Superfluous ideals of N-groups

Abstract

We consider a right nearring N and a module over N (known as, N-group). For an arbitrary ideal (or N-subgroup) $\varOmega$ of an N-group G, we define the notions $\varOmega$-superfluous, strictly $\varOmega$-superfluous, g-superfluous ideals of G. We give suitable examples to distinguish between these classes and the existing classes studied in Bhavanari (Proc Japan Acad 61-A:23–25, 1985; Indian J Pure Appl Math 22:633–636, 1991; J Austral Math Soc 57:170–178, 1994), and prove some properties. For a zero-symmetric nearring with 1, we consider a module over a matrix nearring and obtain one-one correspondence between the superfluous ideals of an N-group (over itself) and those of $M_{n}(N)$-group $N^{n}$, where $M_{n}(N)$ is the matrix nearring over N. Furthermore, we define a graph of superfluous ideals of a nearring and prove some properties with necessary examples.

1 Introduction and Preliminaries

The notion of finite Goldie dimension (denoted by FGD) of a module was defined by Goldie [14] wherein, the key notions for the study of FGD are essential submodules, uniform submodules and complement of a submodule (see, [4, 5]). The dualization of this concept namely, finite spanning dimension (denoted by FSD) in modules over rings was defined by Fleury [13] with the notions such as superfluous submodules, hollow submodules and supplements. Later, these concepts were studied in [1, 2, 7, 18]. The idea of FSD was generalized to module over nearrings (known as N-groups) in [6, 11, 16, 20]. They introduced the notions such as superfluous ideal, hollow ideal and FSD in N-groups and proved the corresponding structure theorems. The motivation of this paper arises from a natural question that what if one substitutes an arbitrary ideal $\varOmega$ in place of an N-group G, so that it generalises the existing study of these notions. The classes of N-groups with these new notions are different from the classes of N-groups studied in [6, 8, 11, 19, 20]. Eventually, in this paper we define $\varOmega$-superfluous ideal of an N-group G, where $\varOmega$ is an ideal of G and obtain some connections to matrix nearrings, and some combinatorial aspects. In section 2, we define g-superfluous and g-supplement ideals of N-groups as a generalization of respective notions defined by [20]. We have constructed examples where G is non-abelian.

In Sect. 3, we introduce superfluous ideals of N-groups and prove some important properties and provide necessary examples. In Sect. 4, we consider the notion strictly superfluous in terms of N-subgroups and gave examples which indicate that the classes of N-groups with superfluous and strictly superfluous are different. Matrix nearrings over arbitrary nearrings were defined in [15] and studied in [23]. In Sect. 5, we introduce the superfluous and g-superfluous ideals in $M_{n}(N)$-group $N^{n}$, and establish a one-one correspondence between superfluous and g-superfluous ideals of N (over itself) and those of $M_{n}(N)$-group $N^{n}$. In Sect. 6, we introduce superfluous ideal graphs of nearrings and prove some properties with examples.

A (right) nearring is a set N together with two binary operations $``+''$ and $``\cdot ''$ such that $(N,+)$ is a group, $(N, \cdot )$ is a semigroup and right distributive law holds. In general, for some $n\in N$, $n\cdot 0 \ne 0$, and so we call N is zero-symmetric if $n\cdot 0 =0$ for all $n\in N$. A normal subgroup I of a nearring N is called an ideal of N (denoted by $I\trianglelefteq N$) if $IN\subseteq I$ and $a(b+i)-ab\in I$ for all $a,b\in N$ and $i\in I$. An additive group G is said to be an N-group if there exists a map $N\times G \rightarrow G$ defined by $(n,g)\longmapsto ng$ satisfying $(n+n_{1})g=ng+n_{1}g$ and $(n\cdot n_{1})g=n(n_{1}g)$ for all $n,n_{1}\in N$ and $g\in G$. Throughout, we use G for an N-group. A subgroup H of G is said to be an N-subgroup (denoted as, $H\le _{N} G$) of G if $NH\subseteq H$; and a normal subgroup I of G is called an ideal (denoted as, $I\trianglelefteq _{N} G$) of G if $n(g+i)-ng\in I$, for all $n\in N$, $g\in G$ and $i\in I$. An ideal S of G is said to be superfluous in G if $S+K=G$ and K is an ideal of G, imply $K=G$, and G is called hollow if every proper ideal of G is superfluous in G. For any two N-subgroups H and K of G, K is said to be a supplement for H if $H+K=G$ and $H+K^{'}\ne G$ for any proper ideal $K^{'}$of K. For any ideals I, J, K of N (or of G), if $K\subseteq I$, then $I\cap (J+K)=(I\cap J)+K$. We use $I\oplus J$ to denote the direct sum of ideals I and J of G. We refer to Pilz [17] and Bhavanari and Kuncham [10] for fundamental literature in nearrings.

We consider simple and finite graphs, whose vertex set is V and edge set is E. A vertex v of a graph is called a universal vertex if degree of v= $|V|-1$. If there exists a path between every two vertices of a graph, then the graph is connected otherwise the graph is disconnected. A graph whose vertex set is empty is called a null graph and a graph having atleast one vertex and empty edge set is called an empty graph.

2 Generalized supplements

The notion of superfluous submodule of module over a ring was studied by [3, 7]. We define generalized superfluous (briefly, g-superfluous) ideal of an N-group G as follows.

Definition 2.1

An ideal K of G is called g-superfluous if $G=K+T$ and $T\le _{e} G$, then $T=G$. We denote this by $K\ll _{gs} G$.

Remark 2.2

Every superfluous ideal of G is g-superfluous.

Example 2.3

Consider the nearring $N=({\mathbb {Z}}_{2}\times {\mathbb {Z}}_{2}\times {\mathbb {Z}}_{2}, +, \cdot )$ with the notation given in page no. 420, (N) of [17]. That is, $(0,0,0)=0$, $(0,0,1)=1$ $(0,1,0)=2$ $(0,1,1)=3$, $(1,0,0)=4$, $(1,0,1)=5$, $(1,1,0)=6$, and $(1,1,1)=7.$ The multiplication table is given below. Let $G=N$ (Table 1).

The proper ideals of G are $I_{1}=\{0,2,4,6\}$ and $I_{2}=\{0,1\}$. It can be seen that $I_{1}$ and $I_{2}$ are g-superfluous but not superfluous in G as $I_{1}+I_{2}=G$

Table 1 Multiplication $\star$ on N

Proposition 2.4

Let K, J be ideals of G such that $K\subseteq J$. If $K\ll _{gs} J$, then $K\ll _{gs} L$ for any ideal L of G with $J\subseteq L$.

Proof

Let $T\le _{e} L$ such that $K+T= L$. We prove $T=L$. Clearly $T\subseteq L$. Since $K+T=L$, we have $(K+T)\cap J=L\cap J$. Since $K\subseteq J$ and modular law, we have $K+(T\cap J)=J$. Since $T\cap J\le _{e}J$ and $K\ll _{gs} J$, we have $(T\cap J)=J$ and so $T\subseteq J$. Since $K\subseteq J\subseteq T$, we get $L=K+T\subseteq J+T=T$. Therefore $L=T$ and hence $K\ll _{gs} L$. $\square$

Proposition 2.5

Let $K_{1}$, $K_{2}$, $G_{1}$, $G_{2}$ be ideals of G such that $K_{1}\subseteq G_{1}$ and $K_{2}\subseteq G_{2}$. If $K_{1}\ll _{gs} G_{1}$ and $K_{2}\ll _{gs} G_{2}$, then $K_{1}+K_{2}\ll _{gs} G_{1}+G_{2}$.

Proof

Since $K_{i}\ll _{gs} G_{i}$, by Proposition 2.4, we have $K_{i}\ll _{gs} G_{1}+G_{2}$ for $i=\{1,2\}$. Let $T\le _{e} G_{1}+G_{2}$ be such that $K_{1}+K_{2}+T=G_{1}+G_{2}$. Since $T\le _{e} G_{1}+G_{2}$, we have $T+K_{2}\le _{e} G_{1}+G_{2}$. Now $K_{1}+(K_{2}+T)=G_{1}+G_{2}$ and $K_{1}\ll _{gs} G_{1}+G_{2}$, imply that $K_{2}+T=G_{1}+G_{2}$. Again, since $K_{2}\ll _{gs} G_{1}+G_{2}$ and $T\le _{e} G_{1}+G_{2}$ implies $T=G_{1}+G_{2}$. $\square$

Note 2.6

Let X, K be ideals of G such that $K\subseteq X$. If $\frac{X}{K}\le _{e} \frac{G}{K}$, then $X\le _{e} G$.

Proposition 2.7

Let U, V and K be ideals of G. If $U\ll _{gs} V$, then $\frac{U+K}{K}\ll _{gs}\frac{V+K}{K}$.

Proof

Let $\frac{T}{K}\le _{e}\frac{V+K}{K}$ be such that $\frac{U+K}{K}+\frac{T}{K}=\frac{V+K}{K}$. Then $U+K+T=V+K$. Since $K\subseteq T$, we get $U+T=V+K$. Now $U\ll _{gs} V$ implies that $U\ll _{gs} V+K$, and $\frac{T}{K}\le _{e} \frac{V+K}{K}$ implies $T\le _{e} V+K$. Since $U\ll _{gs} V+K$, $T\le _{e} V+K$ and $U+T=V+K$, we get $T=V+K$, which implies $\frac{T}{K}=\frac{V+K}{K}$. Therefore, $\frac{U}{K}\ll _{gs} \frac{V+K}{K}$. $\square$

Proposition 2.8

Let J, K, L be ideals of G such that $K\subseteq J$.

• 1. If $J\ll _{gs} G$, then $K\ll _{gs} G$ and $\frac{J}{K}\ll _{gs} \frac{G}{K}$.

• 2. $J+L\ll _{gs}G$ if and only if $J\ll _{gs}G$ and $L\ll _{gs} G$.

Proof

• 1. Suppose $J\ll _{gs} G$. To prove $K\ll _{gs} G$, let $T\le _{e} G$ such that $K+T=G$. Since $K\subseteq J$, we get $J+T=G$. Since $J\ll _{gs} G$, we have $T=G$, shows that $K\ll _{gs} G$.

  Next we prove $\frac{J}{K}\ll _{gs} \frac{G}{K}$. Let $\frac{X}{K}\le _{e} \frac{G}{K}$ such that $\frac{J}{K}+\frac{X}{K}=\frac{G}{K}$. Then $\frac{J+X}{K}=\frac{G}{K}$, implies that $J+X=G$. Since $X\le _{e}G$, we get $X=G$. Therefore $\frac{X}{K}=\frac{G}{K}$.

• 2. Suppose $J+L\ll _{gs} G$. To prove $J\ll _{gs}G$ and $L\ll _{gs} G$, let $T\le _{e} G$ such that $J+T=G$. Then $(J+L)+T=G$. Since $J+L \ll _{gs} G$, we have $T=G$. In a similar way, we get $L\ll _{gs} G$.

  Conversely, suppose that $J\ll _{gs} G$ and $L\ll _{gs} G$. To prove $J+L\ll _{gs} G$, let $T\le _{e} G$ such that $(J+L)+T=G$. This means, $J+(L+T)=G$. Since $L+T\le _{e}G$ and $J\ll _{gs} G$, it follows that $L+T =G$. Again since $L\ll _{gs}G$ and $T\le _{e}G$, we get $T=G$. Therefore $J+L\ll _{gs} G$.

$\square$

Definition 2.9

Let P and Q be N-subgroups of G. Q is said to be a g-supplement of P if $G=P+Q$ and $G=P+T$ with $T\le _{e}Q$ implies that $T=Q$.

An N-group G is called g-supplemented if every ideal of G has a g-supplement.

Remark 2.10

Every supplemented N-group is g-supplemented.

Example 2.11

Consider the nearring given in (K(139), page 418 of [17]). Let $N=(D_{8}, +, \star )$, the dihedral group of order 8 and $G=N$.

The subgroups of $(D_{8}, +)$ are $H_{1}=\langle e\rangle$, $H_{2}=\langle s\rangle$, $H_{3}=\langle sr^{2}\rangle$, $H_{4}=\langle r^{2}\rangle$, $H_{5}=\langle sr^{3}\rangle$, $H_{6}=\langle sr\rangle$, $H_{7}=\langle \{s, r^{2}\}\rangle$, $H_{8}=\langle r\rangle$ and $H_{9}=\langle \{r^{2},sr^{3}\}\rangle$. The N-subgroups are $H_{1}$, $H_{2}$, $H_{3}$, $H_{4}$, $H_{5}$, $H_{7}$ and $H_{9}$, and ideals of $D_{8}$ are $H_{4}$, $H_{7}$ and $H_{9}$. The ideals of $H_{9}$(when it is considered as an N-group) are $H_{4}$ and $H_{5}$ which are not essential in $H_{9}$. Observe that $H_{9}$ is not a supplement of $H_{7}$ as there exists an ideal $H_{5}$ of $H_{9}$ such that $H_{7}+H_{5}=D_{8}$. Furthermore, all ideals of $H_{9}$ are not essential, we do not have any essential ideal I of $H_{9}$ such that $H_{7}+I=D_{8}$. Therefore $H_{9}$ is a g-supplement of $H_{7}$ (Table 2).

Table 2 Multiplication $\star$ on N

Lemma 2.12

Let P, Q be ideals of G. Then Q is a g-supplement of P if and only if $G=P+Q$ and $P\cap Q \ll _{gs} Q$.

Proof

Suppose Q is a g-supplement of P in G. Then $P+Q=G$ and $P+Q^{'}\ne G$ for any essential ideal $Q^{'}$ of Q. We prove $P\cap Q \ll _{gs} Q$. Let $T\le _{e} Q$ such that $(P\cap Q)+T=Q$. Then $G=P+Q=P+(P\cap Q)+T=P+T$, as $(P\cap Q)\subseteq P$. Now $G=P+T$ where $T\le _{e} Q$. Since Q is a g-supplement of P, we get $T=Q$. Therefore $P\cap Q \ll _{gs} Q$.

Conversely, suppose that $G=P+Q$ and $P\cap Q \ll _{gs} Q$. To show Q is a g-supplement of P in G, let $G=P+T$ for some essential ideal T of Q. Now, since $T\subseteq Q$, by modular law we get $Q=Q\cap G=Q\cap (P+T)=(Q\cap P)+T$. Since $(P\cap Q)\ll _{gs}Q$ and $T\le _{e} Q$, we get $T=Q$. Therefore Q is a g-supplement of P in G. $\square$

Proposition 2.13

[12]

• 1. Let G be an N-group and let I, J be the ideals of G with $G=I\oplus J$. Then $a+b=b+a$ for all $a\in I$ and $b\in J$.

• 2. If $N=N_{0}$, $n\in N$, $a\in I$, $b\in J$ and the sum $I\oplus J$ is direct in G, then $n(a+b)=na+nb$.

• 3. Let $N=N_{0}$ and $I\trianglelefteq _{N}G$ be a direct summand. Then each ideal of I is an ideal of G.

Lemma 2.14

Let A, B and C be ideals of G. Then

$$\begin{aligned} A\cap (B+C) \trianglelefteq _{N} B\cap (A+C) + C\cap (A+B). \end{aligned}$$

Proof

We have $A\cap (B+C)\trianglelefteq _{N} G$. Let $p\in A\cap (B+C)$. Then $p\in A$ and $p \in B+C$, which implies $p=b+c$ for some $b\in B$ and $c\in C$. Now, $b=p-c \in A+C$ and $c= -b+p \in B+A= A+B$ and hence $p=b+c \in B\cap (A+C) + C\cap (A+B)$. Therefore $A\cap (B+C)\trianglelefteq _{N} G$, which is contained in $B\cap (A+C) + C\cap (A+B)$ and hence $A\cap (B+C)\trianglelefteq _{N} B\cap (A+C) + C\cap (A+B)$. $\square$

Lemma 2.15

Let N be zero-symmetric and $G_{1}$, U be ideals of G and $G_{1}$ be g-supplemented and a direct summand of G. If $G_{1}+U$ has a g-supplement in G, then U has a g-supplement in G.

Proof

Let X be a g-supplement of $G_{1}+U$ in G. Then by Lemma 2.12, $G_{1}+U+X=G$ and $(G_{1}+U)\cap X\ll _{gs} X$. Since $G_{1}$ is g-supplemented, $(U+X)\cap G_{1}$ has a g-supplement Y in $G_{1}$. That is, $G_{1}\cap (U+X)+Y=G_{1}$ and $G_{1}\cap (U+X)\cap Y \ll _{gs} Y$, by Lemma 2.12. Since $G_{1}$ is a direct summand, $Y\trianglelefteq _{N} G$. This yield,

$$\begin{aligned} G=G_{1}\cap (U+X)+Y+(U+X)=U+X+Y \end{aligned}$$

and

$$\begin{aligned} U\cap (X+Y)&\trianglelefteq _{N} X\cap (U+Y)+ Y\cap (U+X)\\ {}&\trianglelefteq _{N} X\cap (G_{1}+U)+Y\cap G_{1}\cap (U+X) \\&\ll _{gs} X+Y . \end{aligned}$$

Hence, $X+Y$ is a g-supplement of U in G. $\square$

Proposition 2.16

Let G be an N-group. Let K, U and V be ideals of G such that $K\subseteq U$. Let V be a g-supplement of U in G. Then $\frac{V+K}{K}$ is a g-supplement of $\frac{U}{K}$.

Proof

Since V is a g-supplement of U in G, we have $G=U+V$ and $U\cap V\ll _{gs} V$ which implies $\frac{U\cap V+K}{K} \ll _{gs} \frac{V+K}{K}$. Now $\frac{G}{K}=\frac{U+V}{K}=\frac{U}{K}+\frac{V+K}{K}$. Also $\frac{U}{K}\cap \frac{V+K}{K}=\frac{U\cap (V+K)}{K}=\frac{U\cap V +K}{K} \ll _{gs} \frac{V+K}{K}$. Therefore, $\frac{V+K}{K}$ is a g-supplement of $\frac{U}{K}$. $\square$

3 Superfluous ideals

Definition 3.1

Let $\varOmega \trianglelefteq _{N} G$. An ideal (or N-subgroup) H of G is said to be $\varOmega$-superfluous in G if $\varOmega \nsubseteq H$ and for any ideal L of G, $\varOmega \subseteq L+H$ implies $\varOmega \subseteq L$. We denote it by $H \ll _{\varOmega } G$.

Note 3.2

If $\varOmega =G$, then $\varOmega$-superfluous coincides with the notion of superfluous defined by [20]. In this case, we denote $H\ll G$ whenever an ideal H is superfluous in G. Trivially, the ideal (0) is superfluous in G.

Example 3.3

Let $N={\mathbb {Z}},$ the set of integers and $G=({\mathbb {Z}}_{24}, +_{24})$. Then G is an N-group. Let $\varOmega = 8{\mathbb {Z}}_{24}$. Then $6{\mathbb {Z}}_{24}$, $3{\mathbb {Z}}_{24}$, $12{\mathbb {Z}}_{24}$ are $\varOmega$-superfluous, whereas $3{\mathbb {Z}}_{24}$ is not superfluous in G, since $3{\mathbb {Z}}_{24}+2{\mathbb {Z}}_{24}={\mathbb {Z}}_{24}$ but $2{\mathbb {Z}}_{24} \ne {\mathbb {Z}}_{24}$.

Example 3.4

Let $N={\mathbb {Z}}$ and $G={\mathbb {Z}}_{12}$. Then G is an N-group. Let $\varOmega = 4{\mathbb {Z}}_{12}$. Then $6{\mathbb {Z}}_{12}$, $3{\mathbb {Z}}_{12}$ are $\varOmega$-superfluous, whereas $3{\mathbb {Z}}_{12}$ is not superfluous in G, since $3{\mathbb {Z}}_{12}+2{\mathbb {Z}}_{12}={\mathbb {Z}}_{12}$ but $2{\mathbb {Z}}_{12} \ne {\mathbb {Z}}_{12}$.

Example 3.5

Let $N=\begin{pmatrix} 0 &{} {\mathbb {Z}}_{q^m} \\ 0 &{} 0 \end{pmatrix}$ and $G=N$. Then the ideals and N-subgroups are $H_{i} = \{\begin{pmatrix} 0 &{} q^{i}{\mathbb {Z}}_{q^m}\\ 0 &{} 0 \end{pmatrix}: 0\le i \le m \}$. Let $\varOmega = H_{k}$. Then $H_{j}\ll _{\varOmega } G$ for all $j\le k$.

Example 3.6

Consider the N-group given in the Example 2.11.

• 1. $H_{7}\ll _{H_{9}}G$, $H_{4}\ll _{H_{9}} G$, $H_{9}\ll _{H_{7}}G$, $H_{4}\ll _{H_{7}}G$.

• 2. The N-subgroups $H_{2}$, $H_{3}$, $H_{4}$ and $H_{7}$ are $\varOmega$-superfluous in G with $\varOmega =H_{9}$.

• 3. The N-subgroups $H_{4}$, $H_{5}$, $H_{9}$ are $\varOmega$-superfluous in G with $\varOmega =H_{7}$.

Proposition 3.7

Let $\varOmega$ be an ideal and X be an ideal (or N-subgroup) of G. If X is $\varOmega$-superfluous in G, then $X\cap Y$ is $\varOmega$-superfluous in G for any ideal (or N-subgroup) Y of G.

Proof

Suppose X is $\varOmega$-superfluous in G. Let $Y\trianglelefteq _{N} G$. Since $\varOmega \nsubseteq X$, we have $\varOmega \nsubseteq X\cap Y$. On a contrary, suppose $X\cap Y$ is not superfluous in G. Then there exists a proper ideal K of G such that $\varOmega \nsubseteq K$ and $\varOmega \subseteq (X\cap Y)+K$. Now, since $X\cap Y \subseteq X$ we get $\varOmega \subseteq X+K$, a contradiction as $X\ll _{\varOmega } G$. Therefore $X\cap Y\ll _{\varOmega }G$. $\square$

Proposition 3.8

Let $\varOmega$, K be ideals of G. If $K \ll _{\varOmega } G$, then $K\cap \varOmega \ll G$.

Proof

Let $K \ll _{\varOmega } G$. To prove $K \cap \varOmega \ll G$, let $L \trianglelefteq _{N} G$ be such that $(K \cap \varOmega )+L=G$. Now $\varOmega \subseteq (K \cap \varOmega )+L \subseteq K+L$. Now since $K \ll _{\varOmega } G$, we have that $\varOmega \subseteq L$. Also since $K\cap \varOmega \subseteq \varOmega \subseteq L$, it follows that $L=(K\cap \varOmega )+L=G$. Therefore, $K\cap \varOmega \ll G$. $\square$

Proposition 3.9

Let N be zero-symmetric and $\varOmega \trianglelefteq _{N}G$, which is a direct summand, and let $P \trianglelefteq _{N} G$ contained in $\varOmega$. Then $P\ll _{\varOmega } G$ if and only if $P\ll \varOmega$.

Proof

Suppose $P\ll _{\varOmega } G$. Since $P\trianglelefteq _{N}G$ and $P\subseteq \varOmega$, we have $P\trianglelefteq _{N}\varOmega$. To prove $P\ll \varOmega$, let $L \trianglelefteq _{N} \varOmega$ be such that $P+L=\varOmega$. Since $\varOmega$ is a direct summand, by Proposition 2.13(3), $L\trianglelefteq _{N} G$. Now $\varOmega \subseteq P+L$ and $P \ll _{\varOmega } G$, we get $\varOmega \subseteq L$. Since $L\subseteq \varOmega$, it follows that $L=\varOmega$.

Conversely, suppose that $P\ll \varOmega$. Let $L \trianglelefteq _{N} G$ be such that $\varOmega \subseteq P+L$. Now $\varOmega = (P+L)\cap \varOmega =P+ ( L \cap \varOmega )$, by modular law, and since $P\ll \varOmega$, it follows that $\varOmega =L\cap \varOmega$. Hence $\varOmega \subseteq L$. $\square$

Proposition 3.10

Let $K \trianglelefteq _{N}G$ and let P and $\varOmega$ be ideals of G which are contained in K. If $P \ll _{\varOmega } K$, then $P \ll _{\varOmega } G$.

Proof

Suppose that $P \ll _{\varOmega } K$. To prove $P \ll _{\varOmega } G$, let $L \trianglelefteq _{N} G$ be such that $\varOmega \subseteq P+L$. Since $\varOmega \subseteq K$ and by modular law, we have $\varOmega \subseteq (P+L)\cap K=P+(L\cap K)$. Since $L\cap K\trianglelefteq _{N} K$ and $P \ll _{\varOmega } K$, we have $\varOmega \subseteq (L\cap K)$, which implies $\varOmega \subseteq L$. Hence $P\ll _{\varOmega } G$. $\square$

Remark 3.11

It can be easily seen that the Propositions 3.7, 3.8, 3.9 and 3.10 hold for N-subgroups also.

Remark 3.12

The following proposition holds for ideals of G but not for N-subgroups, as sum of two N-subgroups need not be an N-subgroup.

Proposition 3.13

Let $N_{1}$, $N_{2}$, $\varOmega$ be ideals of G. Then $N_{1}\ll _{\varOmega } G$ and $N_{2}\ll _{\varOmega } G$ if and only if $N_{1}+N_{2}\ll _{\varOmega } G$.

Proof

Suppose that $N_{1}\ll _{\varOmega } G$ and $N_{2}\ll _{\varOmega } G$. Let $L\trianglelefteq _{N} G$ be such that $\varOmega \subseteq (N_{1}+N_{2})+L = N_{1}+(N_{2}+L)$. Since $N_{1}\ll _{\varOmega } G$, we have $\varOmega \subseteq N_{2}+L$, and again since $N_{2}\ll _{\varOmega } G$, we get $\varOmega \subseteq L$.

Conversely, suppose $N_{1}+N_{2}\ll _{\varOmega } G$. Let $L\trianglelefteq _{N} G$ be such that $\varOmega \subseteq N_{1}+L \subseteq (N_{1}+N_{2})+L$. Now since $N_{1}+N_{2}\ll _{\varOmega } G$, we get $\varOmega \subseteq L$. Similar assertion proves $N_{2}\ll _{\varOmega } G$. $\square$

Note 3.14

Let N be zero-symmetric and $K_{1}\trianglelefteq _{N} G_{1}\trianglelefteq _{N} G$ and $K_{2}\trianglelefteq _{N} G_{2}\trianglelefteq _{N} G$, $\varOmega \trianglelefteq _{N} G$ such that $G_{1}\oplus G_{2}=G$. Then $K_{1}\ll _{\varOmega } G_{1}$ and $K_{2}\ll _{\varOmega } G_{2}$ if and only if $K_{1}+K_{2} \ll _{\varOmega } G_{1}+G_{2}$.

Proposition 3.15

Let $\varOmega$, K, P be ideals of G such that $K \subset \varOmega$, $K\subseteq P$ and $\varOmega \nsubseteq P$. Then $P\ll _{\varOmega } G$ if and only if $K\ll _{\varOmega } G$ and $\frac{P}{K}\ll _{\frac{\varOmega }{K}} \frac{G}{K}$.

Proof

Suppose $P\ll _{\varOmega } G$. To prove $K\ll _{\varOmega } G$, let $L \trianglelefteq _{N} G$ such that $\varOmega \subseteq K+L$. Since $K\subseteq P$, we get $\varOmega \subseteq P+L$. Since $P \ll _{\varOmega } G$, we have $\varOmega \subseteq L$, and thus $K\ll _{\varOmega } G$.

Now to prove $\frac{P}{K}\ll _{\frac{\varOmega }{K}} \frac{G}{K}$, let $\frac{L}{K} \trianglelefteq _{N} \frac{G}{K}$, where $K\subseteq L\trianglelefteq _{N} G$ such that, $\frac{\varOmega }{K} \subseteq \frac{P}{K} + \frac{L}{K} =\frac{(P+L)}{K}$. Then $\varOmega \subseteq P+L$. Since $P\ll _{\varOmega } G$, we get $\varOmega \subseteq L$, which implies that $\frac{\varOmega }{K} \subseteq \frac{L}{K}$. Hence, $\frac{P}{K}\ll _{\frac{\varOmega }{K}} \frac{G}{K}$.

Conversely, suppose that $K\ll _{\varOmega }G$ and $\frac{P}{K}\ll _{\frac{\varOmega }{K}} \frac{G}{K}$. To prove $P\ll _{\varOmega } G$, let $L\trianglelefteq _{N} G$ such that $\varOmega \subseteq P+L$. Then $\frac{\varOmega }{K} \subseteq \frac{(P+L)}{K} = \frac{P}{K} + \frac{L+K}{K}$. Since $\frac{P}{K}\ll _{\frac{\varOmega }{K}} \frac{G}{K}$, it follows that $\frac{\varOmega }{K} \subseteq \frac{L+K}{K}$, which implies $\varOmega \subseteq L+K$. Since $K\ll _{\varOmega }G$, we get $\varOmega \subseteq L$. Hence, $P\ll _{\varOmega } G$. $\square$

Proposition 3.16

Let $\{\varOmega _{i}\}_{i\in I}$ be a family of ideals of G and $K\trianglelefteq _{N} G$. If for each $i\in I$, $K\ll _{\varOmega _{i}} G$, then $K \ll _{\sum \limits _{i\in I}\varOmega _{i}} G$.

Proof

Suppose $K\ll _{\varOmega _{i}} G$ for each $i\in I$ and $\sum \limits _{i\in I}\varOmega _{i} \subseteq K+L$ where $L\trianglelefteq _{N} G$. Then since $\varOmega _{i}\subseteq \sum \varOmega _{i} \subseteq K+L$ for each $i\in I$ and $K\ll _{\varOmega _{i}} G$, we have $\varOmega _{i} \subseteq L$, which shows that $\sum \varOmega _{i} \subseteq L$. Hence, $K\ll _{\sum \limits _{i\in I}\varOmega _{i}} G$. $\square$

Corollary 3.17

Let $K_{1}$ and $K_{2}$ be ideals of G such that $K_{1}\ll _{K_{2}}G$ and $K_{2}\ll _{K_{1}}G$. Then $K_{1}\cap K_{2} \ll _{K_{1}+K_{2}} G$.

Proof

First we show that $K_{1}\cap K_{2}\ll _{K_{1}} G$. For this, let $K_{1}\subseteq (K_{1}\cap K_{2})+X$, where X is an ideal of G. Now $K_{1}\subseteq K_{2}+X$ and since $K_{2}\ll _{K_{1}}G$ we get $K_{1}\subseteq X$. Therefore, $K_{1}\cap K_{2}\ll _{K_{1}} G$. In a similar way, we get $K_{1}\cap K_{2}\ll _{K_{2}} G$. Hence, by Proposition 3.16, it follows that $K_{1}\cap K_{2} \ll _{K_{1}+K_{2}} G$. $\square$

The converse of the Corollary 3.17 need not be true, as shown in the following example.

Example 3.18

Consider the N-group ${\mathbb {Z}}_{48}$ over ${\mathbb {Z}}$. Let $K_{1}=8{\mathbb {Z}}_{48}$ and $K_{2}=6{\mathbb {Z}}_{48}$. Then $8{\mathbb {Z}}_{48}\cap 6{\mathbb {Z}}_{48} \ll _{8{\mathbb {Z}}_{48}+6{\mathbb {Z}}_{48}} {\mathbb {Z}}_{48}$, whereas $8{\mathbb {Z}}_{48}\ll _{6{\mathbb {Z}}_{48}} {\mathbb {Z}}_{48}$ and $6{\mathbb {Z}}_{48}$ is not $8{\mathbb {Z}}_{48}$-superfluous in ${\mathbb {Z}}_{48}$.

Proposition 3.19

Let K and $\varOmega$ be ideals of G such that $\varOmega \nsubseteq K$. Let $G^{'}$ be an N-group and $f:G\rightarrow G^{'}$ be an epimorphism with $f(\varOmega )\nsubseteq f(K)$. If $K\ll _{\varOmega }G$, then $f(K)\ll _{f(\varOmega )}G^{'}$. The converse holds if f is injective.

Proof

Suppose that $K\ll _{\varOmega }G$. Since f is an epimorphism, we have $f(K)\trianglelefteq _{N}G^{'}$ by Theorem 1.30 of [17]. Let $X\trianglelefteq _{N} G^{'}$ be such that $f(\varOmega )\subseteq f(K)+X$. Then $\varOmega \subseteq K+f^{-1}(X)$. Since $f^{-1}(X)\trianglelefteq _{N}G$ and $K\ll _{\varOmega }G$, it follows that $\varOmega \subseteq f^{-1}(X)$. Hence $f(\varOmega )\subseteq X$.

Conversely, suppose that f is injective and $f(K)\ll _{f(\varOmega )}G^{'}$. Let $X\trianglelefteq _{N}G$ be such that $\varOmega \subseteq K+X$. Then $f(\varOmega )\subseteq f(K+X)= f(K)+f(X)$. Since $f(K)\ll _{f(\varOmega )}G^{'}$, we have $f(\varOmega )\subseteq f(X)$. Therefore, $f^{-1}(f(\varOmega ))\subseteq f^{-1}(f(X))$. Now by 2.17 of [17], $\varOmega +ker~f \subseteq X+ker~f$. As f is injective, we get $\varOmega \subseteq X$. $\square$

Remark 3.20

Unlike in module over rings, the condition f is a homomorphism is not sufficient, as a homomorphic image of an ideal need not be an ideal. So we consider f to be an epimorphism. The following example justifies the condition f is a homomorphism is not sufficient.

Example 3.21

Consider the nearring given in the Example 3.6 and the ideals $H_{9}=\{e, r^{2}, sr^{3}, sr\}$ and $H_{7}=\{e, r^{2}, s, sr^{2}\}$ of G. Let f be an N-endomorphism of G defined by

$$\begin{aligned} f(g)=g\cdot sr~~~\text {for all}~~ g\in G. \end{aligned}$$

Then $f(H_{9})=\{e,sr^{3}\}$ and $f(H_{7})=\{e,r^{2}\}$. It can be seen that $H_{7}\ll _{H_{9}}G$, but $f(H_{7})$ is not $f(H_{9})$ superfluous in G, since $f(H_{9}) \ntrianglelefteq _{N} G$.

Definition 3.22

Let $\varOmega \trianglelefteq _{N}G$. G is said to be $\varOmega$-hollow if every proper ideal of G which does not contain in $\varOmega$ is $\varOmega$-superfluous in G.

Remark 3.23

• 1. Every hollow N-group is $\varOmega$-hollow with $\varOmega =G$.

• 2. $\varOmega$-hollow need not be hollow and we justify this in the following example.

Example 3.24

Consider the Example 3.6 in which $H_{4}$,  $H_{7}$ are $H_{9}$-superfluous in G and $H_{4}$,$H_{9}$ are $H_{7}$-superfluous in G. Hence it is $H_{7}$-hollow as well as $H_{9}$-hollow. However, G is not hollow, since $H_{7}$ is not superfluous in G as $H_{7}+H_{9}=G$ but $H_{9}\ne G$.

Definition 3.25

Let N be zero-symmetric, and let $\varOmega$, H be ideals of G such that $\varOmega \nsubseteq H$. An N-subgroup K of G is said to be an $\varOmega$-supplement of H if $\varOmega \subseteq H+K$ and $\varOmega \nsubseteq H+K^{'}$ for any ideal $K^{'}$ of K.

Example 3.26

Consider the Example 2.11.

Let $\varOmega =H_{7}$. Here $H_{2}$ is an $\varOmega$-supplement of $H_{4}$, but $H_{2}$ is not a supplement of $H_{4}$ as $H_{2}+H_{4}\ne G$.

Example 3.27

$N=D_{8}$ with the multiplication given in the Table 3. Let $G=N$.

The ideals of G are $I_{1}=G$, $I_{2}=\{e, r^{2},r^{3}, r\}$, $I_{3}=\{e, sr^{3}, r^{2}, sr\}$, $I_{4}=\{e,sr^{2},s,r^{2}\}$, $I_{5}=\{e,r^{2}\}$ and $I_{6}=\{e\}$, and N-subgroups are $I_{1}$, $I_{2}$, $I_{3}$, $I_{4}$, $I_{5}$, $X_{1}=\{e,s\}$, $X_{2}=\{e,sr^{2}\}$, $X_{3}=\{e,sr\}$. Let $\varOmega =I_{4}$. Here $I_{3}$ is an $\varOmega$-supplement of $I_{2}$, $X_{1}$ is an $\varOmega$-supplement of $I_{2}$, $I_{3}$ and $I_{5}$. Further, $X_{1}$ is not a supplement of $I_{5}$ as $I_{5}+X_{1}\ne G$.

Table 3 Multiplication $\star$ on N

Note 3.28

If $\varOmega =G$, then $\varOmega$-supplement coincides with the supplement as defined by [20].

4 Strictly superfluous ideals

In case of N-groups, we have substructures namely N-subgroups and ideals, whereas in modules over rings, these concepts coincide. So we consider the notion strictly superfluous in terms of N-subgroups. We provide explicit examples which indicate that the classes superfluous and strictly superfluous are different.

Definition 4.1

An ideal H of G is called strictly superfluous in G (denoted by $H\ll ^{s}G$) if K is any N-subgroup of G such that $H+K=G$, then $K=G$.

Definition 4.2

Let G be an N-group and $\varOmega \le _{N} G$. An ideal H of G is said to be strictly $\varOmega$-superfluous in G if for any N-subgroup L of G, $\varOmega \subseteq L+H$ implies $\varOmega \subseteq L$. We denote this by $H\ll _{\varOmega }^{s} G$.

Example 4.3

Let $N=\Bigg (\ \begin{pmatrix} {\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} {\mathbb {Z}}_{4} \end{pmatrix},~~ +, ~~\cdot \Bigg )$ where ${\mathbb {Z}}_{4}$ is the set of residue classes modulo 4 and $G=N$.

N-subgroups of G are

$$\begin{aligned} H_{1}= & {} \begin{pmatrix} 0 &{} 0 \\ 0 &{} 0 \end{pmatrix}, H_{2}=\begin{pmatrix} 2{\mathbb {Z}}_{4} &{} 0 \\ 0 &{} 0 \end{pmatrix},\\ H_{3}= & {} \begin{pmatrix} 0 &{} 0 \\ 0 &{} 2{\mathbb {Z}}_{4} \end{pmatrix}, H_{4}=\begin{pmatrix} 0 &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 0 \end{pmatrix}, H_{5}=\begin{pmatrix} 0 &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 2{\mathbb {Z}}_{4} \end{pmatrix}, H_{6}=\begin{pmatrix} 0 &{} 2{\mathbb {Z}}_{4} \\ 0 &{} {\mathbb {Z}}_{4} \end{pmatrix},\\ H_{7}= & {} \begin{pmatrix} 2{\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 0 \end{pmatrix}, H_{8}=\begin{pmatrix} 2{\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 2{\mathbb {Z}}_{4} \end{pmatrix}, H_{9}=\begin{pmatrix} 2{\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} {\mathbb {Z}}_{4} \end{pmatrix}, H_{10}=\begin{pmatrix} {\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 0 \end{pmatrix},\\ H_{11}= & {} \begin{pmatrix} {\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} 2{\mathbb {Z}}_{4} \end{pmatrix}, H_{12}=\begin{pmatrix} 2{\mathbb {Z}}_{4} &{} 0 \\ 0 &{} 2{\mathbb {Z}}_{4} \end{pmatrix}, H_{13}=\begin{pmatrix} {\mathbb {Z}}_{4} &{} 0 \\ 0 &{} 0 \end{pmatrix}, H_{14}=\begin{pmatrix} {\mathbb {Z}}_{4} &{} 2{\mathbb {Z}}_{4} \\ 0 &{} {\mathbb {Z}}_{4} \end{pmatrix}. \end{aligned}$$

Ideals are $H_{1}, H_{2}, H_{3}, H_{4}, H_{5}, H_{6}, H_{7}, H_{8}, H_{9}, H_{10}, H_{11}, H_{12}, H_{14}$. Let $\varOmega = H_{3}$. Then $H_{10}\ll _{\varOmega }^{s} G$ but not strictly superfluous in G since $H_{10}+ H_{11}=G$ and $H_{11}\ne G$. $H_{7}$ is not strictly $H_{12}$-superfluous in G as there exists $H_{5}$ such that $H_{12}\nsubseteq H_{5}$ and $H_{12}\subseteq H_{7}+H_{5}$.

Example 4.4

Consider the N-group given in Example 3.6. Then $H_{4}\ll _{H_{5}}^{s}G$, $H_{7}\ll _{H_{5}}^{s}G$. Here $H_{7}$ is not strictly superfluous, since $H_{7}+H_{5}=G$ but $H_{5}\ne G$. Also $H_{7}\ll _{H_{9}} G$ but $H_{7}$ is not strictly $H_{9}$-superfluous since there exists $H_{5}\le _{N} G$ such that $H_{9}\nsubseteq H_{5}$ but $H_{9}\subseteq H_{7}+H_{5}$.

Proposition 4.5

Let $\varOmega \le _{N}G$, $K\trianglelefteq _{N}G$. If $K\ll _{\varOmega }^{s}G$, then $K\cap \varOmega \ll ^{s}G$.

Proof

Let $K \ll _{\varOmega }^{s} G$. To prove $K \cap \varOmega \ll ^{s} G$, let $L \le _{N} G$ be such that $(K \cap \varOmega )+L=G$. Now $\varOmega \subseteq (K \cap \varOmega )+L \subseteq K+L$. Since $K \ll _{\varOmega }^{s} G$, we have that $\varOmega \subseteq L$. Also since $K\cap \varOmega \subseteq \varOmega \subseteq L$, it follows that $L=(K\cap \varOmega )+L=G$. Therefore, $K\cap \varOmega \ll ^{s} G$. $\square$

Proposition 4.6

Let $P\trianglelefteq _{N} G$ and K, $\varOmega$ be N-subgroups of G such that P and $\varOmega$ are contained in K. Then $P \ll _{\varOmega }^{s} K$ implies $P\ll _{\varOmega }^{s} G$.

Proof

Suppose that $P\ll _{\varOmega }^{s} K$. To prove $P\ll _{\varOmega }^{s} G$, let $L\le _{N}G$ be such that $\varOmega \subseteq P+L$. Since $\varOmega \subseteq K$ and by modular law, we get $\varOmega \subseteq (P+L)\cap K=P+(L\cap K)$. Since $L\cap K\le _{N}K$ and $P\ll _{\varOmega }^{s} K$, we conclude that $\varOmega \subseteq L\cap K \subseteq L$. Therefore $P\ll _{\varOmega }^{s} G$. $\square$

The other implication follows when $K=\varOmega$.

Proposition 4.7

Let $P \trianglelefteq _{N} G$ and $\varOmega \le _{N}G$ such that $P\subset \varOmega$. Then $P\ll _{\varOmega }^{s} G$ if and only if $P\ll ^{s} \varOmega$.

Proof

Suppose $P\ll _{\varOmega }^{s} G$. To prove $P\ll ^{s} \varOmega$, let $L \le _{N} \varOmega$ such that $P+L=\varOmega$. Now $\varOmega \subseteq P+L$ and $P \ll _{\varOmega }^{s} G$, we get $\varOmega \subseteq L$. Since $L\subseteq \varOmega$, it follows that $L=\varOmega$. $\square$

Proposition 4.8

Let $N_{1}$, $N_{2}$ be ideals of G. Let $\varOmega \le _{N}G$ such that $\varOmega \nsubseteq N_{1}$, $\varOmega \nsubseteq N_{2}$. Then $N_{1}\ll _{\varOmega }^{s} G$ and $N_{2}\ll _{\varOmega }^{s} G$ if and only if $N_{1}+N_{2}\ll _{\varOmega }^{s} G$.

Proof

The proof is similar to the proof of Proposition 3.13. $\square$

Proposition 4.9

Let N be zero-symmetric and $\varOmega \le _{N}G$. Let K, P be ideals of G such that $K\subseteq P$, $K\subset \varOmega$ and $\varOmega \nsubseteq P$. Then $P\ll _{\varOmega }^{s} G$ if and only if $K\ll _{\varOmega }^{s} G$ and $\frac{P}{K}\ll _{\frac{\varOmega }{K}}^{s} \frac{G}{K}$.

Proof

The proof is similar to the proof of Proposition 3.15. $\square$

In Proposition 4.10 and 4.11, we assume N to be zero-symmetric, so that every ideal can also be considered as an N-group.

Proposition 4.10

Let N be zero-symmetric, $\varOmega$ be an N-subgroup of G, and $\{\Theta _{j}\}_{j\in J}$ be a family of ideals of G. If $K\trianglelefteq _{N}G$ such that $K\ll _{\varOmega }^{s}G$ and $K\ll _{\Theta _{j}}^{s}G$ for all $j\in J$, then $K\ll _{ \varOmega +\sum _{j} \Theta _{j}}^{s} G$.

Proof

Let $K\ll _{\varOmega }^{s}G$ and $K\ll _{\Theta _{j}}^{s}G$ for all $i\in I$ $j\in J$. Let $L\le _{N}G$ be such that $\varOmega +\sum _{j} \Theta _{j} \subseteq K+L$. Now $\varOmega \subseteq \varOmega +\sum _{j} \Theta _{j} \subseteq K+L$. Since $K\ll _{\varOmega }^{s}G$, we get $\varOmega \subseteq L$. Now $\Theta _{j} \subseteq \varOmega +\sum _{j} \Theta _{j} \subseteq K+L$. Since $K\ll _{\Theta _{j}}^{s}G$, we get $\Theta _{j}\subseteq L$. Therefore $\varOmega +\sum _{j} \Theta _{j}\subseteq L$. Hence $K\ll _{ \varOmega +\sum _{j} \Theta _{j}}^{s}G$. $\square$

Proposition 4.11

Let N be zero-symmetric and $K_{1}$, $K_{2}$ be ideals of G. If $K_{1}\ll _{K_{2}}^{s}G$ and $K_{2}\ll _{K_{1}}^{s}G$, then $K_{1}\cap K_{2} \ll _{K_{1}+K_{2}}^{s}G$.

Proof

Suppose $K_{1}\ll _{K_{2}}^{s}G$ and $K_{2}\ll _{K_{1}}^{s}G$. First we show that $K_{1}\cap K_{2}\ll _{K_{1}}^{s} G$. For this, let $K_{1}\subseteq (K_{1}\cap K_{2})+X$, where $X\le _{N}G$. Then $K_{1}\subseteq K_{2}+X$ and since $K_{2}\ll _{K_{1}}^{s}G$ we get $K_{1}\subseteq X$. Therefore $K_{1}\cap K_{2}\ll _{K_{1}}^{s} G$. In a similar way, we get $K_{1}\cap K_{2}\ll _{K_{2}}^{s} G$. Hence, by Proposition 4.10, $K_{1}\cap K_{2} \ll _{K_{1}+K_{2}}^{s} G$. $\square$

Definition 4.12

Let $G_{1}$ and $G_{2}$ be N-groups and $\varOmega \le _{N}G$. An N-epimorphism $f:G_{1}\rightarrow G_{2}$ is called strictly $\varOmega$-superfluous if $ker~f \ll _{\varOmega }^{s} G_{1}$.

Lemma 4.13

Let $K\trianglelefteq _{N}G$ and $\varOmega \le _{N} G$ be such that $\varOmega \nsubseteq K$. Then $K\ll _{\varOmega }^{s}G$ if and only if the natural map $f:G\rightarrow \frac{G}{K}$ is strictly $\varOmega$-superfluous.

Proof

Since $ker~f=\{g\in G:f(g)=0\in \frac{G}{K}\}=K$, the proof is clear. $\square$

Lemma 4.14

Let $K\trianglelefteq _{N}G$ and $\varOmega \le _{N} G$ be such that $\varOmega \nsubseteq K$. Then $K\ll _{\varOmega }^{s}G$ if and only if for every N-group $G_{1}$ and N-homomorphism $h:G_{1}\rightarrow G$ with $\varOmega \subseteq K+Im~h$, $\varOmega \subseteq Im~h$.

Proof

Suppose $K\ll _{\varOmega }^{s}G$. Let $G_{1}$ be an N-group and $h:G_{1}\rightarrow G$ be an N-homomorphism with $\varOmega \subseteq K+Im~h$. Since Im h is an N-subgroup of G and $K\ll _{\varOmega }^{s}G$, we have $\varOmega \subseteq Im~h$.

Conversely, suppose that $\varOmega \subseteq K+X$ where $X\le _{N} G$. Let $i:X\rightarrow G$ be an inclusion map. Clearly i is an N-homomorphism, and so by hypothesis, we can conclude that $\varOmega \subseteq X$. Therefore, $K\ll _{\varOmega }^{s}G$. $\square$

Lemma 4.15

Let $\varOmega$ be an N-subgroup and K be an ideal of G. Let $G^{'}$ be an N-group and $f:G\rightarrow G^{'}$ be an N-epimorphism such that $f(\varOmega )\nsubseteq f(K)$. If $K\ll _{\varOmega }^{s}G$, then $f(K)\ll _{f(\varOmega )}^{s}G$. The converse holds if f is injective.

Proof

Suppose $K\ll _{\varOmega }^{s}G$. Since f is an epimorphism, we have $f(K)\trianglelefteq _{N}G^{'}$. Let $X\le _{N} G^{'}$ be such that $f(\varOmega )\subseteq f(K)+X$. Then $\varOmega \subseteq K+f^{-1}(X)$. Since $f^{-1}(X)\le _{N}G$ and $K\ll _{\varOmega }^{s}G$, we have $\varOmega \subseteq f^{-1}(X)$. Hence $f(\varOmega )\subseteq X$.

Conversely, suppose that $f(K)\ll _{f(\varOmega )}^{s}G^{'}$. Let $X\le _{N}G$ be such that $\varOmega \subseteq K+X$. Then $f(\varOmega )\subseteq f(K+X)=f(K)+f(X)$. Since $f(K)\ll _{f(\varOmega )}^{s}G^{'}$, we have $f(\varOmega )\subseteq f(X)$. Therefore, $f^{-1}(f(\varOmega ))\subseteq f^{-1}(f(X))$ which implies $\varOmega +ker~f \subseteq X+ker~f$. Since f is injective, we get $\varOmega \subseteq X$. $\square$

Example 4.16

Consider the Example 3.21. Then it can be seen that

$H_{9}\ll _{H_{7}}^{s}G$, but $f(H_{9})$ is not strictly $f(H_{7})$ superfluous in G, since

$f(H_{9})$ is not an ideal of G.

Definition 4.17

Let N be zero-symmetric nearring. Let $\varOmega \le _{N} G$ and $H\trianglelefteq _{N}G$ be such that $\varOmega \nsubseteq H$. An N-subgroup K of G is said to be a strictly $\varOmega$-supplement of H if $\varOmega \subseteq H+K$ and $\varOmega \nsubseteq H+K^{'}$ for any ideal $K^{'}$ of K.

The following remark is a straightforward observation.

Remark 4.18

• 1. If N is zero-symmetric and $\varOmega =G$, then every strictly $\varOmega$-supplement is a supplement (defined by [20]).

• 2. Let N be zero-symmetric. Let $H\trianglelefteq _{N} G$ be such that $\varOmega \nsubseteq H$. Then every $\varOmega$-supplement of H is a strictly $\varOmega$-supplement of H.

• 3. If N is zero-symmetric and $\varOmega =G$, then every strictly $\varOmega$-supplement is a supplement.

5 Superfluous ideals of $M_{n}(N)$-group $N^{n}$

For a zero-symmetric right nearring N with 1, let $N^{n}$ be the direct sum of n copies of $(N,+)$. The elements of $N^{n}$ are column vectors and written as $\left( r_1,\cdots ,r_n\right)$. The symbols $i_{j}$ and $\pi _{j}$ respectively, denote the $i^{th}$ coordinate injective and $j^{th}$ coordinate projective maps.

For an element $a\in N$, $i_{i}(a)=(0,\cdots ,\underbrace{a}_{i^{th}},\cdots ,0)$, and $\pi _{j}(a_1,\cdots ,a_n)=a_{j}$, for any $(a_1,\cdots ,a_n)\in N^n$. The nearring of $n\times n$ matrices over N, denoted by $M_n(N)$, is defined to be the subnearring of $M(N^{n})$, generated by the set of functions {$f_{ij}^{a}:N^{n}\rightarrow N^{n}~|~a\in N,1\le i,j\le n$} where $f_{ij}^{a}\left( k_1,\cdots ,k_n\right) :=\left( l_1,l_2,\cdots ,l_n\right)$ with $l_i=ak_j$ and $l_p=0$ if $p\ne i$. Clearly, $f_{ij}^{a}=i_{i}f^{a}\pi _{j}$, where $f^{a}(x)=ax$, for all $a,x\in N$. If N happens to be a ring, then $f_{ij}^{a}$ corresponds to the $n\times n$-matrix with a in position (i, j) and zeros elsewhere.

Notation 5.1

([9], Notation 1.1)

For any ideal ${\mathcal {A}}$ of $M_n(N)$-group $N^n$, we write

$$\begin{aligned} {\mathcal {A}}_{**}=\{a\in N:a=\pi _{j}A,~\text{ for } \text{ some }~A\in {\mathcal {A}},1\le j\le n\}, { \text{ an } \text{ ideal } \text{ of } _{N}{N}}. \end{aligned}$$

We denote $M_n(N)$ for a matrix nearring, $N^n$ for an $M_n(N)$-group $N^n$. We refer to Meldrum & Van der Walt [15] for preliminary results on matrix nearrings.

From [10], for any $s\in G$, the ideal generated by s is denoted by $\langle s \rangle$ and defined as, $\langle s \rangle =\displaystyle \bigcup _{i=1}^{\infty }U_{i+1}$, where $U_{i+1}=U_{i}^{*}\cup U_{i}^{0}\cup U_{i}^{+}$ with $U_{0}=\{s\}$, and $U_{i}^{*}=\{g+y-g: g\in G, y\in U_{i}\},$ $U_{i}^{0}=\{p-q: p,q\in U_{i}\}\cup \{p+q:p,q\in U_{i}\},$ $U_{i}^{+}=\{n(g+a)-ng: n\in N, g\in G, a\in U_{i}\}.$

Theorem 5.2

(Theorem 1.4 of [9]) Suppose $A\subseteq N$.

• 1. If $A^{n}$ is an ideal of ${}_{M_{n}(N)}{N^{n}}$, then $A=(A^{n})_{\star \star }$.

• 2. If A is an ideal of ${}_{N}{N}$ if and only if $A^{n}$ is an ideal of ${}_{M_{n}(N)}{N^{n}}$.

• 3. If A is an ideal of ${}_{N}{N}$, then $A=(A^{n})_{\star \star }$.

Lemma 5.3

(Lemma 1.5 of [9])

• 1. If ${\mathcal {I}}$ is an ideal of ${}_{M_{n}(N)}{N^{n}}$, then $({\mathcal {I}}_{\star \star })^{n}={\mathcal {I}}$.

• 2. Every ideal ${\mathcal {I}}$ of ${}_{M_{n}(N)}{N^{n}}$ is of the form $K^{n}$ for some ideal K of ${}_{N}{N}$.

Note 5.4

(Note 1.7(iii) of [9]) Let A be an ideal of ${}_{N}^{N}$. Then $A\le _{e} {}_{N}{N}$ if and only if $A^{n}\le _{e} {}_{M_{n}(N)}{N^{n}}$.

Theorem 5.5

(Theorem 1.9 [9]) If $l \in N$, then $\langle l\rangle ^{n}=\langle (l,0,\cdots ,0)\rangle$.

Lemma 5.6

If I and J are ideals of N, then $(I+J)^{n}=I^{n}+J^{n}$.

Proof

Clearly, $I\subseteq I+J$ and $J\subseteq I+J$ which implies $I^{n}\subseteq (I+J)^{n}$ and $J^{n}\subseteq (I+J)^{n}$ and so $I^{n}+J^{n}\subseteq (I+J)^{n}$. To prove the other part, let $(x_{1},x_{2},\cdots ,x_{n})\in (I+J)^{n}$. Then $x_{i}\in I+J$ for every $1\le i\le n$ which implies $x_{i}=a_{i}+b_{i}$, where $a_{i}\in I$ and $b_{i}\in J$.

Now,

$$\begin{aligned} (x_{1},x_{2},\cdots ,x_{n})&=(a_{1}+b_{1}, a_{2}+b_{2},\cdots ,a_{n}+b_{n}) \\&=(a_{1},a_{2},\cdots ,a_{n})+(b_{1},b_{2},\cdots ,b_{n})\\&\in I^{n}+J^{n} \end{aligned}$$

Therefore, $(I+J)^{n}\subseteq I^{n}+J^{n}$. Hence, $(I+J)^{n}=I^{n}+J^{n}$. $\square$

Lemma 5.7

$I+J=G$ if and only if $(I+J)^{n}=G^{n}$ if and only if $I^{n}+J^{n}=G^{n}$.

Definition 5.8

An ideal ${\mathcal {A}}$ of $M_{n}(N)$-group $N^{n}$ is said to be superfluous if for any ideal ${\mathcal {K}}$ of $N^{n}$, ${\mathcal {A}}+{\mathcal {K}}=N^{n}$ implies ${\mathcal {K}}=N^{n}$.

Lemma 5.9

Let B be an ideal of ${}_{N}{N}$. If $B\ll {}_{N}{N}$, then $B^{n}\ll {}_{M_{n}(N)}{N^{n}}$.

Proof

Let ${\mathcal {A}}\trianglelefteq {}_{M_{n}(N)}{N^{n}}$ such that $B^{n}+{\mathcal {A}} =N^{n}$. To show ${\mathcal {A}}=N^{n}$. Since ${\mathcal {A}}\trianglelefteq {}_{M_{n}(N)}{N^{n}}$, by Lemma 5.3, we have ${\mathcal {A}}=({\mathcal {A}}_{\star \star })^{n}$, which implies $B^{n}+({\mathcal {A}}_{\star \star })^{n}=N^{n}$. Now using Lemma 5.6, we get $(B+{\mathcal {A}}_{\star \star })^{n}=N^{n}$. Therefore, by Lemma 5.7, $B+{\mathcal {A}}_{\star \star }=N$. Since, $B\ll {}_{N}{N}$, we get ${\mathcal {A}}_{\star \star }=N$. Hence, ${\mathcal {A}}=({\mathcal {A}}_{\star \star })^{n}=N^{n}$. $\square$

Lemma 5.10

If ${\mathcal {A}}\ll {}_{M_{n}(N)}{N^{n}}$, then ${\mathcal {A}}_{\star \star }\ll {}_{N}{N}$.

Proof

Let $B\trianglelefteq {}_{N}{N}$ such that ${\mathcal {A}}_{\star \star }+B=N$. By Lemma 5.7, we have $({\mathcal {A}}_{\star \star }+B)^{n}=N^{n}$. By Lemma 5.6, we have $({\mathcal {A}}_{\star \star })^{n}+B^n=N^{n}$ which implies ${\mathcal {A}}+B^{n}=N^{n}$. Since $B^{n}\trianglelefteq {}_{M_{n}(N)}{N^{n}}$ and ${\mathcal {A}}\ll {}_{M_{n}(N)}{N^{n}}$, we have $B^{n}=N^{n}$. Let $n\in N$. Then $(n,0,\cdots ,0)\in N^{n}=B^{n}$. Therefore, $n\in (B^{n})_{\star \star }=B$ (by Theorem 5.2(3)). Therefore, $B=N$. $\square$

Theorem 5.11

There is a one-one correspondence between the set of superfluous ideals of ${}_{N}{N}$ and those of $M_{n}(N)$-group $N^{n}$.

Proof

Let $P=\{A\trianglelefteq {}_{N}{N}: A\ll {}_{N}{N} \}$. $Q=\{{\mathcal {A}}\trianglelefteq {}_{M_{n}(N)}{N^{n}}:{\mathcal {A}}\ll {}_{M_{n}(N)}{N^{n}} \}$. Define $\Phi :P\rightarrow Q$ by $\Phi (A)=A^{n}$. Then by Lemma 5.9, $A^{n}\ll {}_{M_{n}(N)}{N^{n}}$. Define $\Psi :Q\rightarrow P$ by $\Psi ({\mathcal {A}})={\mathcal {A}}_{\star \star }$. By Lemma 5.10, ${\mathcal {A}}_{\star \star }\ll {}_{N}{N}$. Now $(\Psi \circ \Phi )(A)=\Psi (\Phi (A))=\Psi (A^{n})=(A^{n})_{\star \star }=A$. Therefore, $(\Psi \circ \Phi )=Id_{P}$. Also, $(\Phi \circ \Psi )({\mathcal {A}})=\Phi (\Psi ({\mathcal {A}}))=\Phi ({\mathcal {A}}_{\star \star })=({\mathcal {A}}_{\star \star })^{n}={\mathcal {A}}$, and hence $(\Phi \circ \Psi )=Id_{Q}$. $\square$

Definition 5.12

An ideal ${\mathcal {K}}$ of $M_{n}(N)$-group $N^{n}$ is said to be g-superfluous if for any ideal ${\mathcal {A}}$ of $N^{n}$, ${\mathcal {K}}+{\mathcal {A}}=N^{n}$ and ${\mathcal {A}}\le _{e} N^{n}$ implies ${\mathcal {K}}=N^{n}$.

Lemma 5.13

Let I be an ideal of ${}_{N}{N}$. If $I\ll _{gs} {}_{N}{N}$, then $I^{n}\ll _{gs}{}_{{}_{M_{n}(N)}}{N^{n}}$.

Proof

Let $I\ll _{gs} {}_{N}{N}$. To show $I^{n}\ll _{gs}{}_{M_{n}(N)}{N^{n}}$, let ${\mathcal {K}}$ be an ideal of ${}_{M_{n}(N)}{N^{n}}$ such that $I^{n}+{\mathcal {K}}={}_{M_{n}(N)}{N^{n}}$ and ${\mathcal {K}}\le _{e} {}_{M_{n}(N)}{N^{n}}$. Since ${\mathcal {K}}\trianglelefteq {}_{M_{n}(N)}{N^{n}}$, by Lemma 5.3(2), we have ${\mathcal {K}}=A^{n}$ for some ideal A of ${}_{N}{N}$. Since ${\mathcal {K}}=A^{n}\le _{e} {}_{M_{n}(N)}{N^{n}}$, by Note 5.4, we have $A\le _{e}{}_{N}{N}$. Now, $I^{n}+{\mathcal {K}}=I^{n}+A^{n}=(I+A)^{n}=N^{n}$ which implies $I+A=N$. Since, $I\ll _{gs} {}_{N}{N}$, we get $A=N$. Therefore, ${\mathcal {K}}=A^{n}=N^{n}$. Hence, $I^{n}\ll _{gs} {}_{M_{n}(N)}{N^{n}}$. $\square$

Lemma 5.14

If ${\mathcal {A}}\ll _{gs} {}_{M_{n}(N)}{N^{n}}$, then ${\mathcal {A}}_{\star \star }\ll _{gs} {}_{N}{N}$.

Proof

Let ${\mathcal {A}}\ll _{gs} {}_{M_{n}(N)}{N^{n}}$. To show ${\mathcal {A}}_{\star \star }\ll _{gs} {}_{N}{N}$, let $B\le _{e}{}_{N}{N}$ such that ${\mathcal {A}}_{\star \star }+B={}_{N}{N}$. Since $B\le _{e}{}_{N}{N}$, by Note 5.4, we have $B^{n}\le _{e} {}_{M_{n}(N)}{N^{n}}$. Now, ${\mathcal {A}}_{\star \star }+B=N$ implies $({\mathcal {A}}_{\star \star }+B)^{n}=N^{n}$. By Lemma 5.6, we get $({\mathcal {A}}_{\star \star })^{n}+B^{n}=N^{n}$. Therefore, ${\mathcal {A}}+B^{n}=N^{n}$. Since ${\mathcal {A}}\ll _{gs} N^{n}$, we get $B^{n}=N^{n}$. Now, by Theorem 5.2(3), we get $B=(B^{n})_{\star \star }=(N^{n})_{\star \star }=N$. Therefore, ${mathcal{A}}_{\star \star }\ll _{gs} {}_{N}{N}$. $\square$

Theorem 5.15

There is a one-one correspondence between the set of g-superfluous ideals of ${}_{N}^{N}$ and those of $M_{n}(N)$-group $N^{n}$.

Proof

Let $P=\{A\trianglelefteq {}_{N}{N}: A\ll _{gs}{}_{N}{N}\}$. $Q=\{{\mathcal {A}}\trianglelefteq {}_{M_{n}(N)}{N^{n}}:{\mathcal {A}}\ll _{gs}{}_{M_{n}(N)}{N^{n}} \}$. Define $\Phi :P\rightarrow Q$ by $\Phi (A)=A^{n}$. Then by Lemma 5.13, $A^{n}\ll _{gs}{}_{M_{n}(N)}{N^{n}}$. Define $\Psi :Q\rightarrow P$ by $\Psi ({\mathcal {A}})={\mathcal {A}}_{\star \star }$. By Lemma 5.14, ${\mathcal {A}}_{\star \star }\ll _{gs}{}_{N}{N}$. Now $(\Psi \circ \Phi )(A)=\Psi (\Phi (A))=\Psi (A^{n})=(A^{n})_{\star \star }=A$. $(\Phi \circ \Psi )({\mathcal {A}})=\Phi (\Psi ({\mathcal {A}}))=\Phi ({\mathcal {A}}_{\star \star })=({\mathcal {A}}_{\star \star })^{n}={\mathcal {A}}$. Therefore, $(\Psi \circ \Phi )=Id_{P}$ and $(\Phi \circ \Psi )=Id_{Q}$. $\square$

Definition 5.16

An element $s\in G$ is called hollow if $\langle s\rangle$ is a hollow ideal of G. In this case we call s as an h-element of G.

Example 5.17

• 1. Let $N=({\mathbb {Z}}_{12}, +_{{\mathbb {Z}}_{12}}, \cdot _{{\mathbb {Z}}_{12}})$ and $G=N$. Then $\langle 3\rangle$ is hollow. Therefore, 3 is a hollow element.

• 2. Let $N=(D_{8},+,\cdot )$ given in Example 2.11 and $G=N$. Then $\langle r^{2}\rangle$ is hollow. Therefore, $r^{2}$ is a hollow element.

Proposition 5.18

s is a hollow element of ${}_{N}{N}$ if and only if $(s,0,0,\cdots ,0)$ is a hollow element in $M_{n}(N)$-group $N^{n}$.

Proof

Suppose s is a hollow element then $\langle s\rangle$ is a hollow ideal. To show $\langle (s,0,\cdots ,0)\rangle$ is a hollow ideal of $M_{n}(N)$-group $N^{n}$, let ${\mathcal {I}}$, ${\mathcal {J}}$ be ideals of $\langle (s,0,\cdots ,0)\rangle$ such that ${\mathcal {I}}+{\mathcal {J}}=\langle (s,0,\cdots ,0)\rangle$. Then by Lemma 5.3(1), we have ${\mathcal {I}}=({\mathcal {I}}_{\star \star })^{n}$, ${\mathcal {J}}=({\mathcal {J}}_{\star \star })^{n}$, which implies $({\mathcal {I}}_{\star \star })^{n}+({\mathcal {J}}_{\star \star })^{n}=\langle (s,0,\cdots ,0)\rangle$. Using Lemma 5.6 and by Theorem 5.5, we get $({\mathcal {I}}_{\star \star }+{\mathcal {J}}_{\star \star })^{n}=\langle (s,0,\cdots ,0)\rangle =\langle s\rangle ^{n}$ and so ${\mathcal {I}}_{\star \star }+{\mathcal {J}}_{\star \star }=\langle s\rangle$. Since, $\langle s\rangle$ is hollow, we get either ${\mathcal {I}}_{\star \star }=\langle s\rangle$ or ${\mathcal {J}}_{\star \star }=\langle s\rangle$.

Therefore,

$$\begin{aligned} \mathcal {{\mathcal {I}}}=({\mathcal {I}}_{\star \star })^{n}=\langle s\rangle ^{n}=\langle (s,0,\cdots ,0)\rangle \end{aligned}$$

or

$$\begin{aligned} {\mathcal {J}}=({\mathcal {J}}_{\star \star })^{n}=\langle s\rangle ^{n}=\langle (s,0,\cdots ,0)\rangle . \end{aligned}$$

Conversely, suppose $(s,0,\cdots ,0)$ is hollow in $N^{n}$. Then $\langle (s,0,\cdots ,0)\rangle$ is a hollow ideal of ${}_{M_{n}(N)}^{N^{n}}$, which implies $\langle s\rangle ^{n}$ is a hollow ideal of ${}_{M_{n}(N)}^{N^{n}}$. To show $\langle s\rangle$ is hollow in N, let I and J be two ideals of N contained in $\langle s\rangle$ such that $I+J=\langle s\rangle$. Now, $(I+J)^{n}=\langle s\rangle ^{n}$. Therefore $I^{n}+J^{n}=\langle s\rangle ^{n}$. Since $\langle s\rangle ^{n}$ is hollow, we have $I^{n}=\langle s\rangle ^{n}$ or $J^{n}=\langle s\rangle ^{n}$, and hence, $I=\langle s\rangle$ or $J=\langle s\rangle$. $\square$

Definition 5.19

$X=\{x_{1},x_{2},\cdots ,x_{n}\}\subseteq G$ is said to be a spanning set for G if $\sum _{x_{i}\in X} \langle x_{i}\rangle =G$. If $\{x_{i}: 1\le i\le n\}$ is a spanning set in G, then we say the elements $x_{i},~ 1\le i\le n$ are spanning elements in G.

Theorem 5.20

$\{x_{i}: 1\le i\le n\}$ is a spanning set in ${}_{N}{N}$ if and only if $\{(x_{i},0,\cdots ,0): 1\le i\le n\}$ is a spanning set in $M_{n}(N)$-group $N^{n}$.

Proof

Suppose $\{x_{i}: 1\le i\le n\}$ is a spanning set in ${}_{N}{N}$. Then

$$\begin{aligned} \sum _{1\le i\le n} \langle x_{i}\rangle =N&\Leftrightarrow \langle x_{1}\rangle +\langle x_{2}\rangle +\cdots +\langle x_{n}\rangle =N\\&\Leftrightarrow (\langle x_{1}\rangle +\langle x_{2}\rangle +\cdots +\langle x_{n}\rangle )^{n}=N^{n}\\&\Leftrightarrow \langle x_{1}\rangle ^{n}+\langle x_{2}\rangle ^{n}+\cdots +\langle x_{n}\rangle ^{n}=N^{n}\\&\Leftrightarrow \langle (x_{1},\cdots ,0)\rangle +\langle (x_{2},0,\cdots ,0)\rangle +\cdots +\langle (x_{n},0,\cdots ,0)\rangle =N^{n}\\&\Leftrightarrow \sum _{1\le i\le n}\langle (x_{i},0,\cdots ,0)\rangle =N^{n}. \end{aligned}$$

Therefore $\{(x_{i},0,\cdots ,0): 1\le i\le n\}$ is a spanning set in $M_{n}(N)$-group $N^{n}$. $\square$

Definition 5.21

A subset X of G is said to be a h-spanning set if every element of X is a h-element and X is a spanning set.

Theorem 5.22

Suppose $x_{1}, x_{2},\cdots ,x_{n}\in N$. Then $\{x_{i}:1\le i\le n\}$ is a h-spanning set in N if and only if $\{(x_{i},0,\cdots ,0):1\le i\le n\}$ is a h-spanning set in $M_{n}(N)$-group $N^{n}$.

Proof

$\{x_{i}:1\le i\le n\}$ is a h-spanning set.

$$\begin{aligned} &\Leftrightarrow x_{i},~1\le i\le n \text { are } h\text {-elements and } \sum _{1\le i\le n}\langle x_{i}\rangle =N\\&\Leftrightarrow (x_{i},0,\cdots ,0), ~1\le i\le n \text { are} ~h\text {-elements in } {}_{M_{n}(N)}{N^{n}}\\ {}&~~~~ \text { and } \sum _{1\le i\le n}\langle (x_{i},0,\cdots ,0)\rangle =N^{n}. \end{aligned}$$

Therefore $\{(x_{i},0,\cdots ,0):1\le i\le n\}$ is a h-spanning set in $M_{n}(N)$-group $N^{n}$. $\square$

6 Superfluous ideal graph of a nearring

The authors [22] studied graphs with respect to superfluous elements in a lattice, and in [21] the authors studied the graphs with respect to the dual aspects such as essential elements, complements, etc. Lattice aspects of modules over rings are well-known due to [3, 7]. In this section, we define the superfluous ideal graph of a nearring and study some of its properties.

Definition 6.1

Let N be a nearring. An ideal I of N is said to be superfluous if for any ideal J of N, $I+J=N$ implies $J=N$.

Definition 6.2

The superfluous ideal graph of N, denoted by $S_{N}(G)$, is a graph having set of all non-zero proper ideals of N as vertices and two vertices I and J are adjacent if $I\cap J\ll N$.

Example 6.3

• 1. If N is simple, then $S_{N}(G)$ is a null graph.

• 2. Suppose N is a finitely generated nearring which contains only one non-zero maximal ideal, then every proper ideal of N is superfluous. The vertices of $S_{N}(G)$ are the non-zero proper ideals of N. Since every proper ideal of N is superfluous in N, we have $I\cap J\ll G$ for all proper ideals I, J of N. Therefore $S_{N}(G)$ is a complete graph.

  For example, let $N=(Z_{p^{n}}, +_{p^{n}}, \cdot )$ where p is prime. Then possible ideals are of the form $\langle p^{i}\rangle ,~~i\in \{0,1,\cdots ,n-1\}$. If N is simple, then $S_{N}(G)$ is a null graph. If N is not simple, then N has only one non-zero maximal ideal of the form $\langle p^{k}\rangle$ for some $0\le k\le n-1$. Hence, N is a local nearring. In this case we get a complete graph.

  Consider $R=(Z_{p^{n}}, +_{p^{n}}, \cdot _{p^{n}})$ where addition and multiplication are modulo $p^{n}$. Then R is a ring. In this case, the superfluous ideal graph is a complete graph with $(n-1)$ vertices.

Example 6.4

Let $N=({\mathbb {Z}}_{6}, +_{{\mathbb {Z}}_{6}}, \cdot _{{\mathbb {Z}}_{6}})$. Then $V(S_{{\mathbb {Z}}_{6}}(G))=\{2{\mathbb {Z}}_{6}, 3{\mathbb {Z}}_{6}\}$. Now $2{\mathbb {Z}}_{6}\cap 3{\mathbb {Z}}_{6} =(0)\ll N$. The graph $S_{{\mathbb {Z}}_{6}}(G)$ is shown in Fig. 1.

Fig. 1

$S_{\mathbb {Z}_{6}}(G)$

Fig. 2

$S_{\mathbb {Z}_{12}}(G)$

Example 6.5

Let $N=({\mathbb {Z}}_{12}, +_{{\mathbb {Z}}_{12}}, \cdot _{{\mathbb {Z}}_{12}})$. Non-zero proper ideals of N are $2{\mathbb {Z}}_{12}$, $3{\mathbb {Z}}_{12}$, $4{\mathbb {Z}}_{12}$, $6{\mathbb {Z}}_{12}$ and $6{\mathbb {Z}}_{12}$ is superfluous in ${\mathbb {Z}}_{12}$. Then the corresponding superfluous ideal graph is given in Fig. 2.

Example 6.6

Let $N=({\mathbb {Z}}_{2}\times {\mathbb {Z}}_{2}, +_{{\mathbb {Z}}_{2}}, \cdot _{{\mathbb {Z}}_{2}})$ where addition and multiplication are carried out component-wise. All non-zero proper ideals are of N are $(0)\times {\mathbb {Z}}_{2}$ and ${\mathbb {Z}}_{2}\times (0)$ and $((0)\times {\mathbb {Z}}_{2}) \cap ({\mathbb {Z}}_{2}\times (0))=(0)\ll N$. Therefore, the superfluous ideal graph is given in Fig. 3.

Example 6.7

Let $N=({\mathbb {Z}}_{4}\times {\mathbb {Z}}_{2}, +, \cdot )$ where addition and multiplication are carried out component-wise with the first component modulo 4 and the second component modulo 2. Then the nontrivial ideals are $I_{1}=\{(0,0),(1,0),(2,0),(3,0)\}$, $I_{2}=\{(0,0),(2,0),(0,1),(2,1)\}$, $I_{3}=\{(0,0),(0,1)\}$, $I_{4}=\{(0,0),(2,0)\}$ and $I_{4}$ is a superfluous ideal. The corresponding superfluous ideal graph is given in Fig. 4.

Example 6.8

Consider the nearring given in the Example 2.11. The ideals of N are $H_{4}$, $H_{9}$, $H_{7}$ and it can be seen that $H_{4}$ is superfluous in N. We have $H_{9}\cap H_{7}=H_{4}$, $H_{9}\cap H_{4}=H_{4}$ and $H_{7}\cap H_{4}=H_{4}$. Hence, we get a complete graph given in Fig. 5.

Proposition 6.9

Every non-zero superfluous ideal of N is a universal vertex in $S_{N}(G)$.

Proof

Let X be a non-zero superfluous ideal of N. To prove $XY \in E$ for every $Y \in V$. Let $Y \in V$. By Lemma 3.7, $X\cap Y \ll N$ which implies $XY \in E$. Since Y is arbitrary, X is a universal vertex. $\square$

The converse of the Proposition 6.9 need not be true. We justify this in the following example.

Example 6.10

In Example 6.5, $N=({\mathbb {Z}}_{12}, +_{{\mathbb {Z}}_{12}}, \cdot _{{\mathbb {Z}}_{12}})$. Then $6{\mathbb {Z}}_{12}$ is a non-zero superfluous ideal which is a universal vertex in the corresponding superfluous ideal graph given in Fig. 2. The vertex $3{\mathbb {Z}}_{12}$ is universal but it is not superfluous, as $3{\mathbb {Z}}_{12}+2{\mathbb {Z}}_{12}={\mathbb {Z}}_{12}$, and $2{\mathbb {Z}}_{12} \ne {\mathbb {Z}}_{12}$.

Proposition 6.11

The subgraph $S_{N}[min(N)]$ induced by min(N) is a clique, where min(N) is the set of minimal ideals of N.

Fig. 3

$S_{\mathbb {Z}_{2}\times \mathbb {Z}_{2}}(G)$

Fig. 4

$S_{\mathbb {Z}_{4}\times \mathbb {Z}_{2}}(G)$

Fig. 5

Superfluous ideal graph of the nearring in Example 2.11

Proof

Case 1: Suppose N has exactly one minimal ideal. Then we get a clique $K_{1}$.

Case 2: Suppose N has more than one minimal ideal. Let $M_{1}$, $M_{2}$ be two arbitrary minimal ideals of N. We prove that $M_{1}M_{2}\in E(S_{N}(G))$. Since, $M_{1}$ and $M_{2}$ both are minimal $M_{1}\cap M_{2}=(0)$, which is a superfluous ideal of N, which implies $M_{1}M_{2}\in E(S_{N}(G))$. Since $M_{1}$ and $M_{2}$ are arbitrary, we conclude that there exists an edge between any two minimal ideals. Therefore $S_{N}[min(N)]$ is a clique (Fig. 3). $\square$

Proposition 6.12

$S_{N}(G)$ is an empty graph if and only if N has exactly one non-zero proper ideal.

Proof

If N has exactly one non-zero proper ideal then $S_{N}(G)=K_{1}$. Conversely, suppose $S_{N}(G)$ is an empty graph. We prove that N has exactly one non-zero proper ideal. First we prove that N has exactly one minimal ideal. Suppose on a contrary, N has two minimal ideals $M_{1}$ and $M_{2}$. Then by Proposition 6.11, $M_{1}$ and $M_{2}$ are adjacent in $S_{N}(G)$, a contradiction since $S_{N}(G)$ is an empty graph. Therefore N has a unique minimal ideal say, M. So every non-zero ideal of N different from M contains M. Therefore M is superfluous. We claim that M is the only unique proper ideal of N. On a contrary, suppose $I\ne M$ be a non-zero proper ideal of N. Then $M\subseteq I$, $M\cap I =M$, which is superfluous in N, and we get $MI\in E(S_{N}(G))$, a contradiction, since $S_{N}(G)$ is an empty graph. Therefore M is the unique non-zero proper ideal of N (Fig. 4). $\square$

Definition 6.13

Let I be an ideal of N. The dual annihilator of I, denoted as $ann_{d}(I)$ is the intersection of all ideals J of N such that $I+J=N$. That is, $ann_{d}(I)= \bigcap \limits _{J\trianglelefteq _{N} N,~ I+J=N}J$ (Fig. 5).

Example 6.14

• 1. In the nearring given in Example 2.11, the ideals of N are $H_{7}$, $H_{9}$, $H_{5}$ and $\{e\}$. Therefore $ann_{d}(H_{7})=\cap \{H_{9}, N\}=H_{9}$.

• 2. In the nearring N given in Example 3.27, the ideals of N are N, $I_{2}$, $I_{3}$, $I_{4}$, $I_{5}$ and $\{e\}$. We have $I_{2}+I_{3}=N$ and $I_{2}+I_{4}=N$. Therefore $ann_{d}(I_{2})= \bigcap \{I_{3},I_{4}\}=I_{5}$.

Proposition 1

Let I be any arbitrary ideal of N. Then $I\cap (ann_{d}(I))\ll N$.

Proof

Let $K\trianglelefteq N$ such that $I\cap (ann_{d}(I))+K=N$. Since $I\cap (ann_{d}(I)) \subseteq I$, we have $I+K=N$, which implies $ann_{d}(I)\subseteq K$ and so $I\cap (ann_{d}(I))\subseteq K$. Now $K= K+I\cap (ann_{d}(I))=N$. Therefore, $I\cap (ann_{d}(I))\ll N$. $\square$

7 Conclusion

We have defined the notions superfluous, strictly superfluous (with respect to an ideal $\varOmega$), generalised superfluous, generalised suppements in N-groups. We have proved some properties and exhibited examples which are different from the existing notions. We have defined graph on superfluous ideals of a nearring, and gave some properties. The concepts can be extended to study various finite spanning dimension aspects and related chain conditions in N-groups and those of $M_{n}(N)$-group $N^{n}$.