Journal ArticleDOI

# A complete classification of finite p-groups all of whose noncyclic subgroups are normal

01 Jan 2009-Glasnik Matematicki-Vol. 44, Iss: 1, pp 177-185
TL;DR: In this paper, a complete classification of finite p-groups all of whose noncyclic subgroups are normal is given, which solves a problem stated by Berkovich and solves the problem of non-cyclical subgroups.
Abstract: We give a complete classification of finite p-groups all of whose noncyclic subgroups are normal, which solves a problem stated by Berkovich.

### 1. Introduction and preliminary results

• The authors consider here only finite p-groups and their notation is standard.
• Let G be a finite non-Dedekindian p-group all of whose noncyclic subgroups are normal.
• But l must normalize 〈f〉 and l 2 inverts 〈f〉, a contradiction.

### 2. Proof of Theorem 1.1

• The authors have obtained the group stated in part (v) of their theorem.
• Now suppose that there exist no elements of order p in G−N and consider the abelian group G/W . (ii) Assume that Z(G) is noncyclic.
• Since Ω1(G) = W ∼= Ep2 , Proposition 1.3 shows that G is also metacyclic and the authors have obtained the groups from part (i) of their theorem.
• It is an easy exercise to show that all the groups from Theorem 1.1 satisfy the assumptions of this theorem and the authors are done.

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Content maybe subject to copyright    Report GLASNIK MATEMATI
ˇ
CKI
Vol. 44(64)(2009), 177 185
A COMPLETE CLASSIFICATION OF FINITE p-GROUPS
ALL OF WHOSE NONCYCLIC SUBGROUPS ARE NORMAL
Zdravka Bo
ˇ
zikov and Zvonimir Janko
University of Split, Croatia and University of Heidelberg, Germany
Abstract. We give a complete classiﬁcation of ﬁnite p-groups all of
whose noncyclic subgroups are normal, which solves a problem s tated by
Berkovich.
1. Introduction an d preliminary results
We co nsider here only ﬁnite p-groups and our notation is standard. If in
a p-group all cyclic subgroups are normal, then all subgroups are normal and
we call such a group Dedekindian. A Dedekindian p-gro up G is either abelian
or a Hamiltonian 2 -group, i.e., G = Q × A, where Q
=
Q
8
is the qua ternion
group and A is elementary abelian (Proposition 1.2 ). Here we study non-
Dedekindian p-groups all of whose noncyclic subgroups are normal. Such
groups have been considered by D.S. Passman [6, Pro position 2.9], but he
omitted 2-groups of order 2
7
. In fact here lies the main diﬃculty. In this
paper we reﬁne, improve and extend the arguments of Passman so that a ll
p-groups will be included. It turns out that we get in addition ﬁve exce ptio nal
2-groups: one group of order 2
6
, three groups of order 2
5
and one group of
order 2
4
. More precisely, we prove the following res ult which gives a complete
classiﬁcation of the title groups.
Theorem 1.1. Let G be a nite non-Dedekindian p-group all of whose
noncyclic subgroups are normal. Then G is one of the following groups.
(i) G is metacyclic minimal nonabelian and G is not isomorphic to Q
8
.
2000 Mathematics Subject Classiﬁcation. 20D15.
Key words and phrases. Dedekindian p-groups, Hamiltonian 2-groups, minimal non-
abelian p-groups, central products.
177 178 Z. BO
ˇ
ZIKOV AND Z. JANKO
(ii) G = G
0
Z, the central product of a nonabelian group G
0
of order
p
3
with a cyclic group Z, where G
0
Z = Z(G
0
) and if p = 2, then
|Z| > 2.
(iii) p = 2 and G = Q × Z where Q
=
Q
8
and Z is cyclic of order > 2.
(iv) G is a group of order 3
4
and maximal class with
1
(G) = G
=
E
9
.
(v) G = ha, b | a
8
= b
8
= 1, a
b
= a
1
, a
4
= b
4
i, where |G| = 2
5
, G
=
C
4
,
Z(G)
=
C
4
, G
Z(G)
=
C
2
and
2
(G) is abelian of type (4, 2).
(vi) G
=
Q
16
, the generalized quaternion group of order 2
4
.
(vii) G = D
8
Q
8
, an extraspecial 2-group of order 2
5
and type .
(viii) G = ha, b, c | a
4
= b
4
= [a, b] = 1, c
2
= a
2
, a
c
= ab
2
, b
c
= ba
2
i, where
G is the minimal non-metacyclic group of order 2
5
, G is a special
2-group with G
=
1
(G)
=
E
4
.
(ix)
G = ha, b, c, d | a
4
= b
4
= [a, b] = 1, c
2
= a
2
b
2
, a
c
= a
1
, b
c
= a
2
b
1
,
d
2
= a
2
, a
d
= a
1
b
2
, b
d
= b
1
, [c, d] = 1i,
where G is a special 2- group of order 2
6
with G
=
1
(G)
=
E
4
in which
every maximal subgroup is isomorphic to the minimal non-metacyclic
group of order 2
5
(from (viii)).
Conversely, all the above groups satisfy the assumptions of the theorem.
In the proof o f our theorem we use the following known results.
Proposition 1.2 (Huppert [3, III.7.12]). Let G be a p-group all of whose
subgroups are normal in G. Then G is either abelian or a Hamiltonian 2-
group, i.e., G = Q × A, where Q
=
Q
8
is the quaternion group and A is
elementary abelian.
Proposition 1.3 (L. edei, see [2, Lemma 3.1, 3.2]). A p-group G is
minimal nonabelian if and only if d(G) = 2 (minimal number of generators of
G is 2) and |G
| = p. In that case Φ(G) = Z(G). A p-group G is metacyclic
and m inimal nonabelian if and only if G is minimal nonabelian and |
1
(G)|
p
2
in which case either G
=
Q
8
or
G = ha, b | a
p
m
= b
p
n
= 1, a
b
= a
1+p
m1
, m 2, n 1i.
Proposition 1.4 (see [4, Proposition 1.6]). Let G be a nonabelian 2-
group of order 2
4
with |
2
(G)| 2
3
. Then either
G = ha, t | a
2
n
= t
2
= 1, a
t
= a
1+2
n1
, n 3i = M
2
n+1
or
G = ha, b | a
2
m2
= b
8
= 1, a
2
m3
= b
4
, m 5, a
b
= a
1
i
of order 2
m
.
Proposition 1.5. Let A be an abelian maximal subgroup in a nonabelian
p-group G and let g G A. Then the mapping x x
g
x
1
(x A) is a FINITE p-GROUPS WHOSE NONCYCLIC SUBGROUPS ARE NORMAL 179
homomorphism from A onto G
with kernel Z(G). This gives A/Z(G)
=
G
and |G| = p|Z(G)||G
|.
Proposition 1.6 (see [4, Proposition 1.10]). Let G be a 2-group of order
2
5
such that
2
(G) = ha, bi × hui, where ha, bi = Q
=
Q
8
and u is an
involution. Then G is a uniquely determined group of order 2
5
. Set hzi =
Z(Q) so that a
2
= b
2
= z. There is an element y of order 8 in G
2
(G)
such that y
2
= ua, u
y
= uz, a
y
= a
1
, and b
y
= bu.
Proposition 1.7. (Passman) Let G be a p-group all of whose noncyclic
subgroups are normal. Let H be any non-normal subgroup in G. Then H is
a max imal cyclic subgroup, |G : N
G
(H)| = p and N
G
(H)/H is either cyclic
or p = 2 and N
G
(H)/H
=
Q
8
is quaternion.
Proof. By our ass umption, H is cyclic. Assume that H is not maximal
cyclic and let K > H be a maximal cyclic subgroup containing H. Let L > K
be a subgroup of G such that |L : K| = p. Since L is noncy c lic, we have
d(L) = 2 and L is normal in G. Thus |L : Φ(L)| = p
2
, Φ(L) < K and
Φ(L) = Φ(K). But Φ(L) is normal in G and H Φ(L) and so H is normal
in G, a contradiction. We have proved that H is a maximal cyclic subgroup
in G.
Set N = N
G
(H) so that {1} 6= N/H is Dedekindian. By Proposition 1.2,
N/H is either abelian or Hamiltonian. If N/H has two distinct subgroups
N
1
/H and N
2
/H of order p, then N
1
and N
2
are normal in G and N
1
N
2
= H
so that H is normal in G, a contradiction. Thus N/H has only one s ubgroup
of order p and so N/H is e ither cyclic or p = 2 and N/H
=
Q
8
.
Set M/H =
1
(N/H) so that |M : H| = p and M is normal in G. L e t
H
0
be the subgroup of H with |H/H
0
| = p. By the ﬁr st paragr aph of the
proof, H
0
is normal in G and since M is noncyclic, M/H
0
=
E
p
2
. We act
with G/N on the set of p non-normal subgroups of order p of G/H
0
which
are contained in M/H
0
, where H/H
0
is one of them. This forces |G : N| = p
and we are done.
Proposition 1.8 (). Suppose that G is a 2-group all of whose cyclic
subgroups of order 4 are normal in G. Then either
2
(G) Z(G) or G is
Hamiltonian.
Proof. Set F =
2
(G) so that exp(F ) 4 and suppose that F 6≤ Z(G)
which implies exp(F) = 4. The subgroup F is Dedekindian a nd
1
(F )
Z(G). Since F 6≤ Z(G), there is an element v of order 4 in F and g G such
that v
g
= v
1
= vz, where v
2
= z. Assume tha t o(g) 8. Let hyi be the
cyclic subgroup of order 4 in hgi so that [v, y] = 1 and hvihyi hzi. We have
(vy)
g
= v
1
y = (vy)z. If hvi hyi = hzi, then vy is a noncentral involution,
a contradiction. If hvi hyi = {1}, then o(vy) = 4 and g does not normalize
hvyi. Hence o(g) = 4 and so g F which implies that F is Hamiltonian. 180 Z. BO
ˇ
ZIKOV AND Z. JANKO
Assume that F 6= G and let l be an element of order 8 in G F . Then
hli F = hl
2
i and if u = l
4
, then hui = F
= Φ(F ). Since Z(F ) =
1
(F ),
there is an element f of order 4 in F such that [l
2
, f ] = u and f
2
= u so that
hl
2
, f i
=
Q
8
. But l must normalize hfi and l
2
We have proved that G = F is Hamiltonian.
Proposition 1.9 (Janko [5, Theorem 2.6]). Let G be a nonabelian 2-
group all of whose minimal nonabelian subgroups are isomorphic to H
2
=
ha, b | a
4
= b
4
= 1, a
b
= a
1
i. Assume that exp(G) = 4 and
1
(G)
=
E
4
.
Then G is one of t he following groups.
(a) G
=
H
2
.
(b) G is the minimal non-metacyclic group of order 2
5
:
G = ha, b, c | a
4
= b
4
= [a, b] = 1, c
2
= a
2
, a
c
= ab
2
, b
c
= ba
2
i.
(c)
G = ha, b, c, d | a
4
= b
4
= [a, b] = 1, c
2
= a
2
b
2
, a
c
= a
1
, b
c
= a
2
b
1
,
d
2
= a
2
, a
d
= a
1
b
2
, b
d
= b
1
, [c, d] = 1i,
where G is a special 2-group of order 2
6
with Z(G)
=
E
4
in which
every maximal subgroup is isomorphic to the minimal non-metacyclic
subgroup of order 2
5
(given in (b)).
2. Proof of Theorem 1.1
Let G be a non-Dedekindian p-group a ll of whose noncyclic subgroups are
normal. In particular, G is nonabelian. If G has no normal e le mentary abelian
subgroup of order p
2
, then G is of maximal class. It follows that G
=
Q
16
(part (vi) of our theorem).
In what follows we assume that G has a normal abelian subgroup W of
typ e (p, p). Since each subgroup of G/W is normal, G/W is Dedekindian and
so G/W is either abe lian or Hamiltonian (Proposition 1.2).
Suppo se that p = 2 and there is a normal four-subgroup W such that
G/W is Hamiltonian. In that case
¯
G = G/W =
¯
Q×
¯
A, where
¯
Q is quaternion
and
¯
A is elementary abelian. By Proposition 1.7, d(G) 4 and so |
¯
A| 4
which implies 2
5
|G| 2
7
.
First we consider the case |G| = 2
5
so that G = Q and G/W
=
Q
8
. Set
S/W = Φ(G/W ) so that S is abelian since |G : C
G
(W )| 2. Suppose that S
is elementary abelian. If s S is such that s 6∈ Z(G), then C
G
(s)/hsi contains
the four-subgr oup S/hsi, contrary to Proposition 1.7. Hence S Z(G). Since
G is no t Dedekindian, there is g G S such that g
2
S W and hgi is not
normal in G. But then N
G
(hgi) = hgi×W and N
G
(hgi)/hgi
=
E
4
, contrary to
Proposition 1.7. We have proved that S is abelian of type (4, 2). If g G S, FINITE p-GROUPS WHOSE NONCYCLIC SUBGROUPS ARE NORMAL 181
then g
2
S W and so o(g) = 8. We have proved that
2
(G) = S is of order
8. By P roposition 1.4,
G = ha, b | a
8
= b
8
= 1, a
4
= b
4
, a
b
= a
1
i
since M
2
n+1
(n > 2) does not possess a factor-group is omorphic to Q
8
. We
have obtained the group stated in part (v) of our theorem.
Now we consider the case |G| > 2
5
. We have G = QA, Q A = W ,
Q and A are normal in G, Q/W
=
Q
8
, and A/W is elementary abelian of
order 2 or 4. In particular, exp(G) 8. Suppose that u is an involution in
G with u 6∈ Z(G). By Proposition 1.7, |G : C
G
(u)| = 2 and C
G
(u)/hui is
either cyclic of order 8 or C
G
(u)/hui
=
Q
8
. In any case, |G| 2
5
which
is a contradiction. We have shown that
1
(G) Z(G) and so W Z(G).
Assume that E =
1
(G) is of order 8. Since G is not Dedekindian, G
possesses a cyclic subgroup Z o f composite order which is not normal in G.
Set hzi = Z E so that z is a central involution. Let e, f E hzi such
that he, f, zi
=
E
8
. In this case S
1
= hZ, ei and S
2
= hZ, fi are normal in G.
But S
1
S
2
= Z and so Z is normal in G, a contradiction. It follows that
1
(G) = W Z(G). Set T /W = Φ(Q/W ) so that T is abelian of type (4, 2).
For e ach x Q T , x
2
T W and so o(x) = 8 and
2
(Q) = T is of order 8.
By Proposition 1.4, Q is a metacyclic group isomor phic to the g roup of part
(v) of our theorem and so Z(Q)
=
C
4
, contrary to the fact that W Z(G).
In what follows we may assume that G has a normal abelian subgroup
W of type (p, p) and for each such W , G/W is abelian. This gives that
G
W and so G
is elementary a belian of order p or p
2
. Also, G has no
abelian subgroup of type (p, p, p). Indeed, if E is an abelian subgroup of type
(p, p, p), then considering maximal subgroups of E, we know that they are
normal in G and each of them contains G
which would imply G
= {1}.
(i) Assume that Z(G) is cyclic. Set Z =
1
(Z(G)) so that |Z| = p and let
J be another subgroup of order p. Then J is not normal in G and W = Z × J
is a normal ab elian subgroup of type (p, p). If N = N
G
(J), then |G : N | = p
and N/J 6= {1 } is either cyclic or p = 2 and N/J
=
Q
8
(Proposition 1.7).
First suppose that N/J
=
Q
8
so that |G| = 2
5
. We have W = Z(N)
and since G/N acts faithfully on W , we get Z = Z(G). If X/J is a max imal
subgroup o f N/J, then X/J
=
C
4
and so X is abelian. Hence N ha s at least
three abelian maximal subgroups which implies |N
| = 2 and N
< W so
that N
= Z = Z(G). Let n N W so that n
2
W J. Suppose that
n
2
6∈ Z so that hn
2
i is not no rmal in G. By Proposition 1.7, hn
2
i must be a
maximal cyclic subgroup in G, which is not the case. Thus, n
2
Z and so
Φ(N) = N
= Z. Let Q
= ha, bi be a maximal subgroup of N which does not
contain J = hui so that Q
=
Q
8
and N = J × Q
. Suppose that there is an
element x G N such that o(x) 4. Then x
2
W and D = hW, xi
=
D
8
.
There is i D N such that i is a noncentral involution. By Proposition 1.7,
|G : C
G
(i)| = 2 and the fact that [i, J] 6= {1} gives that C
N
(i) covers N/J so

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