1. Introduction
Let
Q be a finitely generated nilpotent group. The Mislin genus,
, is defined to be the set of isomorphism classes of finitely generated nilpotent groups
R such that for every prime
p, the
p-localizations
and
are isomorphic. Mislin, in [
1], gave a description of the genus set
if
Q is finitely generated with a finite commutator subgroup. In particular, he showed how to compute the order of the genus set
. Furthermore, the authors in [
2] showed that the genus set
, which is finite, admits an abelian group structure with
Q as its identity element.
Let
G be any group. We define the noncancellation set, denoted by
, to be the set of all isomorphism classes of groups
K such that
. Suppose that
G is a finitely generated infinite nilpotent group with a finite commutator subgroup. Warfield’s result [
3], asserts that
.
In this study, all groups considered are supposed to be nilpotent. We therefore study the class
of all finitely generated infinite nilpotent groups with finite commutator subgroups. In particular, we are interested in the class
of semidirect products of the form
, where
T is a finite abelian group and
. Many computations of the genus of nilpotent groups which belong to the class
can be found in the literature [
2] and [
4]. We note that there is no general method for the computation of
when
. Hence, we describe the following subclass of
.
Let
Q be a nilpotent group. Consider the short exact sequence
where
is the torsion subgroup of
Q and
is the torsion-free quotient. We have that
if and only if
is finite and
is free abelian of finite rank. The authors in [
5] defined a subclass
if the following additional conditions hold.
- 1.
and are commutative;
- 2.
The sequence splits for the action ;
- 3.
, where is the center of .
In the presence of (1), condition (2) is equivalent to requiring that for each
there exists
such that
for all
, as observed in [
5].
For a group
, the authors in [
5] gave a general method for the calculation of the genus set
. Let
d be the height of
in
; that is,
, and let
be the multiplicative group of units of
. The authors showed that
In the case where
or 2, the genus of the group
Q is trivial. If
has a nontrivial genus, then
is cyclic. The only calculations of the genus of a group in
are those groups in the class
or the direct products of groups in
[
2,
4,
6,
7,
8].
For any
m prime to
d, there is a group
in the genus of
Q. The groups
are all supposed to be finitely generated infinite nilpotent. The authors in [
5] proved that
, and that
consists of isomorphism classes of groups
. That is, there is a correspondence to
, and this correspondence provides an isomorphism by (
1). The groups
have the same properties as those of
Q.
The calculation of the Mislin genus was extended from the class
to a subclass in
of direct products of
k copies of
Q, where
. If the direct product
of groups
, where
involves a group
with a noncyclic torsion-free quotient
, then the genus,
is trivial. The direct product of groups
, each with a cyclic torsion-free quotient
, is a group
with a noncyclic torsion-free quotient
. The authors in [
4] proved that
need not be trivial. Let
, and
be the
direct power of
Q, where
. The authors in [
5] proved that there is a surjective homomorphism
given by
, where
. Suppose that
p is a prime. Let the torsion subgroup
be a
p-group, and
be cyclic. The authors further showed that
is an isomorphism.
If , then for we have that if and only if Q is itself commutative since, in general, does not inherit condition (3) from Q.
Mislin, in [
1], defined the genus set for groups in the class
, of all finitely generated infinite nilpotent groups with finite commutator subgroups. This paper is a further contribution to the notion of the Mislin genus of groups in the class
. For further studies regarding the genus set of a group not in the class
, see [
9]. In
Section 2, we discuss the group structure of the genus of a group
computed by the author in [
10]. We construct a pullback
of
l-equivalences
and
,
, where for a prime
p and
, we have
. We compute the group structure of the genus of the group
by studying the relationship between
and
.
Let
be a relatively prime pair of integers. Let
and
be
-groups in
. In
Section 3, we prove that the group of the form
is a subgroup of the direct product
. In this case, the group
L is a pullback. Moreover, we compute the structure of the genus set
by a slight generalization of the results from [
10]. Further, we describe the group
such that
but
; and show that
consists of groups
.
Section 4 then closes with exact calculations for our general construction of
from
Section 3. We show that
and that
is a surjective group homomorphism, which then allows us to determine the exact structure and order of the genus group of
L since we can calculate
.
2. Relations between Genus Groups
We discuss the genus set
for
H, a finitely generated infinite nilpotent group with finite commutator subgroup. Let us first recall some notation from [
2] and [
1]. Denote by
the center of
H and by
the torsion subgroup of
. The group
is the free center of
H. In particular, the quotient
H/
FZH denotes the group
QH. Then
QHab is the abelianization group of
QH.
Now let
p be a prime and let
. Suppose
, where
and
is prime to
. Let the cyclic group
generated by
act on
by
where
Then, we consider
Let be the order of . Note that the case where is excluded, since fails to be the order of . It is clear to see that the torsion subgroup is finite and the torsion-free quotient is free abelian of finite rank. Therefore, .
In view of Proposition 3.2 of [
11], we have the following proposition.
Proposition 1. Let , then the p-part of the exponent of is given by
Proof.
Assume that L nilpotent, then we find the following:
, where ,
,
and
.
Thus, it follows that for all i. □
Let
P be the set of prime divisors of the exponent of the torsion subgroup of
H. Let
denote the semigroup of
P-automorphisms of
H. Since
, the exact sequence [
2]
is used in the computation of
.
Let
be represented by
. The author in [
1] defined a surjective group homomorphism
by
, where
. Now suppose that
is a
P-equivalence; Theorem 1.4 of [
2] shows that
if and only if
. This implies that
. Using this fact, and given that
is multiplicative, we have that
is a subgroup of
. We can also show that
Thus,
. It is then easy to prove that
. Since
is a surjective group homomorphism, by the first isomorphism theorem, we have that
It is clear that the genus
is a finite group; although, to deduce more about its structure, one must examine the image of
, which, by [
10], has order
. Thus, the genus group of
H is a cyclic group of order
, as shown by the author in [
10].
Now, we use the notion of pullback in the category of nilpotent groups to describe the structure of the group
. We start by constructing a pullback
from the
l-equivalences
and
, where
. Let
, and
H be nilpotent groups. Consider the following commutative diagram [
2] of group homomorphisms.
We prove the following proposition.
Proposition 2. Let H, , and be nilpotent groups. Then with is a nilpotent group and the diagram in Figure 1 is a pullback. Proof.
Let
. Then
is a subgroup of
.
Figure 2 is a pullback diagram of the diagram in
Figure 1. We show here that
is an isomorphism. [
12]. The two morphisms
and
define a unique homomorphism
such that
and
. We must prove that
is an isomorphism. Let
e,
,
, and
be the identity elements of
,
,
, and
, respectively. Let
. Then, we have that
Suppose that
is injective. Then, since the diagram in
Figure 1 is commutative, we have that
. Thus, by Lemma 1.4 (iii) of [
13] we have that
and
are injective. Therefore,
and so
. Hence,
, and
is injective.
Now let . Since , we have that . By assumption, we can write for some a in . Now, suppose that for some . Then . Thus, is surjective and is consequently an isomorphism. □
Theorem 1. Let and let . Then is an isomorphism.
Proof. We have that
. This implies that the order of
is
s. Recall that the genus group of
H is a cyclic group of finite order
s. Thus,
is a bijection since
. Now, we have that
and
are cyclic groups; hence, we may find generators
and
X, respectively. Define the bijection
by
, for some
. Let
. Since
generates
, there exist
such that
. Thus,
Thus, is a bijective homomorphism and is therefore an isomorphism. □
3. Group Structure on the Genus of A Group
Let act on , that is, we have a nontrivial group homomorphism . Suppose that L is given by , where the action is given by . Then the group L is expressible as a short exact sequence where is the torsion subgroup of L and is the torsion-free quotient. Since is finite and is free abelian of finite rank, we have that . We want to interpret the structure of the group . We will begin by proving some results on the group L.
Proposition 3. Let be a nontrivial group homomorphism given by , . Let K, G, and M be groups such that and . Let Figure 3 be a commutative diagram of group homomorphisms. Define a subset L of byThen L is a subgroup of . Proof.
Let be a nontrivial group homomorphism given by , . Let . Then, by definition, .
The multiplication on the set
L is given by
. Then we have
Thus, . Therefore,
.
Therefore, . Note that L is closed under the operation of . Thus, the result follows. □
Proposition 4. Let be a relatively prime pair of integers. Let be a group action given by , . Let , and . Suppose that α is injective. Then, the commutative diagram Figure 4 is a pullback if and only if the following conditions hold: - (1)
;
- (2)
;
- (3)
.
Proof.
The two morphisms ϕ and ψ define a unique homomorphism . If the diagram is a pullback, then γ is an isomorphism and condition (1) and (2) are satisfied. Now we prove condition (3). Let . By assumption, for some . Then . Since α is injective, , and so . Thus .
Conversely, let . Then there is such that . Since α is injective, we have that . Thus, .
If the three conditions hold, then γ is injective by condition (1). Now let us take . Then, since , we have that , so for some by condition (2). Now, suppose that , for some . Then . Thus γ is surjective. Therefore, the diagram is a pullback. □
We fix some notation.
Let
be generated by
and suppose that
. Then
L is given by
. Note that the subgroup
is normal in
L, and so
, and similarly
. It is given that the group
L is non-abelian, hence
and
. We define two groups by
and
where
. We also have that
The actions of these groups are given by
and
, where
,
and
u is relatively prime to
n of order
d, respectively. We note that
.
The properties of the group
were discussed in the papers [
5,
14,
15,
16]. Here, our focus is to study the relationship between the groups
and
L. The author in [
14] proved the following theorems.
Theorem 2. Let . Then .
Theorem 3. Let and let be a cyclic group. Then
Let denote the order of . Set , , and let be the order of , where for all i. We prove the following theorem.
Theorem 4. Let and be abelian groups. The group is nilpotent.
Proof.
Let , with and be the order of where for all i. Let us write and . Let , be the terms of the lower central series of the group L, where . Then .
Therefore, for sufficiently large r, we have that if and only if for some i. Therefore, the group L is nilpotent. □
Remark 1. Since , it is clear to see that the commutators are powers of . Thus, , and is obviously finite.
We note that the torsion subgroup of L is noncyclic. Therefore, condition (3) of the definition of is not automatically satisfied. Hence, we prove the following proposition.
Proposition 5. Let with and commutative. Then L is contained in .
Proof.
We only need to prove condition (3) of the definition of . Let be a group homomorphism given by with , for all and . The image is the inner automorphism group since it is a subgroup of . Since is abelian, then . Therefore, . This implies that . The identity element commutes with every element of a group. Hence, must commute with every element of . Therefore, is contained in the center of . Henceforth, . □
By Proposition 1, we note that the exponent of the group
is given by
. We now proceed with the analysis of the structure of the genus
. Recall the exact sequence (
2)
Theorem 5. Let with a cyclic torsion-free quotient, . Then is a finitely generated cyclic group.
Proof.
Suppose that the sequence (
4) is exact, which implies that
is a surjective group homomorphism. In addition, we have that
. Indeed, since
δ is a surjective group homomorphism, we have that
is a subgroup of
. Further, we have that
θ is multiplicative, so that
is a subgroup of
, consisting of integers
m prime to
such that
for all
i, where
is the order of
. Therefore,
, and since we can conclude that
, then by the first isomorphism theorem,
□
Since is a cyclic group, we can find a generator . Let c be the least exponent of h such that , where c = pvi−1(p−1)/2.
This gives us the description of the genus group of L,(L). However, we still do not know the exact order of this genus group. In the next section, we compute this order.
In the following theorem we want to prove that for each prime p, the two groups L and Lm such that L ≇ Lm are p−equivalent. We will denote by P′ the set of primes not in P.
Definition 1. Let L be a nilpotent group. We call the group L P-local if and only if for any , is a bijection for all .
Theorem 6. Let . Then .
Proof.
Let
. Suppose that there exists a positive integer
m such that
and
. Let
and let
be defined by
,
for all
. We claim that
is a group homomorphism. Indeed, in
we have that
, so that
, but since
, then
. Thus,
ϕ is a group homomorphism and yields a commutative diagram [
5]
Figure 5.
Since , then is P-local, and hence . If we localize at p, then ϕ is a isomorphism. Thus, ϕ is a equivalence, and therefore L is P-equivalent to . □
The authors in [
5] used Theorem 6 and Theorem 5 to prove the following corollary.
Corollary 1 (Corollary 2.3 [
5]).
The group consists of isomorphism classes of groups . Remark 2. If L and are nilpotent groups, then they are in the same Mislin genus.
4. A Cyclic Genus of Finite Order
Consider the groups
,
, and
, where
is a nontrivial homomorphism given by
, and
is a nontrivial homomorphism given by
, respectively. Consider the pullback diagram of group homomorphisms
Figure 6.
Proposition 6. Let . Let and be generated by and , respectively. Suppose that . Then the generators of G are subject to the relations and .
Proof.
Let
and
. Let
be a nontrivial group homomorphism. Then
G is generated by
. We have that the subgroup
is normal in
G; therefore,
. Note that since
G is a non-abelian group, we have that
. Thus, we have that
where
. □
Proposition 7. Let , for all . Let and suppose that . Then K is generated by the same generators as that of the groups .
Proof.
Let and . Suppose that is a nontrivial group homomorphism. Similarly to proposition 6, the group has generators and relations x and y satisfying and . Furthermore, since K is the direct product of each , for all , we have that K is generated by the same generators. □
The torsion subgroup of
K is finite and the torsion-free quotient is free abelian of finite rank, therefore we have that
. Let
, with
acting on
by
where
u is relatively prime to
.
Proposition 8. Let . Then the genus, , is trivial.
Proof.
Let , where each . Since the direct product of two or more infinite cyclic groups is not cyclic, then the torsion-free quotient is not cyclic. This implies that the genus of K is trivial. Furthermore, the torsion-free quotient of each is cyclic. Therefore, need not be trivial. However, since the torsion subgroup is a cyclic p-group, where p is a prime, and the torsion-free quotient is cyclic, then must be isomorphic to . Thus, is also trivial. □
We observe that if we take the direct product , then . From here, we can discuss the genus of this direct product. In the direct product of K and G, the torsion-free quotient is noncyclic; thus, the genus of this direct product, , is trivial. From this, we can deduce that is an isomorphism. In addition, we have that .
Proposition 9. Let with a cyclic torsion-free quotient . Then the genus of G is trivial.
Proof.
It is sufficient for us to show that or 2. Let and let , , . We have that if or if . In the second case, since , we have that , and so . Therefore, ; thus, the genus is trivial. □
In the next theorem, we determine the order of the genus of the group L. In particular, we prove that is nontrivial. For the proof of the theorem we need the following proposition, but let us first fix some notation.
Notation. Let a be an integer. We denote by the class which contains all elements a c, where c is the order of the genus group, .
Proposition 10. Let . Suppose that L is a subgroup of . Then is a surjective group homomorphism.
Proof.
Let
and let
be defined by
, for all
, where
c is the order of
. Let
, then
Hence,
is a homomorphism and is obviously surjective since every element in
is mapped to the identity element. □
Theorem 7. Let with a cyclic torsion-free quotient group. Then the group is a cyclic group of order 2.
Proof.
Since is mapped to the identity element of , the kernel, must be the group itself. This implies that . In Theorem 5, we showed that is a cyclic group. Therefore, the group can only have one subgroup of order in , which must be the group itself. Then must have only two subgroups: the trivial subgroup and the group itself. Therefore, we can conclude that the genus group of L is cyclic of order 2. □
Example 1. Let with . Let the action of on be given by and , for all . Thus, and . It is clear that . We claim that is nontrivial. Indeed, . Thus, .
In Proposition 4, we proved that the group is a pullback, where is a nontrivial homomorphism given by . We then proceeded to show that the pullback L is nilpotent with a finite commutator subgroup. In particular, we proved in Proposition 5 that L is contained in the class .
We then considered the groups K and G, where . We proved that is contained in . Since is not abelian, K is not contained in . In this case, the condition that , where is the center of and is the action of on , fails to hold; in general, direct products do not inherit this condition. Therefore, we see the importance of this condition in obtaining the genus of a group in . Propositions 8 and 9 show that the groups and are both trivial, respectively, and therefore are isomorphic to each other.
Finally, in Proposition 10, we showed that there is a surjective homomorphism between the groups and . This surjective homomorphism enabled us to prove that the genus of L, is nontrivial. In particular, we showed in Theorem 7 that is a cyclic group of order 2.
It is clear that the group is an abelian group of finite order. We, therefore, realize an abelian group as a Mislin genus. Now that we have proven that the genus of a pullback L is abelian, we want to further study its subgroups in more detail.