Sets of prime power order generators of finite groups
A subset X of prime power order elements of a finite group G is called pp-independent if there is no proper subset Y of X such that 〈Y,Ф(G)〉 = 〈X,Ф(G)〉, where Ф(G) is the Frattini subgroup of G. A group G has property Bpp if all pp-independent generating sets of G have the same size. G has the pp-ba...
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Цитувати: | Sets of prime power order generators of finite groups / A. Stocka // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 1. — С. 129–138. — Бібліогр.: 12 назв. — англ. |
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irk-123456789-1885082023-03-04T01:27:05Z Sets of prime power order generators of finite groups Stocka, A. A subset X of prime power order elements of a finite group G is called pp-independent if there is no proper subset Y of X such that 〈Y,Ф(G)〉 = 〈X,Ф(G)〉, where Ф(G) is the Frattini subgroup of G. A group G has property Bpp if all pp-independent generating sets of G have the same size. G has the pp-basis exchange property if for any pp-independent generating sets B₁,B₂ of G and x ∈ B₁ there exists y ∈ B₂ such that (B₁ \ {x}) ∪ {y} is a pp-independent generating set of G. In this paper we describe all finite solvable groups with property Bpp and all finite solvable groups with the pp-basis exchange property. 2020 Article Sets of prime power order generators of finite groups / A. Stocka // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 1. — С. 129–138. — Бібліогр.: 12 назв. — англ. 1726-3255 DOI:10.12958/adm1479 2010 MSC: Primary 20D10; Secondary 20F05 http://dspace.nbuv.gov.ua/handle/123456789/188508 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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A subset X of prime power order elements of a finite group G is called pp-independent if there is no proper subset Y of X such that 〈Y,Ф(G)〉 = 〈X,Ф(G)〉, where Ф(G) is the Frattini subgroup of G. A group G has property Bpp if all pp-independent generating sets of G have the same size. G has the pp-basis exchange property if for any pp-independent generating sets B₁,B₂ of G and x ∈ B₁ there exists y ∈ B₂ such that (B₁ \ {x}) ∪ {y} is a pp-independent generating set of G. In this paper we describe all finite solvable groups with property Bpp and all finite solvable groups with the pp-basis exchange property. |
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Stocka, A. |
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Stocka, A. Sets of prime power order generators of finite groups Algebra and Discrete Mathematics |
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Stocka, A. |
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Stocka, A. |
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Sets of prime power order generators of finite groups |
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Sets of prime power order generators of finite groups |
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Sets of prime power order generators of finite groups |
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Sets of prime power order generators of finite groups |
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Sets of prime power order generators of finite groups |
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sets of prime power order generators of finite groups |
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Інститут прикладної математики і механіки НАН України |
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2020 |
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http://dspace.nbuv.gov.ua/handle/123456789/188508 |
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Sets of prime power order generators of finite groups / A. Stocka // Algebra and Discrete Mathematics. — 2020. — Vol. 29, № 1. — С. 129–138. — Бібліогр.: 12 назв. — англ. |
series |
Algebra and Discrete Mathematics |
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AT stockaa setsofprimepowerordergeneratorsoffinitegroups |
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2025-07-16T10:36:01Z |
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2025-07-16T10:36:01Z |
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1837799479049191424 |
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“adm-n1” — 2020/5/14 — 19:35 — page 129 — #137
© Algebra and Discrete Mathematics RESEARCH ARTICLE
Volume 29 (2020). Number 1, pp. 129–138
DOI:10.12958/adm1479
Sets of prime power order generators
of finite groups∗
A. Stocka
Communicated by I. Ya. Subbotin
Abstract. A subset X of prime power order elements of
a finite group G is called pp-independent if there is no proper subset
Y of X such that 〈Y,Φ(G)〉 = 〈X,Φ(G)〉, where Φ(G) is the Frattini
subgroup of G. A group G has property Bpp if all pp-independent
generating sets of G have the same size. G has the pp-basis exchange
property if for any pp-independent generating sets B1, B2 of G and
x ∈ B1 there exists y ∈ B2 such that (B1 \ {x}) ∪ {y} is a pp-
independent generating set of G. In this paper we describe all finite
solvable groups with property Bpp and all finite solvable groups with
the pp-basis exchange property.
1. Introduction
Throughout this paper, all groups are finite. Let G be a group. We
denote by Φ(G) the Frattini subgroup of G and we call a group with
the trivial Frattini subgroup a Frattini-free group. For other notation,
terminology and results one can consult for example [3, 4].
In this paper, our purpose is to extend the famous theorem of Burnside
known as Burnside basis theorem. This theorem provides that the Frattini
quotient of every p-group is an elementary abelian p-group. Hence it can
∗This article has received financial support from the Polish Ministry of Science and
Higher Education under subsidy for maintaining the research potential of the Faculty
of Mathematics and Informatics, University of Białystok.
2010 MSC: Primary 20D10; Secondary 20F05.
Key words and phrases: finite groups, independent sets, minimal generating
sets, Burnside basis theorem.
https://doi.org/10.12958/adm1479
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130 Sets of prime power order generators
be view as a vector space over the field of order p. So generating sets of
p-groups share properties with generating sets of vector spaces. However
generating sets outside the class of p-groups do not have such properties,
even generating sets of cyclic groups whose order is divisible by at least
two different primes.
Obviously all elements of p-groups have prime power orders. So also in
arbitrary groups we want to consider sets of prime power order generators.
In this purpose we introduce the concept of a pp-element which simplifies
our considerations. So we say that an element g ∈ G is a pp-element if it
has prime power order, while by p-element, as usual, we mean an element
of order being a power of a prime p. Many authors have studied similar
problems concerning sets of not only pp-generators, see for instance [1, 8,
9, 11] and the reference therein. In particular in [1] groups in which all
minimal generating sets have the same size are classified.
A subset X of pp-elements of a group G will be called pp-independent
if 〈Y,Φ(G)〉 6= 〈X,Φ(G)〉 for every Y ⊂ X and a pp-base of G if X is
a pp-independent generating set of G. We say that a finite group G
• has property Bpp (is a Bpp-group for short) if all pp-bases of G have
the same size;
• has the pp-embedding property if every pp-independent set of G can
be embedded to a pp-base of G;
• has the pp-basis exchange property if for any two pp-basis B1, B2 and
x ∈ B1 there exists y ∈ B2 such that (B1 \ {x}) ∪ {y} is a pp-base
of G.
• is a pp-matroid group if G has property Bpp and the pp-embedding
property.
In view of the above definitions, Burnside basis theorem provides that
all finite p-groups are pp-matroid and have the pp-basis exchange property.
Another example, outside the class of p-groups, is a group called a scalar
extension. After [6] we say that G is a scalar extension if G = P ⋊ Q,
where P is an elementary abelian p-group, Q is a non-trivial cyclic q-group
for distinct primes p 6= q such that Q acts faithfully on P and the Fp[Q]-
module P is a direct sum of isomorphic copies of one simple module. This
construction will be constantly use in our further considerations. A scalar
extension is not always a pp-matroid group (only if Q has prime order,
see [11]) but every scalar extension is a Bpp-group (see [8]).
Our focus of interest is to study the structure of groups which have
one of the properties listed above. Solvable groups with the pp-embedding
property were studied in [10, 11]. In [7] all pp-matroid groups were de-
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A. Stocka 131
scribed. Moreover in [7] it was proved that pp-matroid groups have the
pp-basis exchange property.
The properties of pp-matroid groups imply that every maximal pp-
independent set of a pp-matroid group G is a pp-base of G. Let I be
the family of all pp-independent sets of G. Then the pair (I, G) forms
a matroid where every pp-base of G is a base of a matroid (I, G) (see [12]).
Hence pp-matroid groups can be view as a generalization of p-groups in
the sense of generating sets. Thus the aim of this paper is to describe
the structure of solvable groups with property Bpp and the structure of
solvable groups with the pp-basis exchange property.
By [6, Theorem 4.2], we know that every pp-independent set (pp-base)
of G/Φ(G) may be lifted to a pp-independent set (pp-base) of G. Hence
using properties of the Frattini subgroup we obtain the following
Theorem 1.1. A group G has property Bpp, the pp-embedding property,
the pp-basis exchange property if and only if G/Φ(G) has, respectively,
property Bpp, the pp-embedding property and the pp-basis exchange property.
In particular G is pp-matroid if and only if G/Φ(G) is pp-matroid.
Based on the above theorem we may restrict our consideration to
Frattini-free groups. The structure of the paper is as follows. We present
our concepts and main results in Sections 1. In Section 2 we present
the classification of all solvable groups with property Bpp. The proof of
Theorem 1.2 is presented in Section 3.
Theorem 1.2. Let G be a Frattini-free solvable group. Then G has prop-
erty Bpp if and only if it is one of the following groups:
1) an elementary abelian p-group;
2) a scalar extension;
3) a direct product of groups given in (1) and (2) with pairwise coprime
orders.
Using the above theorem we describe in Section 3 solvable groups with
the pp-basis exchange property. The proof of Theorem 1.3 is presented in
Section 4.
Theorem 1.3. Let G be a Frattini-free solvable groups. Then G has the
pp-basis exchange property if and only if G is a Bpp-group such that all
pp-elements of G have prime orders.
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132 Sets of prime power order generators
2. Groups with property Bpp
In this section we present the classification of groups with property
Bpp. First results concerning a Bpp-groups appear in [6,8]. We recall some
of these results which we will apply in further proofs.
Theorem 2.1 ([6]). Let G = P ⋊Q be a non-trivial semidirect product,
where P is a p-group and Q is a cyclic q-group, for distinct primes p 6= q.
Then the following conditions are equivalent:
1) G is a Bpp-group.
2) G/Φ(G) is a scalar extension.
Furthermore, suppose that the above conditions hold. Then all minimal
generating sets of G have the same size.
Theorem 2.2 ([8]). Let G be a group and G/Φ(G) be a scalar extension.
Then
1) G has a unique Sylow p-subgroup P ;
2) G = P ⋊Q for a Sylow q-subgroup Q and all Sylow q-subgroups of
G are cyclic;
3) Φ(G) = Φ(P )× 〈xq
m
〉, where x is a generator of Q. Moreover, xq
m
centralizes P.
Theorem 2.3 ([6]). If G is a Bpp-group, then every homomorphic image
of G is also a Bpp-group.
Theorem 2.4 ([6]). Let G1 and G2 be groups with coprime orders. Then
G1 and G2 are Bpp-groups if and only if G1 ×G2 is a Bpp-group.
3. Solvable Bpp-groups
In this section we investigate the finite solvable Bpp-groups. The
following lemmas will be needed for proving Theorem 1.2.
Remark 3.1. Let G be a solvable group and G = P ⋊ H, where P is
a minimal normal subgroup of G and CH(P ) = 1. Assume that d > 2
is a size of a minimal generating set of G. Theorem 7 of [2] follows
that there exists a minimal generating set {h1, . . . , hd} of H such that
〈hx1
1
, . . . , hxd
d 〉 = G for some x1, . . . , xd ∈ P. We say then, after the authors,
that (P,H) does not satisfy the strong complement property.
Lemma 3.2. Let G be a solvable group with a minimal normal p-subgroup
P, and let H be a complement to P in G, where p is a prime and p does
not divide |H|. Assume that H is a non-cyclic q-group for some prime q
“adm-n1” — 2020/5/14 — 19:35 — page 133 — #141
A. Stocka 133
or H/Φ(H) is a scalar extension. If H acts non-trivially on P, then G is
not a Bpp-group.
Proof. By assumption, H is a q-group or H/Φ(H) is a scalar extension.
Hence, by Theorem 2.1, all minimal generating sets of H have the same
size, say d. Assume d > 2. Since H acts non-trivially on P, by Re-
mark 3.1 there exists a minimal generating set {h1, . . . , hd} of H such that
〈hx1
1
, . . . , hxd
d 〉 = G for some x1, . . . , xd ∈ P. Observe that {hx1
1
, . . . , hxd
d }
is a generating set of pp-elements of G. Hence there exists a pp-base
B′ ⊆ {hx1
1
, . . . , hxd
d } of G such that |B′| 6 n.
On the other hand P = 〈a〉H , for every 1 6= a∈P. Hence {a, h1, . . . , hn}
is a pp-base of G. Thus G is not a Bpp-group.
Lemma 3.3. Let G be a solvable group with a minimal normal p-subgroup
P, and let H be a nilpotent complement to P in G, where p is a prime
and p does not divide |H|. If G is an indecomposable Bpp-group, then H
is a cyclic q-group for some prime divisor q of |H|.
Proof. Assume that H = P1 × . . .× Pn, where Pi is a Sylow pi-subgroup
of H and [P, Pi] 6= 1, for i = 1, . . . , n. Let 1 6= a ∈ P and Bi be a pp-
base of Pi, for i = 1, . . . , n. Then {a} ∪ B1 ∪ . . . ∪ Bn is a pp-base
of G. Moreover assume that B1 = {x1, . . . , xk} and B2 = {y1, . . . , yl}.
Hence there exist c1, . . . , ck ∈ P such that xa1 = c1x1, . . . , x
a
k = ckxk.
Observe that (xai )
yj (xai )
−1 = (cixi)
yj (cixi)
−1 = c
yj
i c−1
i 6= 1 for at least one
j ∈ {1, . . . , l}. So {xa1, . . . , x
a
k, y1, . . . yl} ∪B3 ∪ . . . ∪Bk is a pp-base of G,
a contradiction. Hence only one Pi acts non-trivially on P. Without loss
of generality we can set i = 1. Then G = (P ⋊ P1)× (P2 × . . .× Pn), in
contradiction to our assumption. This contradiction implies that H = P1.
Thus, by Lemma 3.2, H is a cyclic q-group with q = p1.
Lemma 3.4. Let G be a solvable group with a minimal normal p-subgroup
P1, where p is a prime and let H be a complement to P in G. Assume
that H = Q⋊ P2, where Q is a q-group for some prime q 6= p and P2 is
a cyclic p-group such that H/Φ(H) is a scalar extension. Then G is not
a Bpp-group.
Proof. Since P1 is a minimal normal subgroup of G and G is solvable, P1
is elementary abelian and 〈g〉H = P1 for all 1 6= g ∈ P1. Let P2 = 〈y〉
and {x1, . . . , xn} ⊆ Q be a minimal set such that 〈x1, . . . , xn〉
P2 = Q.
Hence, by the assumption, {g, x1, . . . , xn, y} is a pp-base of G. We need
to consider the following cases:
1. [P1, Q] 6= 1, [P1, P2] 6= 1;
“adm-n1” — 2020/5/14 — 19:35 — page 134 — #142
134 Sets of prime power order generators
2. [P1, Q] = 1, [P1, P2] 6= 1;
3. [P1, Q] 6= 1, [P1, P2] = 1;
4. [P1, Q] = 1, [P1, P2] = 1.
1. In this case there exists a ∈ P1 such that Qa 6= Q. From [4, Theorem
2.3], we know that P1 = CP1
(Q)× [P1, Q]. Assume first that CP1
(Q) 6= 1.
Since P1 is a minimal normal subgroup of G, there exists a ∈ CP1
(Q)
such that ay
−1
/∈ CP1
(Q). Set b = ay
−1
. Then by = a ∈ CP1
(Q) and
b−1by
−1
/∈ CP1
(Q). It follows that Qb 6= Qby
−1
. Since H/Φ(H) is a scalar
extension,
Q/Φ(Q) = Q1Φ(Q)/Φ(Q)× . . .×QnΦ(Q)/Φ(Q),
where QiΦ(Q)/Φ(Q) is a simple Fq[Q]-module. Hence 〈xi〉
Q = Qi, for
every xi ∈ Qi \ Φ(Q). Moreover Q = Q1 · . . . · Qn. It follows that there
exists Qi such that Qb
i 6= Qi, for some i = 1, . . . , n. Thus at least one
element, say xi, satisfies xbi /∈ Qi. Consider the set X = {y, xb1, . . . , x
b
n}.
Observe that
xbyi = xyy
−1by = (xyi )
by .
Since xyi ∈ Q and by ∈ CP1
(Q), we have xbyi = xyi ∈ 〈X〉. Hence xi ∈ 〈X〉
and further c = x−1
i xbi ∈ P1 ∩ 〈X〉, where c 6= 1. It follows that G = 〈X〉
and X is a generating set of pp-elements of G. So there exists a pp-base
B ⊆ X of G such that |B| < n+ 2 = |{g, x1, . . . , xn, y}|.
Assume now that CP1
(Q) = 1 and CP1
(P2) 6= 1. So P1 = [P1, Q] and
hence there exists c ∈ CP1
(P2), where c = [a, x−1
1
], for some a ∈ P1 and
x1 ∈ Q\Φ(Q). Thus there exist x2, . . . , xn ∈ Q such that 〈x1, . . . , xn〉
P2 =
Q. Let X = {xa1, . . . , x
a
n, y}. Since xa1 = [a, x−1
1
]x1 = cx1 for some 1 6=
c ∈ P1, we have (xa1)
−1(xa1)
y = x−1
1
xy
1
. Moreover 1 6= x−1
1
xy
1
∈ 〈X〉. Hence
〈x1〉
P2 ⊆ 〈X〉 and x1 ∈ 〈X〉. So 1 6= x−1
1
xa1 ∈ 〈X〉 ∩ P1 and G = 〈X〉. It
follows that {a, x1, . . . , xn, y} and {y, xa1, . . . , x
a
n} are a pp-base of G.
2. Now there exists a ∈ P1 and at least one xi, say x1 such that
[a, y] 6= 1 6= [x1, y]. Then ya(yx1)−1 = a−1ay
−1
xy
−1
1
x1 6= 1. It follows that
{ya, yx1 , x2, . . . , xn} is a pp-base of G.
3. Since [P1, Q] 6= 1, there exists a ∈ P1 such that Q 6= Qa. Moreover
P1 = [P1, Q]. Otherwise CP1
(Q) is a normal subgroup of G, contradicting
the minimality of P1. Hence there exist c1, . . . , cn ∈ P1 such that xa1
1
=
c1x1, . . . , x
a
n = cnxn. So we obtain (xai )
−1(xai )
y = x−1
i xyi 6= 1. It follows
that {xa1, . . . , x
a
n, y} is a pp-base of G.
4. In this case {g, x1, x2 . . . , xn, y} and {gx1y, x2 . . . , xn, y} are pp-
bases of P ×H.
Hence G is not a Bpp-group in all the cases. So the proof is complete.
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A. Stocka 135
Remark 3.5. Let G be a solvable group with a minimal normal p-
subgroup P, where p is a prime and let H be a complement to P in
G. Assume that H/Φ(H) is a scalar extension. It follows, by Theorem 2.1,
that H is a Bpp-group and we may assume that d is the size of every
pp-bases of H, for some positive integer d. Then from proofs of Lem-
mas 3.2, 3.4 we immediately deduce that there exist pp-bases B1 and B2
of G such that |B1| = n+ 1 and |B2| < n+ 1.
Proof of Theorem 1.2. Let G be a Frattini-free solvable group with prop-
erty Bpp. We use induction on |G|. Let P = Op(G) be a maximal normal
p-group of G, for some prime p. Hence Φ(P ) 6 Φ(G) = 1 and P is an
elementary abelian p-group. By [3, Theorem 10.6], there exists a subgroup
H of G such that G = P ⋊ H. From Theorem 2.3, H is a Bpp-group.
So by the induction assumption H = H1 × . . . × Hk, where Hi/Φ(H)i
is an elementary abelian q-group or a scalar extension for i = 1, . . . , k.
By [3, Theorem 10.6], P = P1 × . . .× Pn, where Pi is a minimal normal
subgroup of G, for i ∈ {1, . . . , n}.
Let ai ∈ Pi be a non-trivial element for i ∈ {1, . . . , n} and {h1, . . . , hr}
be a pp-base of H. Then B = {a1, . . . , an, h1, . . . , hr} is a pp-base of G.
Assume that Pi and Pj are not isomorphic as Fp[H ]-module for some i 6= j.
Then B′ = (B \ {ai, aj}) ∪ {aiaj} is a pp-base of G. Since |B′| = |B| − 1,
G is not a Bpp-group, a contradiction. So all the Pi are isomorphic to each
another as Fp[H ]-modules. In particular, this implies that CH(Pi) = 1 for
each Pi.
Again, by Theorem 2.3, we may suppose that P is a minimal normal
subgroup of G. Thus G = P ⋊ (H1 × . . .×Hk) and P = 〈a〉H , for every
1 6= a ∈ P. Assume that Bi is a pp-base of Hi for i = 1, . . . , k. Then
{a} ∪B1 ∪ . . . ∪Bk is a pp-base of P ⋊H, as (|Hi|, |Hj |) = 1, for i 6= j.
If for some i ∈ {1, . . . , k}, Hi is a q-group, then q 6= p, by the choice of
P. So suppose that H1/Φ(H1) is a scalar extension and [P,H1] 6= 1. Let
P1 = 〈a〉H1 6 P, for some a ∈ P1. Then {a} ∪B1 is a pp-base of P1 ⋊H1.
Moreover, by Remark 3.5, there exists a pp-base, say B of P1 ⋊H1, such
that |B| < |B1|+ 1. Observe that B ∪ B2 ∪ . . . ∪ Bk is a generating set
of pp-elements of G. So there exists a pp-base C ⊆ B ∪B2 ∪ . . . ∪Bk of
G such that |C| < |{a} ∪ B1 ∪ . . . ∪ Bk|, a contradiction. Hence either
[P,H1] = 1 or H1 is a p-group. If [P,H1] = 1 and P and H have not
coprime orders, then by Case 4. of Lemma 3.4 and analogous consideration
as the above, we obtain that G is not a Bpp-group.
It follows that if Hi is not a q-group, then Hi centralises P. It implies
that G = [P⋊(H1×. . .×Hr)]×Hr+1×. . .×Hk. Moreover Hi is a qi- group,
“adm-n1” — 2020/5/14 — 19:35 — page 136 — #144
136 Sets of prime power order generators
where [Hi, P ] 6= 1 and (qi, p) = 1 for i = 1, . . . , r while (|Hi|, |p|) = 1, for
i = r + 1, . . . , k.
Further, by Theorem 2.3, P ⋊ (H1× . . .×Hr) is a Bpp-group. Then, by
Lemma 3.3, only one Hi acts non-trivially on P and such Hi is cyclic. It
follows that G = (P ⋊H1)×H2 × . . .×Hk, where H1 is a cyclic q-group,
and (|P ⋊H1||, Hj |) = 1 for j = 2, . . . , n. Moreover, by Theorems 2.1, 2.3,
P ⋊H1 is a scalar extension or is abelian.
Conversely, let G = G1 × . . .×Gn, where Gi is either an elementary
abelian p-group or a scalar extension and (|Gi|, |Gj |) = 1 for i 6= j.
Then, by Theorem 2.1, every direct factor of G is a Bpp-group. Hence, by
Theorem 2.4, G is a Bpp-group. So the proof is complete.
4. Solvable groups with the pp-basis exchange property
In this section we investigate the structure of finite solvable groups
which have the pp-basis exchange property. We start from the statement
analogous to Theorem 2.4.
Proposition 4.1. Let G1 and G2 be groups with coprime orders. Then
G1 and G2 have the pp-basis exchange property if and only if G1 ×G2 has
the pp-basis exchange property.
Proof. Since G1 and G2 have coprime orders, an element g = g1g2 ∈
G1 × G2, where g1 ∈ G1, g2 ∈ G2, is a pp-element if and only if g = g1
or g = g2. Hence B is a pp-base of G1 ×G2 if and only if B = B1 ∪B2,
where B1, B2 are pp-bases of G1, G2, respectively. From here the result
follows immediately.
Proposition 4.2. Let G be a Frattini-free solvable group. If G has the
pp-basis exchange property, then G has property Bpp and all pp-elements
of G have prime orders.
Proof. Assume that B1, B2 are two pp-bases of G such that |B1| < |B2|.
We choose B1 and B2 with the property that |B2 \ B1| is minimal. Let
x ∈ B2 \ B1. Since G has the pp-basis exchange property, there exists
y ∈ B1 \ B2 such that (B2 \ {x}) ∪ {y} is a pp-base of G. Moreover
|(B2 \ {x}) ∪ {y} \ B1| < |B2 \ B1|. This contradicts the minimality of
|B1 \B2|. So G is a Bpp-group.
Now, by Theorem 1.2, G = G1× . . .×Gk where Gi is either an elemen-
tary abelian p-group or a scalar extension. If Gi is elementary abelian, then
obviously all elements have prime orders. So assume that Gi = P ⋊ 〈x〉 is
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A. Stocka 137
a scalar extension and x is a q-element. Moreover assume xq /∈ CGi
(P ).
Let a1, . . . , as ∈ P be a minimal set such that 〈a1, . . . as〉
〈x〉 = P. Hence
B1 = {a1x, a2, . . . , as, x
q} and B2 = {a1, . . . as, x} are pp-bases of Gi.
Moreover 〈(B1 \ {a1x}) ∪ {y}〉 6= Gi for every y ∈ B2. Hence Gi has not
the pp-basis exchange property. So, by Theorem 4.1, G also has not the
pp-basis exchange property, a contradiction. Hence xq ∈ CGi
(P ). Since
G is Frattini-free, it follows, by Theorem 2.2, that all pp-elements of Gi
have prime orders.
Lemma 4.3. Let G = P ⋊Q be a scalar extension. If all pp-elements of
G have prime orders, then G has the pp-basis exchange property.
Proof. Assume that |Q| = q, then all pp-elements of G have prime orders
and all pp-basis have n elements . Let B1 = {x1, . . . , xn} and B2 =
{y1, . . . , yn} be pp-bases of G. Assume that x1 /∈ B2 and H = 〈x2, . . . , xn〉.
We show that H is a maximal subgroup of G. In this purpose we consider
two cases:
1. x1 ∈ P. Then 〈x1〉
Q is a minimal normal subgroup of G and H =
P/〈x1〉
Q ⋊Qa, where a ∈ P. So H is a maximal subgroup of G.
2. x1 /∈ P. Then x1 is a q-element, where q is a prime and Q is a q-
group. Since H * P there exists in H another q-element, say x2. We
may assume that x1 = xa1 and x2 = xa2 , where a1a
−1
2
/∈ CP (Q). Hence
xa1x−a2 = c ∈ P and c /∈ H. Indeed, if c ∈ H and xa2 ∈ H, then xa1 ∈ H,
a contradiction. It follows, by analogous as in (1), that H is a maximal
subgroup in G.
By assumption, there exists yi /∈ H for some i ∈ {1, . . . , n}. Since H is
a maximal subgroup of G, 〈H, yi〉 = G. It follows that 〈(B1\{x1})∪{y1}〉 =
G and |(B1 \ {x1})∪{y1}| = n. Since G is a Bpp-group, (B1 \ {x1})∪{y1}
is a pp-base of G. The proof is complete.
Proof of Theorem 1.3. It follows immediately from Proposition 4.1, Propo-
sition 4.2 and Lemma 4.3.
Using Theorem 1.2 we obtain
Corollary 4.4. Let G be a Frattini-free solvable group. Then G has the
pp-basis exchange property if and only if it is one of the following groups:
1) an elementary abelian p-group;
2) a scalar extension P ⋊Q, where P is an elementary abelian p-group,
Q has order q for distinct primes p 6= q;
3) a direct product of groups given in (1) and (2) with pairwise coprime
orders.
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138 Sets of prime power order generators
Remark 4.5. By [5], we know that every simple group is generated by
an involution and an element of prime order. So a simple group has
a 2-element pp-base. On the other hand, by the Classification of Finite
Simple Group, we know that every simple group is generated by at least
three involutions, so every simple group has a pp-base which has at least
3 elements. It implies that all simple groups do not have property Bpp.
By the first part of the proof of Proposition 4.2, we may deduced that if
a simple group has not property Bpp, then it has not the pp-basis exchange
property.
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Contact information
Agnieszka Stocka Faculty of Mathematics University of Białystok
K. Ciołkowskiego 1M 15-245 Białystok
E-Mail(s): stocka@math.uwb.edu.pl
Received by the editors: 17.10.2019
and in final form 17.12.2019.
A. Stocka
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