The lower and upper approximations and homomorphisms between lower approximations in quotient groups

Abstract

Rough sets are an emerging mathematical tool in the field of data analysis. The relation between rough sets and algebraic systems (groups, rings and modules etc.) has already been identified. In this paper we studied the lower and upper approximations of a rough set in a quotient group. It can be proved that under certain assumptions the lower approximation is a normal subgroup of the quotient group; however this result is not valid for the upper approximation. We developed several homomorphisms between two lower approximations of the quotient groups.

Keywords

Rough sets lower approximation upper approximation quotient groups group homomorphisms

1 Introduction

Theory of rough sets was proposed by Pawlak [16]. The theory of rough sets is an extension of the set theory. A rough set is a subset of a universal set, defined by a pair of ordinary sets called the lower and upper approximations. These approximations help us in extraction of useful information hidden in the unprocessed data for decision making. Its valuable features made its successful implementation to many real life applications. In this theory equivalence relations play a key role in treating abstraction and imprecision in the data. This theory has been successfully implemented to many real life problems. Most noteworthy amongst its various applications are data mining, and pattern recognition. In the recent years, this theory witnessed appreciable development. Researchers foresee this as an alternate to statistical approach for resolving complex problems in data analysis. Most of the sets are based on the imprecise information. In the analysis of such kind of information, mathematical logics are extremely helpful.

Some researchers have studied the rough algebraic structures. In [2, 8] and [15] the concept of rough groups, rough quotient groups is established. In [14] Kuroki and Wang introduced the notion of the lower and upper approximations with respect to a normal subgroup in a group. Moreover they also defined the lower and upper approximations with respect to a t-level subset of a fuzzy normal subgroup.

In [12], Kuroki set up the rough left (resp., right and bi-) ideals in a semigroup. He has also defined the lower and upper approximations of a quotient semigroup with respect to a congruence relation over a semigroup. Kuroki verified that these are left (resp., right and bi-) ideals in the quotient semigroup. Xiao and Zhang [19] has established the relations between the upper (lower) rough prime ideals and rough fuzzy prime ideals in a semigroup.

After that Zhan and Davvaz come up with the idea of rough rings and ideals (see [5] and [20]). They introduced the notion of rough sub-ring (resp., ideal) with respect to an ideal of a ring which is an extended notion of a sub-ring (resp., ideal) in a ring. Also they confirmed several properties of the lower and upper approximations with respect to an ideal in a ring.

In [6], Davvaz defined the upper rough ideal in a ring R with respect to a t-level congruence relation of a fuzzy ideal on R. In [1], Bi-cai introduced anti-homomorphism of a group. Isaac and Neelima [11] have studied some properties related to rough ring homomorphism and anti-homomorphism. Additionally, Han et al. [10] defined the rough ring in an approximation space. In 2006, Davvaz and Mahdavipour introduced the rough modules [7]. They studied some properties of the lower and the upper approximations in a rough module. In [21] and [22] the roughness in soft hemirings and n-ary semigroups is discussed.

In parallel we introduced the notion of the lower and upper approximations in a quotient group. We studied several properties such as intersection, union and product. Our study revealed that the lower approximation is a normal subgroup of the quotient group. Moreover it is shown that the lower and upper approximations of a normal subgroup of the quotient group do not provide us any new information. We were able to establish some homomorphisms between the two lower approximation spaces.

2 Lower and upper approximations in a quotient group

First of all in this section we will recall the notation of the rough sets.

Definition 2.1. Let ∅ ≠ U be a universe and θ an equivalence relation over U. Then the pair (U, θ) is called an approximation space.

Definition 2.2. If (U, θ) is an approximation space, then the mapping Apr : P (U) → P (U) × P (U) defined by: $Apr (X) = (\underline{X}, \bar{X}) for all X \in P (U),$ is called a rough approximation operator. Here $\underline{X} : {x \in U : {[x]}_{θ} \subseteq X}$ and $\bar{X} : {x \in U : {[x]}_{θ} \cap X \neq \emptyset}$ are called the lower and upper rough approximations of X in (U, θ) respectively.

Note that it is clear from the definition that $\underline{X} \subseteq X \subseteq \bar{X}$ .

Definition 2.3. For a given approximation space (U, θ), a pair (A, B) ∈ P (U) × P (U) is called a rough set iff (A, B) = Apr (X) for some X ∈ P (U).

Throughout this paper G will be denoted as a group under multiplication with identity element e. Let N be any normal subgroup of G. Now we define a relation θ over G/N as follows: $\begin{matrix} xN θ yN \Leftrightarrow xN = (aN) (yN) (aN)^{- 1} \\ = ({aya}^{- 1}) N for some a \in G . \end{matrix}$

Note that xNθyN if and only if they are conjugates in G/N. It is well known that θ is an equivalence relation over the quotient group G/N. Hence (G/N, θ) is an approximation space. We will denote the equivalence class of xN ∈ G/N by [xN] _θ. So [xN] _θ is the set of all conjugates of xN in G/N. That is we have: $\begin{matrix} {[xN]}_{θ} & = & {(aN) (xN) (aN)^{- 1} : aN \in G / N} \\ = & {({axa}^{- 1}) N : a \in G} . \end{matrix}$

First of all we will prove the following lemma:

Lemma 2.4.With the previous notation we have: ${[x_{1} x_{2} N]}_{θ} = {[(x_{1} N) (x_{2} N)]}_{θ} \subseteq [x_{1} N]_{θ} [x_{2} N]_{θ}$ for all x₁N, x₂N ∈ G/N.

Proof. By normality of N the first equality is trivial. Now let gN ∈ [x₁x₂N] _θ . Then gN = (ax₁x₂a^-1) N for some a ∈ G. By normality of N we get that $gN = ({ax}_{1} x_{2} a^{- 1}) N = (({ax}_{1} a^{- 1}) N) (({ax}_{2} a^{- 1}) N)) .$

By definition (ax_ia^-1) N ∈ [x_iN] _θ for all i = 1, 2. It follows that gN ∈ [x₁N] _θ [x₂N] _θ. It provides the following inclusion: $[x_{1} x_{2} N]_{θ} \subseteq [x_{1} N]_{θ} {[x_{2} N]}_{θ} . □$

Let H be any subset of G and N a normal subgroup of G such that N ⊆ H. Then H/N ⊆ G/N is the set of all those elements aN ∈ G/N such that a ∈ H. Moreover if K ⊆ G with N ⊆ K, then (H/N) (K/N) is the following set: $\begin{matrix} (H / N) (K / N) & = & {(hN) (kN) : h \in H, k \in K} \\ = & {hkN : h \in H, k \in K} . \end{matrix}$

It is clear that (H/N) (K/N) = (HK)/N. Now we define the following sets of G/N: ${\underline{Apr}}_{G / N} (H / N) : = {xN \in G / N : [xN]_{θ} \subseteq H / N}$ and ${\bar{Apr}}_{G / N} (H / N) : = {xN \in G / N : [xN]_{θ} \cap H / N \neq \emptyset} .$

Then ${\underline{Apr}}_{G / N} (H / N)$ (resp., ${\bar{Apr}}_{G / N} (H / N)$ ) is called the lower (resp., upper) approximation of H/N with respect to N in the approximation space (G/N, θ).

Example 2.5. Let U = G/N = {u₁, …, u₆} be the universe set, where G denotes the permutation group over four elements and $N = {I, (12) (34), (13) (24), (14) (23)} .$

Suppose that P is the following subset of attributes: $P = {Normal, Headache, Flu, Temperature} .$

Note that we can make the following identifications: u₁ = N, u₂ = (1234) N, u₃ = (1324) N, u₄ = (1243) N, u₅ = (123) N and u₆ = (132) N . Let θ be the conjugacy relation over U. Suppose that x, y ∈ U have the same disease if and only if xθy. Then we have the following information system (Table 1):

Table 1
Information system

Normal Headache Flu Temperature

u ₁ Yes No No No

u ₂ No Yes Yes No

u ₃ No Yes Yes No

u ₄ No Yes Yes No

u ₅ No No Yes Yes

u ₆ No No Yes Yes

	Normal	Headache	Flu	Temperature
u ₁	Yes	No	No	No
u ₂	No	Yes	Yes	No
u ₃	No	Yes	Yes	No
u ₄	No	Yes	Yes	No
u ₅	No	No	Yes	Yes
u ₆	No	No	Yes	Yes

Let X = {u₂, u₅, u₆}. Then we have ${\underline{Apr}}_{U} (X) = {u_{5}, u_{6}}$ and ${\bar{Apr}}_{U} (X) = {u_{2}, u_{3}, u_{4}, u_{5}, u_{6}} .$

Lemma 2.6.With the above notation suppose that H_i is a non-empty subset of G such that N ⊆ H_i for all i = 1, 2. If H₁/N ⊆ H₂/N, then ${\underline{Apr}}_{G / N} (H_{1} / N) \subseteq {\underline{Apr}}_{G / N} (H_{2} / N)$ and ${\bar{Apr}}_{G / N} (H_{1} / N) \subseteq {\bar{Apr}}_{G / N} (H_{2} / N) .$

Proof. Let $xN \in {\bar{Apr}}_{G / N} (H_{1} / N) .$ Then [xN] _θ and H₁/N have a non-empty intersection. Since H₁/N ⊆ H₂/N so it follows that $[xN]_{θ} \cap H_{2} / N \neq \emptyset .$

By definition of the upper approximation we conclude that $xN \in {\bar{Apr}}_{G / N} (H_{2} / N)$ . Hence ${\bar{Apr}}_{G / N} (H_{1} / N)$ is a subset of ${\bar{Apr}}_{G / N} (H_{2} / N)$ . By the similar arguments we can prove the other inclusion. □

Proposition 2.7.Suppose that N is a normal subgroup of G and H_i ⊆ G such that N ⊆ H_i for all i = 1, 2. Then the following conditions hold:

$\emptyset \neq {\underline{Apr}}_{G / N} (H_{1} / N) \subseteq H_{1} / N \subseteq {\bar{Apr}}_{G / N} (H_{1} / N) .$

${\bar{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N) = {\bar{Apr}}_{G / N} (H_{1} / N) \cup {\bar{Apr}}_{G / N} (H_{2} / N) .$

${\underline{Apr}}_{G / N} (H_{1} / N) \cup {\underline{Apr}}_{G / N} (H_{2} / N) \subseteq {\underline{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N)$ .

${\bar{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N) = {\bar{Apr}}_{G / N} ((H_{1} / N) \cup (H_{2} / N))$ .

${\underline{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N) = {\underline{Apr}}_{G / N} ((H_{1} / N) \cup (H_{2} / N))$ .

If H₁ ⊆ H₂, then ${\underline{Apr}}_{G / N} (H_{1} / N) \subseteq {\underline{Apr}}_{G / N} (H_{2} / N)$ .

If H₁ ⊆ H₂, then ${\bar{Apr}}_{G / N} (H_{1} / N) \subseteq {\bar{Apr}}_{G / N} (H_{2} / N)$ .

Proof. (1) It is obvious. Note that e ∈ N ⊆ H₁ and [N] _θ = {N} ⊆ H₁/N .

(2) Since N ⊆ H_i for all i = 1, 2, so it follows that N ⊆ H₁ ∪ H₂. Then $\begin{matrix} xN & \in & {\bar{Apr}}_{G / N} (H_{1} / N) \cup {\bar{Apr}}_{G / N} (H_{2} / N) \\ \Leftrightarrow & xN \in {\bar{Apr}}_{G / N} (H_{1} / N) \\ or xN \in {\bar{Apr}}_{G / N} (H_{2} / N) . \\ \Leftrightarrow & [xN]_{θ} \cap H_{1} / N \neq \emptyset or [xN]_{θ} \cap H_{2} / N \neq \emptyset . \\ \Leftrightarrow & There exists yN \in G / N such that \\ yN \in H_{1} / N or yN \in H_{2} / N and xN θ yN . \\ \Leftrightarrow & yN = hN for some h \in H_{1} \\ or h \in H_{2} and xN θ yN . \\ \Leftrightarrow & yN = hN for some h \in H_{1} \cup H_{2} \\ and xN θ yN . \\ \Leftrightarrow & yN \in (H_{1} \cup H_{2}) / N and xN θ yN . \\ \Leftrightarrow & [xN]_{θ} \cap (H_{1} \cup H_{2}) / N \neq \emptyset . \\ \Leftrightarrow & xN \in {\bar{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N) . \end{matrix}$

This proves the equality.

(3) Since H_i ⊆ H₁ ∪ H₂ for all i = 1, 2, it implies that H_i/N ⊆ (H₁ ∪ H₂)/N for all i = 1, 2. Then by Lemma 2.6 we have ${\underline{Apr}}_{G / N} (H_{i} / N) \subseteq {\underline{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N)$ for all i = 1, 2. Hence $\begin{matrix} {\underline{Apr}}_{G / N} (H_{1} / N) & \cup & {\underline{Apr}}_{G / N} (H_{1} / N) \\ \subseteq & {\underline{Apr}}_{G / N} ((H_{1} \cup H_{2}) / N) . \end{matrix}$

Note that (4) and (5) are obvious in view of the fact that (H₁ ∪ H₂)/N = (H₁/N) ∪ (H₂/N) .

(6) Since H₁ ⊆ H₂, it implies that H₁/N ⊆ H₂/N. Then by Lemma 2.6 we have ${\underline{Apr}}_{G / N} (H_{1} / N) \subseteq {\underline{Apr}}_{G / N} (H_{2} / N) .$

(7) It can be proved by the same arguments as we used in (6). □

In the following we will show that ${\underline{Apr}}_{G / N} (H_{1} / N)$ and ${\bar{Apr}}_{G / N} (H_{1} / N)$ are not subgroups of G/N in general.

Example 2.8. Let G = Q₈ = {±1, ± i, ± j, ± k}, the quaternion group with the following relations: $i^{2} = j^{2} = k^{2} = - 1, ij = k, jk = i and ki = j .$

Let N = {±1} and H = N ∪ {i, j}. Then it is clear that N is a normal subgroup of G such that N ⊆ H. Moreover H is not a subgroup of G. Then we have $G / N = {N, iN, jN, kN} and H / N = {N, iN, jN} .$

Hence one can see that ${\underline{Apr}}_{G / N} (H / N) = H / N = {\bar{Apr}}_{G / N} (H / N) .$

Therefore ${\underline{Apr}}_{G / N} (H_{1} / N)$ and ${\bar{Apr}}_{G / N} (H_{1} / N)$ are not subgroups of G/N.

Proposition 2.9. Suppose that H_i ⊆ G with N ⊆ H_i for all i = 1, 2, where N is a normal subgroup of a group G. Then we have:

${\bar{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \subseteq {\bar{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \subseteq {\bar{Apr}}_{G / N} (H_{1} / N) \cap {\bar{Apr}}_{G / N} (H_{2} / N)$ .

${\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \subseteq {\underline{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) = {\underline{Apr}}_{G / N} (H_{1} / N) \cap {\underline{Apr}}_{G / N} (H_{2} / N)$ .

Proof. Note that (H₁ ∩ H₂)/N ⊆ (H₁/N) ∩ (H₂/N) ⊆ H_i/N for all i = 1, 2. By Lemma 2.6 it follows that $\begin{matrix} {\bar{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ \subseteq {\bar{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \\ \subseteq {\bar{Apr}}_{G / N} (H_{i} / N) for all i = 1, 2 . \end{matrix}$

Then we have $\begin{matrix} {\bar{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ \subseteq {\bar{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \\ \subseteq {\bar{Apr}}_{G / N} (H_{1} / N) \cap {\bar{Apr}}_{G / N} (H_{2} / N) . \end{matrix}$

By the above Similar arguments we can prove that $\begin{matrix} {\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ \subseteq {\underline{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \\ \subseteq {\underline{Apr}}_{G / N} (H_{1} / N) \cap {\underline{Apr}}_{G / N} (H_{2} / N) . \end{matrix}$

Now let xN be an arbitrary element of ${\underline{Apr}}_{G / N} (H_{1} / N) \cap {\underline{Apr}}_{G / N} (H_{2} / N)$ . Then $xN \in {\underline{Apr}}_{G / N} (H_{i} / N)$ for all i = 1, 2. It implies that [xN] _θ ⊆ H_i/N for all i = 1, 2. Then [xN] _θ ⊆ (H₁/N) ∩ (H₂/N). It follows that $xN \in {\underline{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N))$ . This finishes the proof of Proposition. □

Corollary 2.10. 12 With the notation of Proposition 2.9 assume in addition that H₁ and H₂ are subgroups of G. Then the following are true: $\begin{matrix} {\bar{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ = {\bar{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \end{matrix}$ and $\begin{matrix} {\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ = {\underline{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) . \end{matrix}$

Proof. Suppose that H₁ and H₂ are subgroups of G. We claim that (H₁/N) ∩ (H₂/N) = (H₁ ∩ H₂)/N. Note that (H₁ ∩ H₂)/N ⊆ (H₁/N) ∩ (H₂/N). Let xN ∈ (H₁/N) ∩ (H₂/N) . Then $xN = h_{1} N = h_{2} N, where h_{i} \in H_{i} for all i = 1, 2 .$

It implies that $h_{1} h_{2}^{- 1} \in N \subseteq H_{i}$ for all i = 1, 2 . Since H₁ and H₂ are subgroups, so h₁, h₂ ∈ H_i for all i = 1, 2 . It follows that xN ∈ (H₁ ∩ H₂)/N. This proves the claim. Moreover it also provides us the required equalities. □

Note that in Corollary 2.10 the condition of H₁ and H₂ are subgroups of G is necessary. In the next we will give an example such that the results in Corollary 2.10 are not true if we skip this condition.

Example 2.11. Let G = A₄, the alternating group and N = {I, (12) (34) , (13) (24) , (14) (23)}. Then N is a normal subgroup of G. Let H₁ = N ∪ {(123) , (124)} and H₂ = N ∪ {(132) , (142)}. Note that none of the H_i’s is a subgroup of G. Moreover we have $\begin{matrix} G / N & = & {N, (123) N, (132) N} \\ = & {N, (142) N, (124) N} \end{matrix}$ and $H_{1} / N = H_{2} / N = G / N .$

Now H₁ ∩ H₂ = N . Then (H₁ ∩ H₂)/N ≠ (H₁/N) ∩ (H₂/N) . Since N and G both are normal subgroups of G, so by the next Lemma 2.12 it follow that $\begin{matrix} {\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \\ = N / N = {\bar{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) \end{matrix}$ and $\begin{matrix} {\underline{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) \\ = G / N = {\bar{Apr}}_{G / N} ((H_{1} / N) \cap (H_{2} / N)) . \end{matrix}$

This proves that the equality does note hold in the statement of the last Corollary 2.10.

In the following we will show that if N and H both are normal subgroups of G such that N ⊆ H, then the lower and upper approximations of H/N does not provide us any new information.

Lemma 2.12. Suppose that N and H are normal subgroups of G with N ⊆ H . Then the following result hold: ${\underline{Apr}}_{G / N} (H / N) = H / N = {\bar{Apr}}_{G / N} (H / N) .$

Proof. To prove the result let $xN \in {\bar{Apr}}_{G / N} (H / N)$ . Then there exists yN ∈ G such that yN = hN ∈ H/N for some h ∈ H and xNθyN. It implies that [xN] _θ = [yN] _θ = [hN] _θ . Since H is normal, so it follows that aha^-1N ∈ H/N for all a ∈ G and [hN] _θ ⊆ H/N. So we get that $xN \in {\underline{Apr}}_{G / N} (H / N) .$ Hence ${\bar{Apr}}_{G / N} (H / N) \subseteq {\underline{Apr}}_{G / N} (H / N)$ . This proves the result in view of Proposition 2.7(1).

Theorem 2.13.Let N be a normal subgroup of G. Then for any non-empty subsets H₁and H₂of G containing N the following statements hold:

${\bar{Apr}}_{G / N} ((H_{1} H_{2}) / N) \subseteq {\bar{Apr}}_{G / N} (H_{1} / N) {\bar{Apr}}_{G / N} (H_{2} / N) .$

${\underline{Apr}}_{G / N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N) \subseteq {\underline{Apr}}_{G / N} ((H_{1} H_{2}) / N) .$

Proof. For the proof of (1) let $xN \in {\bar{Apr}}_{G / N} ((H_{1} H_{2}) / N)$ . Then [xN] _θ∩ (H₁H₂)/N ≠ ∅. Then there exists yN ∈ G/N such that yN ∈ (H₁H₂)/N and xNθyN. It implies that yN = (h₁h₂) N and xN = (aya^-1) N where a ∈ G and h_i ∈ H_i for all i = 1, 2 . Since N is normal, so we conclude that $\begin{matrix} xN & = & ({aya}^{- 1}) N = (aN) (yN) (a^{- 1} N) \\ = & (aN) ((h_{1} h_{2}) N) (a^{- 1} N) \\ = & {ah}_{1} h_{2} a^{- 1} N = ({ah}_{1} a^{- 1}) ({ah}_{2} a^{- 1}) N \\ = & ({ah}_{1} a^{- 1} N) ({ah}_{2} a^{- 1} N) . \end{matrix}$

Since h_iN ∈ H_i/N for all i = 1, 2, so it induces the following fact: $[{ah}_{i} a^{- 1} N]_{θ} \cap H_{i} / N = [h_{i} N]_{θ} \cap H_{i} / N \neq \emptyset$ for all i = 1, 2. Then by definition ${ah}_{i} a^{- 1} N \in {\bar{Apr}}_{G / N} (H_{i} / N)$ for all i = 1, 2. So we get that xN belongs to the set ${\bar{Apr}}_{G / N} (H_{1} / N) {\bar{Apr}}_{G / N} (H_{2} / N)$ . It proves that ${\bar{Apr}}_{G / N} ((H_{1} H_{2}) / N)$ is a subset of ${\bar{Apr}}_{G / N} (H_{1} / N) {\bar{Apr}}_{G / N} (H_{2} / N) .$ This finishes the proof of (1).

(2) Suppose that $xN \in {\underline{Apr}}_{N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N)$ . Then it implies that $xN = (y_{1} N) (y_{2} N) = y_{1} y_{2} N,$ where $y_{i} N \in {\underline{Apr}}_{G / N} (H_{i} / N) for all i = 1, 2 .$ Then [y_iN] _θ ⊆ H_i/N for all i = 1, 2. By Lemma 2.4 it follows that $[xN]_{θ} = [y_{1} y_{2} N]_{θ} \subseteq [y_{1} N]_{θ} [y_{2} N]_{θ} .$

Now [y₁N ] _θ [y₂N] _θ ⊆ ( H₁/ N ) (H₂/ N) = (H₁H₂)/ N . It implies that $xN \in {\underline{Apr}}_{G / N} ((H_{1} H_{2}) / N)$ . Hence this proves the claim in (2). □

In the next example we will show that ${\underline{Apr}}_{G / N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N)$ is not equal to ${\underline{Apr}}_{G / N} ((H_{1} H_{2}) / N) .$

Example 2.14. 13 Let G = S₄, the permutation group over the set {1, 2, 3, 4}. Let $N = {I, (12) (34), (13) (24), (14) (23)},$

H₁ = N ∪ {(1324)} and H₂ = N ∪ {(3412) , (1243)}. Then N is a normal subgroup such that N ⊆ H_i for all i = 1, 2. Since |G/N|=6 so it follows that $\begin{matrix} H_{1} / N & = & {N, (1324) N}, \\ H_{2} / N & = & {N, (3412) N, (1243) N} \end{matrix}$ and $\begin{matrix} G / N & = & {N, (1234) N, (1324) N, (1243) N, \\ (123) N, (132) N} . \end{matrix}$

Moreover we have ${\underline{Apr}}_{G / N} (H_{i} / N) = {N}$ for all i = 1, 2. Also note that $(H_{1} H_{2}) / N = {N, (1324) N, (3412) N, (1243) N} .$

It implies that $\begin{matrix} {\underline{Apr}}_{G / N} ((H_{1} H_{2}) / N) \\ = {N, (1324) N, (3412) N, (1243) N} . \end{matrix}$

Hence it follows that $\begin{matrix} {\underline{Apr}}_{G / N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N) \\ \neq {\underline{Apr}}_{G / N} ((H_{1} H_{2}) / N) . \end{matrix}$

Definition 2.15. Let N ⊆ G be a normal subgroup and H ⊆ G containing N. Suppose that ${Apr}_{G / N} (H / N) : = ({\underline{Apr}}_{G / N} (H / N), {\bar{Apr}}_{G / N} (H / N))$ and ${\underline{Apr}}_{G / N} (H / N) \neq {\bar{Apr}}_{G / N} (H / N)$ . Then Apr_G/N (H/N) is called a rough set in the approximation space (G/N, θ).

Apr_G/N (H/N) is called an upper rough subgroup (resp., normal subgroup) of G/N if ${\bar{Apr}}_{G / N} (H / N)$ is a subgroup (resp., normal subgroup) of G/N.

Apr_G/N (H/N) is called a lower rough subgroup (resp., normal subgroup) of G/N if ${\underline{Apr}}_{G / N} (H / N)$ is a subgroup (resp., normal subgroup) of G/N.

Apr_G/N (H/N) is called a rough subgroup (resp., normal subgroup) of G/N if it is both upper and lower rough subgroup (resp., normal subgroup).

As we have shown in Example 2.8 that the lower and upper approximations are not subgroups. But in the next result we will prove that the lower approximation of H/N is a normal subgroup of G/N provided that H is a subgroup of G.

Proposition 2.16.Let N be a normal subgroup and H a subgroup of G with N ⊆ H. Then Apr_G/N (H/N) is a lower rough normal subgroup of G/N.

Proof. By Proposition 2.7(1) ${\underline{Apr}}_{G / N} (H / N) \neq \emptyset$ . Since H/N is a subgroup, so [N] _θ = {N} ⊆ H/N and hence $N \in {\underline{Apr}}_{G / N} (H / N) .$ It shows that the identity element exists in ${\underline{Apr}}_{G / N} (H / N) .$

Now let $x_{i} N \in {\underline{Apr}}_{G / N} (H / N)$ for all i = 1, 2. Then by definition we have [x_iN] _θ ⊆ H/N for all i = 1, 2. By Lemma 2.4 it follows that $[x_{1} x_{2} N]_{θ} \subseteq [x_{1} N]_{θ} [x_{2} N]_{θ} \subseteq H / N .$

Recall that H/N is a subgroup. So we have $(x_{1} N) (x_{2} N) = x_{1} x_{2} N \in {\underline{Apr}}_{G / N} (H / N) .$ This proves that the closure law holds in ${\underline{Apr}}_{G / N} (H / N) .$

Now let $(g_{1} x_{1}^{- 1} g_{1}^{- 1}) N \in [x_{1}^{- 1} N]_{θ}$ be an arbitrary element, where g₁ ∈ G. Since N is normal, so we have $(g_{1} x_{1}^{- 1} g_{1}^{- 1}) N = ((g_{1} x_{1} g_{1}^{- 1}) N)^{- 1} .$

Since H/N is a subgroup and $(g_{1} x_{1} g_{1}^{- 1}) N \in H / N$ , so it follows that $(g_{1} x_{1}^{- 1} g_{1}^{- 1}) N \in H / N$ . This proves that $[x_{1}^{- 1} N]_{θ} \subseteq H / N$ . Then $x_{1}^{- 1} N \in {\underline{Apr}}_{G / N} (H / N)$ . Moreover the associative law holds in G/N, so it is also hold in ${\underline{Apr}}_{G / N} (H / N)$ . Hence ${\underline{Apr}}_{G / N} (H / N)$ is a subgroup of G/N .

To prove normality let gN ∈ G/N and $xN \in {\underline{Apr}}_{G / N} (H / N)$ be any elements. Then it follows that ${gxg}^{- 1} N \in [{gxg}^{- 1} N]_{θ} = [xN]_{θ} \subseteq H / N .$

Since N is normal, so by definition it implies that $(gN) (xN) (g^{- 1} N) = {gxg}^{- 1} N \in {\underline{Apr}}_{G / N} (H / N)$ . Consequently, this proves the normality of ${\underline{Apr}}_{G / N} (H / N)$ . Hence Apr_G/N (H/N) is a lower rough normal sub-group of G/N. □

The next example shows that Apr_G/N (H/N) is not an upper rough sub-group of G/N even H is a subgroup of G. Moreover it is also shown that the lower approximation of a subgroup H/N is a proper normal subgroup of H/N.

Example 2.17. Let G = S₄, $N = {I, (12) (34), (13) (24), (14) (23)}$ and $H = N \cup {(12), (34), (1324), (4231)} .$

Then N is a normal subgroup and H is a subgroup of G such that N ⊆ H. Moreover H is not normal in G. Since |H/N|=2 and |G/N|=6 so it follows that $\begin{matrix} G / N & = & {N, (1234) N, (1324) N, (1243) N, \\ (123) N, (132) N} \end{matrix}$ and $H / N = {N, (1324) N} .$

Then we conclude that ${\bar{Apr}}_{G / N} (H / N) = {N, (1234) N, (1324) N, (1243) N} .$

It implies that $| {\bar{Apr}}_{G / N} (H / N) | = 4$ which does not divide 6. So ${\bar{Apr}}_{G / N} (H / N)$ is not a subgroup of G/N . Therefore Apr_G/N (H/N) is not an upper rough sub-group of G/N. Moreover ${\underline{Apr}}_{G / N} (H / N) = {N}$ is a proper normal subgroup of H/N .

Remark 2.18. If H is a subgroup of G contains a normal subgroup N. By Example 2.17 it follows that ${\bar{Apr}}_{G / N} (H / N)$ does not satisfy the closure law.

The next Corollary shows that the intersection and product of the lower approximations are also normal subgroups.

Corollary 2.19.Let N and M be two normal subgroups of G. Let H_i be a subgroup of G such that N ⊆ H_i for all i = 1, 2. Suppose that M ⊆ H₁ . Then each of the following is a normal subgroup of G/N:

${\underline{Apr}}_{G / N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N) .$

${\underline{Apr}}_{G / N} (H_{1} / N) \cap {\underline{Apr}}_{G / N} (H_{2} / N) .$

${\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N) .$

${\underline{Apr}}_{G / NM} (H_{1} / NM)$ .

${\underline{Apr}}_{G / N \cap M} (H_{1} / N \cap M)$ .

Proof. Note that ${\underline{Apr}}_{G / N} (H_{i} / N)$ is a normal subgroup of G/N for all i = 1, 2 (see Proposition 2.16). Then it follows that ${\underline{Apr}}_{G / N} (H_{1} / N) \cap {\underline{Apr}}_{G / N} (H_{2} / N)$ and ${\underline{Apr}}_{G / N} (H_{1} / N) {\underline{Apr}}_{G / N} (H_{2} / N),$ both are normal subgroups of G/N. Also note that H₁ ∩ H₂ is a subgroup of G containing N. Again Proposition 2.16 implies that ${\underline{Apr}}_{G / N} ((H_{1} \cap H_{2}) / N)$ is a normal subgroup of G/N.

Furthermore note that NM and N ∩ M both are normal subgroups of G contained in H₁. So it follows that ${\underline{Apr}}_{G / NM} (H_{1} / NM)$ and ${\underline{Apr}}_{G / N \cap M} (H_{1} / N \cap M)$ are normal subgroups of G/N. □

3 Homomorphism between lower approximations

Let N and M be any two normal subgroups of G. Throughout this section we will denote θ₁ and θ by the conjugacy relations over G/M and G/N respectively. Let H be any subset of G containing N and M. Here we will relate the lower and upper approximations of H/N and H/M. Moreover we are able to develop some homomorphisms between the lower approximations of H/N and H/M.

Theorem 3.1. Suppose that N ⊆ M are normal subgroups of G and H ⊆ G such that M ⊆ H. Let x ∈ G be any fixed element. If $xN \in {\underline{Apr}}_{G / N} (H / N)$ (resp., $xN \in {\bar{Apr}}_{G / N} (H / N)),$ then $xM \in {\underline{Apr}}_{G / M} (H / M)$ (resp., $xM \in {\bar{Apr}}_{G / M} (H / M)) .$

Proof. First of all let N ⊆ M and $xN \in {\underline{Apr}}_{G / N} (H / N)$ . Let gxg^-1M ∈ [xM] _{θ
₁} be an arbitrary element with g ∈ G. Since gxg^-1N ∈ [xN] _θ ⊆ H/N. So there exists h ∈ H such that gxg^-1N = hN. It implies that ${gxg}^{- 1} h^{- 1} \in N \subseteq M .$

Since M is a subgroup, so we have gxg^-1M = hM. Then gxg^-1M ∈ H/M. But gxg^-1M ∈ [xM] _{θ
₁} was arbitrary, so [xM] _{θ
₁} ⊆ H/M. By definition of the lower approximation of H/M we have $xM \in {\underline{Apr}}_{G / M} (H / M)$ .

Now let $xN \in {\bar{Apr}}_{G / N} (H / N)$ . Then [xN] _θ ∩ H/N ≠ ∅ . There exist g, y ∈ G with yN ∈ H/N and xN = gyg^-1N. Then yN = hN where h ∈ H. It implies that yh^-1 ∈ N ⊆ M and x (gyg^-1) ^-1 ∈ N ⊆ M . Then $yM = hM and xM = {gyg}^{- 1} M .$

So we have yM ∈ H/M such that xMθ₁yM. By definition of the upper approximation we get that [xM] _{θ
₁}∩ H/M ≠ ∅ and $xM \in {\bar{Apr}}_{G / M} (H / M)$ . This completes the proof. □

The next Proposition shows that the converse of Theorem 3.1 also holds under the additional assumption of H is a subgroup of G.

Proposition 3.2.Let N and M be two normal subgroups of G and H a subgroup of G containing N and M. Then for any x ∈ G the following conditions are equivalent:

$xN \in {\underline{Apr}}_{G / N} (H / N)$ .

$xM \in {\underline{Apr}}_{G / M} (H / M) .$

$xNM \in {\underline{Apr}}_{G / NM} (H / NM) .$

$x (N \cap M) \in {\bar{Apr}}_{G / N \cap M} (H / N \cap M) .$

Proof. (1) ⇒ (2). Suppose that $xN \in {\underline{Apr}}_{G / N} (H / N)$ . Then it follows that [xN] _θ ⊆ H/N. Let gxg^-1M by any element of [xM] _{θ
₁}, where g ∈ G. Note that there exists h ∈ H such that gxg^-1N = hN. It implies that ${gxg}^{- 1} h^{- 1} \in N \subseteq H .$

Since H is a subgroup of G and h ∈ H, it implies that gxg^-1 ∈ H. So we have gxg^-1M ∈ H/M. This proves that [xM] _{θ
₁} is a subset of H/M. Therefore $xM \in {\underline{Apr}}_{G / M} (H / M) .$ By interchanging the role of N and M we can prove that (2) implies (1).

Note that by the above same arguments we can prove that (1) is also equivalent to (3) and (4). This completes the proof. □

Corollary 3.3.Let N and M be two normal subgroups and H any subset of G containing both N and M. Let x ∈ G be a fixed element. Then the following conditions hold:

Suppose that NM ⊆ H. If $xN \in {\underline{Apr}}_{G / N} (H / N)$ or $xM \in {\underline{Apr}}_{G / M} (H / M),$ then $xNM \in {\underline{Apr}}_{G / NM} (H / NM) .$

Suppose that NM ⊆ H. If $xN \in {\bar{Apr}}_{G / N} (H / N)$ or $xM \in {\bar{Apr}}_{G / M} (H / M),$ then $xNM \in {\bar{Apr}}_{G / NM} (H / NM) .$

If $x (N \cap M) \in {\underline{Apr}}_{G / N \cap M} (H / N \cap M),$ then $xN \in {\underline{Apr}}_{G / N} (H / N) and xM \in {\underline{Apr}}_{G / M} (H / M) .$

If $x (N \cap M) \in {\bar{Apr}}_{G / N \cap M} (H / N \cap M),$ then $xN \in {\bar{Apr}}_{G / N} (H / N) and xM \in {\bar{Apr}}_{G / M} (H / M) .$

Proof. It is straightforward in view of the Theorem 3.1. Note that NM and N ∩ M both are normal subgroups of G. □

Before proving the next result we need some preparation. Let N and M be two normal subgroups of G and H a subgroup of G containing N and M. Since ${\underline{Apr}}_{G / N} (H / N)$ and ${\underline{Apr}}_{G / M} (H / M)$ are normal subgroups of G/N and G/M respectively (see Proposition 2.16). So by one-one correspondence Theorem ${\underline{Apr}}_{G / N} (H / N)$ and ${\underline{Apr}}_{G / M} (H / M)$ are of the form K/N and T/M respectively. Here K and T are normal subgroups of G such that N ⊆ K ⊆ H and M ⊆ T ⊆ H.

Theorem 3.4. With the above notation suppose that N ⊆ M. Then the following results are true:

There is a group isomorphism $\frac{G / N}{K / N} \to \frac{G / M}{T / M}, xN (K / N) \mapsto xM (T / M) .$ In particular G/K is isomorphic to G/T.

There is a group isomorphism $\frac{H / N}{K / N} \to \frac{H / M}{T / M}, xN (K / N) \mapsto xM (T / M) .$ In particular H/K is isomorphic to H/T.

There is an onto group homomorphism $K / N \to T / M, xN \mapsto xM .$ with kernel is equal to M/N. In particular we have the following inclusions: $M / N \subseteq K / N \subseteq H / N and M \subseteq K \subseteq H .$

K/M is isomorphic to T/M.

Proof. (1) Since N ⊆ M, so it induces an onto group homomorphism f : G/N → G/M, xN ↦ xM. We claim that f (K/N) ⊆ T/M. If xM ∈ f (K/N) , then there exists yN ∈ K/N such that yM = f (yN) = xM. Since H is a subgroup of G and yN ∈ K/N, so it follows that yM ∈ T/M (see Proposition 3.2). This proves that xM ∈ T/M and f (K/N) ⊆ T/M. So we have proved the claim. Then it is well known that f induces the following group homomorphism $ψ : \frac{G / N}{K / N} \to \frac{G / M}{T / M}$ with ψ (xN (K/N)) = f (xN) (T/M) = xM (T/M). Since f is onto, so it follows that ψ is onto. We only need to prove that ψ is injective. Let xN ∈ ker(ψ). Then $ψ (xN (K / N)) = xM (T / M) = T / M .$

It implies that xM ∈ T/M. By Proposition 3.2 we get that xN ∈ K/N. Recall that H is a subgroup of G. This proves that ψ is an isomorphism. Moreover by second isomorphism Theorem we get that G/K is isomorphic to G/T. By following the same steps we can prove the isomorphisms in (2).

(3) Let φ : K/N → T/M be defined as φ (xN) = xM for all xN ∈ K/N. Firstly we will show that φ is well-defined. For this let x₁N = x₂N with x_iN ∈ K/N for all i = 1, 2 . By Theorem 3.1 it implies that $x_{1} x_{2}^{- 1} \in N \subseteq M$ such that x_iM ∈ T/M for all i = 1, 2 . So we conclude that $φ (x_{1} N) = x_{1} M = x_{2} M = φ (x_{2} N) .$

This proves that φ is well-defined. Moreover $\begin{matrix} φ ((x_{1} N) (x_{2} N)) & = & φ (x_{1} x_{2} N) = x_{1} x_{2} M \\ = & (x_{1} M) (x_{2} M) \\ = & φ (x_{1} N) φ (x_{2} N), \end{matrix}$ since N and M are normal. So φ is a group homomorphism. By Proposition 3.2 it is easy to see that φ is onto. Now we calculate ker(φ) as follows: $\begin{matrix} ker (φ) & = & {xN : φ (xN) = M} \\ = & {xN : xM = M} \\ = & {xN : x \in M} = M / N . \end{matrix}$

This proves the claim in (2) in view of one-one correspondence Theorem. Moreover (4) follows from (3) in view of the second isomorphism Theorem. □

Corollary 3.5.Let N and M be two normal subgroups of G contained in a subgroup H. Then we have:

There is an onto group homomorphism $\begin{matrix} {\underline{Apr}}_{G / (N \cap M)} (H / (N \cap M)) \to {\underline{Apr}}_{G / N} (H / N), \\ x (N \cap M) \mapsto xN . \end{matrix}$ with kernel is equal to N/(N ∩ M) .

There is an onto group homomorphism $\begin{matrix} {\underline{Apr}}_{G / N} (H / N) \to {\underline{Apr}}_{G / NM} (H / NM), \\ xN \mapsto x (NM) . \end{matrix}$ with kernel is equal to NM/N .

There is an onto group homomorphism $\begin{matrix} {\underline{Apr}}_{G / (N \cap M)} (H / (N \cap M)) \\ \to {\underline{Apr}}_{G / NM} (H / NM), \\ x (N \cap M) \mapsto xNM . \end{matrix}$ with kernel is equal to NM/(N ∩ M) .

Proof. Since (N ∩ M), NM both are normal subgroups of G contained in H and (N ∩ M) ⊆ N ⊆ NM. Then all the claims of Corollary are obvious in view of Theorem 3.4(2). □

4 Conclusions

Rough sets are the technical tool for modeling the incomplete information system. The group structure plays an important role in which many of the physical and real world problems are modeled. In this paper we studied the roughness in quotient groups via conjugacy relation. We confirmed some fundamental properties of the lower and upper approximations of the subsets of quotient groups. Also we have proved that for a normal subgroup N and a subgroup H of a group G with N ⊆ H the lower approximation of H/N is a normal subgroup of G/N. We successfully established relation among the lower approximations of two different quotient groups. Several homomorphisms between these approximations are developed. Real life applications call for more advanced extensions of this theory and our results can contribute significantly to it. We are hopeful that this new idea will be helpful to develop other classical algebraic systems to the rough algebraic systems.

Footnotes

Acknowledgment

The authors are grateful to the reviewers for suggestions to improve the manuscript.

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The lower and upper approximations and homomorphisms between lower approximations in quotient groups

Abstract

Keywords

1 Introduction

2 Lower and upper approximations in a quotient group

Table 1 Information system Normal Headache Flu Temperature u 1 Yes No No No u 2 No Yes Yes No u 3 No Yes Yes No u 4 No Yes Yes No u 5 No No Yes Yes u 6 No No Yes Yes

4 Conclusions

Footnotes

Acknowledgment

References

Table 1
Information system

Normal Headache Flu Temperature

u ₁ Yes No No No

u ₂ No Yes Yes No

u ₃ No Yes Yes No

u ₄ No Yes Yes No

u ₅ No No Yes Yes

u ₆ No No Yes Yes