/(p−1) where r is the rank of G/G′ = Z/pn1Z×· · ·×Z/pnrZ.
3) Nc(G) = pn−1 +pn−2−pn−3 and each nontrivial conjugacy class has
p elements.
The proof of Theorem 3.1 results from the following Lemmas.
Lemma 3.2. Let G be an abelian finite p− group, then the number of
subgroups of order p equals (pr − 1)/(p− 1) where r is the rank of G.
Proof of Lemma 3.2. Since rk(G) = r, G is isomorphic with the group
Z/pn1Z×· · ·×Z/pnrZ. If g is an element of order p inG, then g = g1g2 . . . gr
such that gi has order p or 1 and g 6= 1. The number of such elements g
equals
(p− 1) · C1r + (p− 1)2C2r + · · ·+ (p− 1)rCrr = pr − 1.
Since a group of order p has p − 1 elements of order p, the number of
subgroups of order p is (pr − 1)(p− 1). 
Lemma 3.3. Let G be a p−group satisfying |G : Z(G)| = pm, then |G′| ≤
pm(m−1)/2.
Proof. By induction on m. 
Proof of Theorem 3.1. 1) Let H be an abelian maximal subgroup of G,
then H does not satisfy the condition “N”, so Z(G) ⊂ H. Consequently
H/Z(G) is a subgroup of order p of the elementary p− group G/Z(G) '
Z/pZ×Z/pZ by Lemma 3.2, there is (p2−1)(p−1) = p+1 such subgroups.
2) By Lemma 3.3, |G : Z(G)| = p2 implies that |G′| = p. Let H be a
maximal subgroup of G, then G′ ⊂ H ⊂ G implies that H/G′ is a maximal
subgroup of G/G′. By using the known result of Steinitz [5]: The number
of subgroups of order pk equals the number of subgroups of order pn−k
in a finite abelian group of order pn, we conclude that the number of the
maximal subgroups H is equal to the number of subgroups of order p of
G/G′. If rk(G/G′) = r, then the number of maximal subgroups satisfying
271
L. Noui
the condition “N” is (pr − 1)(p − 1) − (p + 1) = (pr − p2)(p − 1) by
lemma 3.2.
3) If x ∈ Z(G), the conjugacy class of x is trivial, i.e, xG = {x}. If
x /∈ Z(G), then Z(G)  CG(x)  G therefore
∣∣∣xG∣∣∣ = |G : CG(x)| = p. Let
k the number of nontrivial conjugacy classes, then |G| = pn = pn−2 + kp.
Consequently k = pn−1 − pn−3 and Nc(G) = pn−1 + pn−2 − pn−3, so the
proof is complete. 
Corollary 3.4. Let G be a p−group of order pn such that |G : Z(G)| = p2.
If G/G′ is elementary p−group, then G has exactly (pn−1 − 1)(p − 1)
maximal subgroups.
Proof. This follows easily from Theorem 3.1. 
4. Examples
The following examples illustrate some applications of the previous results.
Example 4.1. Let G be a p−group of order p3, then
1) The number N of maximal subgroups is given in the following table
G Z
p3Z
Z
pZ
× Z
p2Z
Z
pZ
× Z
pZ
× Z
pZ
G is nonabelian
N 1 1 + p 1 + p+ p2 1 + p
2) The number of conjugacy classes is Nc(G) = p2 + p − 1 (if G is
nonabelian).
To prove the result 1) we consider the two cases
a) If G is abelian, we apply Lemma 2.5.
b) If G is nonabelian, |G : Z(G)| = p2 because the index of center does
not equal to a prime, so G′ = Z(G) and G is extra-special, [3]. Now
G′ ⊂ Frat(G) ⊂ G implies that |Frat(G)| = p2 or p, the first case
is impossible because, by Burnside Basis Theorem, [3], |G : Frat(G)| = p
implies that G is generated by one element, that is, G is cyclic. Hence G′ =
Z(G) = Frat(G). IfG/G′ is not elementary p−group, then by Theorem 3.1
G has one maximal subgroup, so |Frat(G)| = p2, a contradiction. Thus
G/G′ is elementary p−group and the result is an immediate consequence
of Corollary 3.4.
272
Subgroups not containing their centralizers
2) Since |G : Z(G)| = p2, to calculate Nc(G) it is enough to apply
Theorem 3.1 for n = 3.
Example 4.2. Let G be a p−group of order p4 such that |G : Z(G)| = p2.
1) If G/G′ is an elementary p−group, then G has exactly p+ 1 abelian
maximal subgroups and p2 nonabelian maximal subgroups.
2) If G/G′ is not an elementary p− group, G has p + 1 maximal sub-
groups, all abelian.
In order to prove this, we consider two cases.
1) In the first place, if G/G′ is an elementary p− group, then G has
p3−1
p−1 = p
2 + p + 1 maximal subgroups by Corollary 3.4. To calculate
the number of nonabelian maximal subgroups we can apply the second
assertion of Theorem 3.1.
2) Now assume that G/G′ is not an elementary p−group, then the rank
rk(G/G′) = 1 or 2, the first case implies that |G : Frat(G)| = p1 and G
is cyclic. Hence rk(G/G′) = 2, by Theorem 3.1, G has not a nonabelian
maximal subgroup and it has exactly p+1 maximal subgroups, all abelian,
as required.
Example 4.3. Let G be a p−group of order p4 such that |G : Z(G)| = p3.
Then
1) G has one maximal abelian subgroup and p nonabelian maximal
subgroups.
2) G has exactly 2p2 − 1 conjugacy classes.
In order to prove this result, we first note that every element x of G
belongs to a maximal subgroup. Second, we establish two Lemmas.
Lemma 4.4. Let G be a p−group of order p4 such that |G : Z(G)| = p3.
Then ∣∣∣xG∣∣∣ = p⇐⇒ x ∈M − Z(G),
where M is an abelian maximal subgroup of G.
Proof of Lemma 4.4. If
∣∣∣xG∣∣∣ = p, then |CG(x)| = p3, let M = CG(x).
Since G is a finite nilpotent group, M C G and G/M is abelian, conse-
quently G′ ⊂M . If G/Z(G) is abelian, then G′ ⊂ Z(G)⇒ G′ = Z(G), so
G is extra-special and |G| = p2k+1, [3], a contradiction. Hence G/Z(G) is
not abelian, since G is nilpotent, G′ ∩ Z(G) 6= {1}, so Z(G)  G′  M
and |G′| = p2. Assume that M is not abelian, then Z(G) ⊂ Z(M) ⊂ M .
273
L. Noui
Since the index of center does not equal to a prime, Z(G) = Z(M). If
y ∈ M , yx = xy so x ∈ Z(M) and we reach the contradiction x ∈ Z(G).
HenceM is abelian. Conversely, letM be an abelian maximal subgroup of
G and x ∈M−Z(G). If
∣∣∣xG∣∣∣ = p2, then |CG(x)| = p2. SinceM is abelian,
M ⊂ CG(x)  G, so M = CG(x) and |CG(x)| = p3, by this contradiction
we obtain
∣∣∣xG∣∣∣ = p. 
Lemma 4.5. Let G be a p−group of order p4 such that |G : Z(G)| = p3
and let M be a maximal subgroup of G. Then
1) IfM is abelian,M contains exactly p2−1 nontrivial conjugacy classes
which has p elements.
2) IfM is not abelian,M−G′ contains exactly p−1 nontrivial conjugacy
classes which has p2 elements.
Proof of Lemma 4.5. 1) Assume that M is abelian. Let x ∈ M − Z(G),
since M C G, xG ⊂ M . By Lemma 13,
∣∣∣xG∣∣∣ = p, consequently G has
p3−p
p = p
2 − 1 nontrivial conjugacy classes which has p elements.
2) If M is not abelian, let x ∈M −G′. From Lemma 13 it follows that∣∣∣xG∣∣∣ = p2, soM−G′ has exactly p3−p2p2 = p−1 nontrivial conjugacy classes
which has p2 elements. 
We will prove the last result as following. IfM1 andM2 are two maximal
subgroups in G, it is clear that M1 ∩M2 = G′. We denote by ka ( respec-
tively kna ) the number of abelian (respectively nonabelian) maximal sub-
groups in G. If x ∈ G′−Z(G), xG ⊂ G′, so
∣∣∣xG∣∣∣ = p and |CG(x)| = p3, we
have shown in the proof of Lemma 4.4 that CG(x) is abelian, consequently
ka 6= 0.
LetM1,M2 be two abelian maximal subgroups of G. Let x ∈ G′−Z(G),
then G′ ⊂M1  G and by Lemma 4.4,
∣∣∣xG∣∣∣ = p.
Let x ∈ M1. Since M1 is abelian, M1 ⊂ CG(x)  G, so M1 = CG(x).
By the same way we obtain M2 = CG(x). Hence M1 = M2 and ka = 1.
Each maximal subgroup M satisfy G′ ⊂ M ⊂ G, so MG′ is a subgroup of
order p of the group G/G′ ' Z/pZ × Z/pZ. By Lemma 3.2, G has p + 1
maximal subgroups, so kna = p.
2) By using Lemma 4.4 and Lemma 4.5 and the first assertion of Ex-
ample 4.2, we obtain Nc(G) = p+ (p2 − 1) + p(p− 1) = 2p2 − 1.
274
Subgroups not containing their centralizers
Remark 4.6. In [1], M. Reid proved that if G is a finite group whose order
is not divisible by 3, and G has m conjugacy classes, then the congruence
|G| ≡ m mod 3 holds. With the hypotheses of Theorem 3.1, we have the
congruence |G| ≡ Nc(G) mod 6 because
|G| −Nc(G) = pn − pn−1 − pn−2 + pn−3 = pn−3(p− 1)2(p+ 1).
Acknowledgements. I would like to thank the referee for his comments
and remarks.
References
[1] M. Reid – The number of conjugacy classes, Amer. Math. Monthly
105 (1998), no. 4, p. 359–361.
[2] D. J. S. Robinson – Finiteness conditions and generalized soluble
groups. Part 1, Springer-Verlag, New York, 1972, Ergebnisse der Math-
ematik und ihrer Grenzgebiete, Band 62.
[3] , A course in the theory of groups, second éd., Graduate Texts
in Mathematics, vol. 80, Springer-Verlag, New York, 1996.
[4] G. Sherman – A lower bound for the number of conjugacy classes in
a finite nilpotent group, Pacific J. Math. 80 (1979), no. 1, p. 253–254.
[5] N. Watson – Subgroups of finite abelian groups, 1995, Summer re-
search paper, Haverford College.
Lemnouar Noui
Department of Mathematics
Faculty of Science
University of Batna,
Algeria
nouilem@yahoo.fr
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