# Groups of Negations on the Unit Square

Authors: Jiachao Wu
Journal: The Scientific World Journal
 (2014)

---

## Abstract

The main results are about the groups of the negations on the unit square, which is considered as a bilattice. It is proven that all the automorphisms on it form a group; the set, containing the monotonic isomorphisms and the strict negations of the first (or the second or the third) kind, with the operator “composition,” is a groupG2 (or G3 or G4, correspondingly). All these four kinds of mappings form a group G5. And all the groups Gi,i=2,3,4 are normal subgroups of G5. Moreover, for G5, a generator set is given, which consists of all the involutive negations of the second kind and the standard negation of the first kind. As a subset of the unit square, the interval-valued set is also studied. Two groups are found: one group consists of all the isomorphisms on LI, and the other group contains all the isomorphisms and all the strict negations on LI, which keep the diagonal. Moreover, the former is a normal subgroup of the latter. And all the involutive negations on the interval-valued set form a generator set of the latter group.


---

## Body

## 1. Introduction

Negations, as a basic operation, play important roles in logic. In [1], the groups of the negations on the unit interval are studied. And in this paper, groups of negations on the bilattice [0,1]2 and on the interval-valued set LI, which can be seen as a sublattice of the unit square, will be studied.Bilattices, introduced by Ginsberg [2] as a uniform framework for inference in artificial intelligence, are algebraic structures that proved useful in many fields [3–5]. And the unit square [0,1]2 is a very special bilattice, which is a subset of the real plane. Thus it is of particular interest. In [6, 7], three kinds of negations on bilattices are introduced. Then one problem arises: are the properties of these negations similar to the negations on [0, 1]? In the third section of this paper, some results will be given.All the vague sets, the interval-valued sets, the intuitionisitic fuzzy sets and the grey sets are isomorphic to sublatticeLI={(x1,x2)∈[0,1]2∣x1≤x2}, with the natural order (x1,x2)≤k(y1,y2) if and only if x1≤y1, x2≤y2. Their properties could be found in [3, 7–23], and so forth. For sometimes, the sets are characterized as L*={(x1,x2)∈[0,1]2∣x1+x2≤1}, with the order (x1,x2)≤t(y1,y2) if and only if x1≤y1, x2≥y2. Both LI and L* are sublattices of the bilattice [0,1]2, with orders ≤k and ≤t. Therefore, similar to [1], the groups of negations on the interval-valued sets can be discussed.The contents will be arranged as follows. In the next section, some basic notions will be given. In the third section, the groups of the strict negations will be studied. And in the fourth section, we will study the negations on the interval-valued sets. At the end of this paper, a section of conclusion is shown.


## 2. Negations on the Bilattice[0,1]2

In [6, 7], three kinds of negations on the bilattice are given. As a particular bilattice, the unit square [0,1]2 also has these kinds of negations. These negations are based on two kinds of orders on the unit square [0,1]2, ≤t and ≤k, defined as, for any x=(x1,x2), y=(y1,y2)∈[0,1]2,
(1)(x1,x2)≤k(y1,y2)ifx1≤y1,x2≤y2,(x1,x2)≤t(y1,y2)ifx1≤y1,x2≥y2,
in which ≤ is the natural order on [0, 1]. Actually, ≤k is the natural order on the unit square.Definition 1.
The first kind negationN¬, called reflection, is a unary operator on the square, satisfying the following properties: ∀x=(x1,x2), y=(y1,y2)∈[0,1]2,(1)
ifx≤ky, then N¬(x)≥kN¬(y);(2)
ifx≤ty, then N¬(x)≤tN¬(y);(3)
N¬(0,0)=(1,1) and N¬(1,1)=(0,0).
The second kind negationN-, named conflation, is a unary operator satisfying(1′)
ifx≤ky, then N-(x)≥kN-(y);(2′)
ifx≤ty, then N-(x)≥tN-(y);(3′)
N-(0,0)=(1,1) and N-(1,1)=(0,0).
The last kind negationN~ is a unary operator satisfying(1′′)
ifx≤ky, then N~(x)≤kN~(y);(2′′)
ifx≤ty, then N~(x)≥tN~(y);(3′′)
N~(0,0)=(0,0) and N~(1,1)=(1,1).For convenience, in this paper, these three kinds of negations are collectedly called negationsN on the unit square, as a bilattice ([0,1]2,≤k,≤t).The composition of two negationsN, N′ is defined as (N∘N′)(x1,x2)=N(N′(x1,x2)). Then, it is not hard to check that N¬∘N- and N-∘N¬ are negations of the third kind; N¬∘N~ and N~∘N¬ are the second kind negations, and N~∘N- and N-∘N~ are of the first kind.Similar to the definition of strict interval-valued negations in [11], the strict negations on the unit square are given as follows.Definition 2.
A negation on the unit square is called strict, if it is continuous and both≤k and ≤t are strict, that is, in Definition 1, if both ≤k and ≤t are replaced by <k and <t in the premise, then the conclusion will be changed to <k and <t correspondingly.Example 3.
LetN1(x1,x2)=(n1(x1),n2(x2)) and N2(x1,x2)=(n1(x2),n2(x1)), with
(2)ni(a)={0,ifa=1,ci,ifa∈(0,1),1,ifa=0,
in which ci, i=1,2 are constants in [0, 1]. Then both N1 and N2 are negations, but neither of them is strict.Definition 4.
If a negationN satisfies N(N(x))=x, then it is called an involutive negation.The mappingNs,1 defined as Ns,1(x1,x2)=(1-x2,1-x1), for all (x1,x2)∈[0,1]2, is the standard negation of the first kind and named as the first standard negation. The mapping Ns,2 defined as Ns,2(x1,x2)=(1-x1,1-x2), for all (x1,x2)∈[0,1]2, is called the second standard negation. And Ns,3(x1,x2)=(x2,x1) is called the third standard kind negation. It is obvious that Ns,1, Ns,2, and Ns,3 are involutive.Obviously, each involutive negation is strict, but the converse is not valid.Example 5.
The following negation is strict but not involutive:(3)N3(x1,x2)=(1-x22,1-x1).Lemma 6 (see [6]).
Each of the first kind strict negations could be characterized asN¬(x1,x2)=(n1(x2),n2(x1)), with n1, n2 being strict negations on [0, 1], and every second kind strict negation could be characterized as N-(x1,x2)=(n1(x1),n2(x2)), with n1 and n2 being strict negations on [0, 1].
Each of the first kind involutive negations could be characterized asN¬(x1,x2)=(n1(x2),n2(x1)), with n1=n2-1, negations on [0, 1], and every second kind involutive negation could be characterized as N-(x1,x2)=(n1(x1),n2(x2)), with n1 and n2 being involutive negations on [0, 1].Similarly, each of the third kind negations could be characterized as follows.Lemma 7.
N
~ is a strict negation of the third kind, if and only if there are two isomorphisms ϕ1 and ϕ2 on [0, 1], s.t. 							(4)N~(x1,x2)=(ϕ1(x2),ϕ2(x1)).Proof.
Since eachN~ could be represented as N~=N¬∘N-, in which N¬ is the first kind and N- is the second kind. Then from Lemma 6, the result could be obtained.From Lemmas6 and 7, since ni and ϕi, i=1,2 are bijections on [0, 1], we can know each strict negation N is a bijection on [0,1]2.Note that, the composition of two involutive negations of different kinds is still a negation, but it may not be involutive. The following example shows it.Example 8.
N
4
(
x
1
,
x
2
)
=
(
1
-
x
2
2
,
1
-
x
1
) is an involutive negation of the first kind. The composition of it and the second standard negation N2,s,
(5)(N4∘N2,s)(x1,x2)=(1-(1-x2)2,1-(1-x1))=(2x2-x22,x1),
is a strict but not involutive negation of the third kind.Definition 9.
A continuous mappingΦ:[0,1]2→[0,1]2 is a monotonic isomorphism on the unit square, if it is bijective and preserves both the orders ≤k and ≤t.Obviously, each monotonic isomorphism is continuous. DefineN∘Φ and Φ′∘Φ as (N∘Φ)(x1,x2)=N(Φ(x1,x2)) and (Φ′∘Φ)(x1,x2)=Φ′(Φ(x1,x2)).Lemma 10.
IfΦ is a monotonic isomorphism on the unit square, then there are isomorphisms ϕ1 and ϕ2 on the unit interval [0, 1] such that
(6)Φ(x1,x2)=(ϕ1(x1),ϕ2(x2)).Proof.
Firstly, let us showΦ(1,1)=(1,1), Φ(0,1)=(0,1), Φ(1,0)=(1,0), and Φ(0,0)=(0,0). For any (x1,x2)∈[0,1]2, (x1,x2)≤k(1,1). Since Φ preserves ≤k, we can know Φ(x1,x2)≤kΦ(1,1). Thus, the greatest element of (Φ([0,1]2),≤k) is Φ(1,1). Because Φ is a bijection, we have Φ(1,1)=(1,1). Similarly, Φ(0,0)=(0,0), Φ(0,1)=(0,1), and Φ(1,0)=(1,0) can be proven.
Now, let us show thatΦ∘Ns,1 is a strict negation of the first kind.
Suppose(x1,x2)≤k(y1,y2)∈[0,1]2, then Ns,1(x1,x2)≥kNs,1(y1,y2). Because Φ preserves ≤k, we have
(7)(Φ∘Ns,1)(x1,x2)=Φ(Ns,1(x1,x2))≥kΦ(Ns,1(y1,y2))=(Φ∘Ns,1)(y1,y2).
It shows that Φ∘Ns,1 reverses the order ≤k. Similarly, since both Φ and Ns,1 keep the order ≤t, we can get that Φ∘Ns,1 preserves the order ≤t. Since Φ is a bijection, Φ strictly preserves the orders ≤k and ≤t. Similarly, it can be proven that Φ∘Ns,1 strictly preserves the order ≤t and strictly reverses the order ≤k.
Since bothΦ and Ns,1 are continuous, Φ∘Ns,1 is continuous.
SinceΦ(1,1)=(1,1), Φ(0,0)=(0,0), and Ns,1(1,1)=(0,0), Ns,1(0,0)=(1,1), the following hold:
(8)(Φ∘Ns,1)(0,0)=Φ(Ns,1(0,0))=(1,1),(Φ∘Ns,1)(1,1)=Φ(Ns,1(1,1))=(0,0).
These two formulas, together with the facts that Φ∘Ns,1 is continuous, strictly reverses the order ≤k, and strictly preserves the order ≤t, show that Φ∘Ns,1 is a strict negation of the first kind, denoted by N¬=Φ∘Ns,1.
Then for any(x1,x2)∈[0,1]2, (N¬∘Ns,1)(x1,x2)=([Φ∘Ns,1]∘Ns,1)(x1,x2)=Φ(Ns,1(Ns,1(x1,x2)))=Φ(x1,x2), that is, Φ=N¬∘Ns,1. From Lemma 6, N¬(x1,x2)=(n1(x2),n2(x1)), with n1, n2 being negations on the unit interval [0, 1]. Since Ns,1(x1,x2)=(1-x2,1-x1), we could get Φ(x1,x2)=(n1(1-x1),n2(1-x2)). Because n1(1-x1) and n2(1-x2) are isomorphisms on the unit interval, Φ can be characterized as Φ(x1,x2)=(ϕ1(x1),ϕ2(x2)), in which ϕ1(x1)=n1(1-x1), ϕ2(x2)=n2(1-x2) are isomorphisms on the unit interval.Lemma 11.
IfΦ is a monotonic isomorphism and Ns is a standard negation, then Φ∘Ns and Ns∘Φ are strict negations and of the same kind as Ns. Conversely, for each strict negation N, there exists a monotonic isomorphism Φ and a standard negation Ns of the same kind with N, s.t. N=Φ∘Ns.Proof.
IfNs=Ns,1, the proof of Lemma 10 has already shown that Φ∘Ns,1 is a strict negation. And by Lemma 10, (Ns,1∘Φ)(x1,x2)=(1-ϕ(x2),1-ϕ(x1)). It shows that Ns,1∘Φ is a continuous bijection on [0,1]2, which strictly preserves the order ≤k and strictly reverses ≤t and satisfies (Ns,1∘Φ)(1,1)=(0,0) and (Ns,1∘Φ)(0,0)=(1,1). Therefore, Ns,1∘Φ is a strict negation of the first kind.
Similar forNs=Ns,2 or Ns=Ns,3. Therefore, Φ∘Ns and Ns∘Φ are strict negations and of the same kind as Ns, for Ns=Ns,i,i=1,2,3.
Next, let us show the second part. SupposeN is a negation and Ns is the standard negation of the same kind as N. Then, N∘Ns is a monotonic isomorphism Φ. As a result, for any (x1,x2)∈[0,1]2,
(9)(Φ∘Ns)(x1,x2)=([N∘Ns]∘Ns)(x1,x2)=N[Ns(Ns(x1,x2))]=N(x1,x2),
that is, N=Φ∘Ns.


## 3. Groups of the Negations on the Unit Square

Based on the notions in the above section, the groups of the negations on the unit square will be discussed.In [1], the following theorem is obtained.Theorem 12 (see [1, Theorem 4]).
(
1
) For every strict negation n on the unit interval, there exist three involutive negations n1, n2 and n3 s.t. n=n1∘n2∘n3.
(
2
) For every isomophism ϕ on the unit interval, there exist four involutive negations n1, n2, n3 and n4 s.t. ϕ=n1∘n2∘n3∘n4.This theorem shows that under the operator “composition,” all the involutive negations and isomorphisms on the unit interval cannot form a group, since it is not closed under the operator “composition.” But the set of all the strict negations and isomorphisms, together with the operator “composition,” is a group [1].From this theorem and Lemmas6 and 10, we could get the following result on the unit square.Theorem 13.
(
1
) For every strict negation N of the second kind on the unit square, there exist three involutive negations N1, N2, and N3 of the second kind, s.t. N=N1∘N2∘N3.
(
2
) For every monotonic isomorphism Φ on the unit square, there exist four involutive negations N1, N2, N3, and N4 of the second, s.t. Φ=N1∘N2∘N3∘N4.For the other two kinds of negations, we have not got similar results, since the composition operator is not commutative.Next, let us discuss the groups of the negations on the unit square. The following sets are denoted byS1={Φ:Φ is a monotonic isomorphism on the unit square.}S2=S1∪{N¬:N¬ is a first kind strict negation on the unit square.}S3=S1∪{N-:N- is a second kind strict negation on the unit square.}S4=S1∪{N~:N~ is a third kind strict negation on the unit square.}S5=S2∪S3∪S4.Theorem 14.
G
i
=
(
S
i
,
∘
), i=1,2,3,4,5 are groups, with the same unit element id. Moreover, G1◃G2,3,4,5◃G5, that is, G1 is a normal subgroup of Gi, i=2,3,4,5, and G2,3,4,5 are normal subgroups of G5.Proof.
Obviously, for any elementf in Gi, i=1,…,5, id∘f=f∘id=f, that is, id is the unit element of Gi, i=1,…,5.
For any two monotonic isomorphismsΦ1, Φ2, the composition of them is still a monotonic isomorphism. Thus G1 is closed under the operator “∘”. For two strict negations N1, N2 of the same kind, the composition of them is also a monotonic isomorphism. The composition of a strict negation N and a monotonic isomorphism Φ is still a strict negation of the same kind with N. Therefore, G2, G3, and G4 are closed under the operator “∘”. For any two strict negations N1, N2 of different kind, the composition is a strict negation of the other kind. So G5 is closed.
Now, let us prove the associativity ofG5. This proof also shows the associativity of Gi, i=1,2,3,4.
From Lemmas6, 7, and 10, each of the negations and the monotonic isomorphisms can be characterized as F(x1,x2)=(f1(x1),f2(x2)) or F(x1,x2)=(f1(x2),f2(x1)), in which both of f1 and f2 are isomorphisms on [0, 1] or both are strict negations on [0, 1]. Let F1, F2, F3 be negations or monotonic isomorphisms on [0,1]2. Then,
(10)(F1∘[F2∘F3])(x1,x2)=F1[(F2∘F3)(x1,x2)]=F1[F2(F3(x1,x2))]=[F1∘F2](F3(x1,x2))=([F1∘F2]∘F3)(x1,x2),
that is, F1∘[F2∘F3]=[F1∘F2]∘F3, which shows the associativity of G5.
For any monotonic isomorphismΦ and any strict negation N, obviously Φ-1 and N-1 exist. Moreover, Φ-1 is also a monotonic isomorphism and N-1 is a negation of the same kind as N, that is, the inverse about the composition operator “∘” exists. Therefore, Gi, i=1,…,5 are groups.
SinceS1⊂S2,3,4⊂S5, G1<G2,3,4<G5, we only need to prove that they are normal subgroups.
Firstly, let us showG1◃G2. For any Φ∈S1, obviously, Φ∘G1∘Φ-1⊂G1, because G1 is a group. For N∈S2∖S1, from Lemma 6, we know N(x1,x2)=(n1(x2),n2(x1)) and N-1(x1,x2)=(n1-1(x2),n2-1(x1)). For any Φ(x1,x2)=(ϕ1(x1),ϕ2(x2))∈S1,
(11)(N∘Φ∘N-1)(x1,x2)=(n1(ϕ2(n2-1(x1))),n2(ϕ1(n1-1(x2)))).
Because both n1∘ϕ2∘n2-1 and n2∘ϕ1∘n1-1 are isomorphisms on the unit interval [0, 1], N∘Φ∘N-1 is in S1. Thus, N∘G1∘N-1⊂G1, that is, G1 is a normal subgroup of G2.
Similarly, we could proveG1◃G3,4. Since, S5=S2∪S3∪S4, we could know G1◃G5.
Now, let us showG2◃G5. Because G2 is a group, for all f∈G2 (f is a strict negation of the first kind or a monotonic isomorphism on [0,1]2), we have f∘G2∘f-1⊂G2. For any N∈G5∖G2, N is a negation of the second kind or the third kind. If N is of the second kind, then N-1 is also of the second kind. Thus for all Φ∈G2, N∘Φ∘N-1∈G2, and for all N′∈G2, N∘N′∘N-1 is of the first kind, since N′∘N-1 is of the third kind. If N is of the third kind, then N-1 is also of the third kind. Thus for all Φ∈G2, N∘Φ∘N-1∈G2, and for all N′∈G2, N∘N′∘N-1 is of the first kind, since N′∘N-1 is of the second kind. Therefore, for all f∈G5, f∘G2∘f-1⊂G2, that is, G2 is a normal subgroup of G5. Similarly, G3◃G5, G4◃G5 could be proven.From Theorem13, we immediately get the following result.Theorem 15.
The setSI,2 of all the involutive negations of the second kind could generate the group G3, that is, for any f∈G3, there is some involutive negations Ni, i=1,…,m of the second kind, such that f=N1∘N2∘⋯∘Nm. And if f is a negation, m could be 3; if f is an isomorphism, m could be 4.This theorem shows the relation between the involutive negations of the second kind and the groupG3. However, the following problem is still unproven.Problem 16.
CouldG2 (or G4) be generated by the set of all the involutive negations of the first (or the third) kind?A generator set ofG5 is given in the following theorem.Theorem 17.
The setSI,2∪{Ns,1} is a generator set of the group G5. And SI,2∪{Ns,3} is also a the generator set of G5.Proof.
By Lemma11, for any strict negation N of the first or the third kind, there exists a monotonic isomorphism Φ, s.t. N=Ns,1∘Φ or N=Ns,2∘Φ. Since Ns,1∘Ns,2=Ns,3, from Theorem 15, SI,2∪{Ns,1} generates the group G5.
BecauseNs,2∘Ns,3=Ns,1, SI,2∪{Ns,3} is also a the generator set of G5.


## 4. Negations on the Interval-Valued Set

The interval-valued setLI is a sublattice of the unit square [0,1]2, so the negations of the interval-valued set could be defined as the restriction of the negations on the unit square.Definition 18.
A negationNI on the interval-valued set is the restriction of some negation N of the first kind on the unit square, and satisfies that
(12)∀x∈LI,NI(x)=N|LI(x)∈LI.Definition 19.
An interval-valued negation is strict, if it is the restriction of some strict negations of the first kind on the unit square and satisfies (12).
An interval-valued negationNI is involutive, if it satisfies
(13)NI(NI(x))=x,∀x∈LI.From this definition, each strict interval-valued negation is an injection and each involutive negation is strict [6]. Also, in [6], it is proven that for each involutive interval-valued negation NI, it keeps the diagonal Δ={(x1,x2)∈LI:x1=x2}, that is,
(14)∀x∈Δ,NI(x)∈Δ.Definition 20.
A mappingΦI on the interval-valued set is an isomorphism, if it is a bijection and keeps the natural order.Actually, ifΦI is an isomorphism on the interval-valued set, then ΦI keeps both the orders ≤k and ≤t [6]. Thus, there exist some isomorphisms Φ on the unit square, s.t. ΦI=Φ|LI. Moreover, every Φ keeps the diagonal Δ [6].Different from the unit square, all the strict negations and the isomorphisms on the interval-valued set, together with the composition operator, do not consist of a group. The following is a counter example.Example 21.
There is no strict negation on the interval-valued set, which is an inverse of the following negationN1I:
(15)N1I(x1,x2)=(1-x2,1-x12),∀(x1,x2)∈LI.It is not hard to check thatN1I is a strict negation on the interval-valued set.It seems that there is a mappingF1(x1,x2)=(1-x22,1-x1), s.t. F1 is its inverse. However, not all of the points of the interval-valued set are well defined under F1, such as the point (0.5, 0.5). The “image” of it is (0.75, 0.5), which is out of the interval-valued set.Also, the following mappingF2 is also not the inverse of N1I, because it is not an injection, thus not a strict negation on LI. Consider
(16)F2(x1,x2)={(1-x22,1-x1),if(1-x1)2≥1-x2,(x1,x1),otherwise.Now, let us give the proof of Example21.SupposeN2I is the inverse negation of N1I. Clearly, N1I maps the interval-valued set LI to the set S={(x1,x2):0≤1-x2≤(1-x1)2≤1}. Then (N2I)|S is a surjection S to LI. Since for the points in LI∖S, their images under N2I also should be in LI, we can get that N2I is not an injection, thus not a strict negation on LI. So N1I has no inverse.Denote thatS1I={ΦI:Φisanisomorphismon⁡LI.}S2I=S1I∪{NI:NI is a strict negation on LI and keeps the diagonal Δ.}Then we have the following theorem.Theorem 22.
G
1
I
=
(
S
1
I
,
∘
) and G2I=(S2I,∘) are groups, with ∘ the composition of the mappings. Moreover, G1I◃G2I.Proof.
Obviously, the unit element is the identity mapping id, and the operation is closed and associative.
SupposeΦI is an isomorphism in S1I. Then ΦI can be represented as ΦI(x1,x2)=(ϕ(x1),ϕ(x2)), with ϕ an isomorphism on the unit interval [0, 1], because ΦI keeps the diagonal. Define the mapping (ΦI)-1 as (ΦI)-1(x1,x2)=(ϕ-1(x1),ϕ-1(x2)). Then (ΦI)-1 is also an isomorphism on LI, that is, (ΦI)-1∈SI, and
(17)[(ΦI)-1∘ΦI](x1,x2)=(x1,x2)=[ΦI∘(ΦI)-1](x1,x2),
that is, (ΦI)-1 is the inverse of ΦI. Thus, G1I is a group.
LetNI be a strict negation on LI, which keeps the diagonal Δ. From Lemma 6, there exist some strict negations n1,n2 on [0, 1], s.t. NI(x1,x2)=(n1(x2),n2(x1)). From (14), n1=n2. Then (NI)-1(x1,x2)=(n1-1(x2),n1-1(x1)) is also a strict negation on LI and keeps the diagonal. Also we can check that
(18)[(NI)-1∘NI](x1,x2)=(x1,x2)=[NI∘(NI)-1](x1,x2);
that is, (NI)-1 is the inverse of NI. Thus, G2I is a group.
Similar to the proof of Theorem14, G1I◃G2I could be proven.This theorem could be extended to the unit square.Theorem 23.
(
1
) All the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called G6.
(
2
) All the strict negations of the first kind and the monotonic isomorphisms on the unit square, which keep the diagonal, form a group, called G7.The proof is similar to Theorem22.From Theorems12, 22, and 23, we could obtain the following theorem.Theorem 24.
(
1
) The set of all the involutive negations on LI generates the group G2I.
(
2
) The set of all the involutive negations of the first kind on the unit square, which keep the diagonal, generates the group G7.


## 5. Conclusion

In this paper, we firstly study the negations on the unit square. The main results are Theorems14 and 23, which show the groups that are formed by the strict negations and the monotonic isomorphisms. Then we discuss the negations on the interval-valued set. The main result is Theorem 22, that is, all the strict negations and isomorphisms on LI, which keep the diagonal, form a group. Moreover, some generator sets of the groups are given.



---

*Source: 917432-2014-08-13.xml*