On equivalence of cyclic and dihedral zero-divisor codes having nilpotents of nilpotency degree two as generators

Abstract

Zero-divisor codes are codes constructed using group rings where their generators are zero-divisors. Generally, zero-divisor codes can be equivalent despite their associated groups are non-isomorphic, leading to the proposed conjecture “Every dihedral zero-divisor code has an equivalent form of cyclic zero-divisor code”. This paper is devoted to study equivalence of zero-divisor codes in $F_2G$ having generators from the 2-nilradical of $F_2G$, consisting of all nilpotents of nilpotency degree 2 of $F_2G$. Essentially, algebraic structures of 2-nilradicals are first studied in general for both commutative and non-commutative $F_2G$ before specialized into the case when G is cyclic and dihedral. Then, results are used to study the conjecture above in the cases where the codes generators are from their respective 2-nilradicals.

1 Introduction

Codes defined by non-trivial ideals of group algebras are called group codes where their generators are essentially zero-divisors [1, 5]. Generally, a group code in $F_2G$ has the form of $F_2G\lbrace u_1,u_2,\dots ,u_k\rbrace$ where $\lbrace u_1,u_2,\dots ,u_k\rbrace \subseteq F_2G$ is called a set of generators of the group code. In 2009, Ted and Paul Hurley introduced a new encoding method using group rings, where the resultant codes need not have ideal structures and are called group ring codes [4]. A group ring code in $F_2G$ is defined as $Wu=\lbrace wu\vert w\in W\rbrace$, where W is a $F_2$-submodule of $F_2G$ and $u\in F_2G$. In addition, if the generator $u\in F_2G$ is a zero-divisor or unit, then Wu is called a zero-divisor code or unit-derived code respectively. For every $u=\sum \limits _{g\in G}a_gg\in F_2G$, define its support $supp(u)=\lbrace g\in G\,\vert \, a_g\ne 0\rbrace$ and weight $wt(u)=\vert supp(u)\vert$. Details about group rings can be found in [6].

It is instructive to note that zero-divisor codes are of single generator whereas group codes can have multiple generators in general. Moreover, any zero-divisor code Wu possesses module structure and is in fact a subcode of the corresponding group code $F_2Gu$. Note that when $W=F_2G$, the resultant zero-divisor code Wu possess left ideal structure.

Two length n binary codes are said to be equivalent if one can be obtained from another via permuting the n digits of the codewords. The main study of equivalent codes is to develop conditions which classify codes that look different as a set but perform alike as codes from both theoretical and practical points. Equivalent codes exhibit similar algebraic structures as well as always having identical parameters.

Some results on conditions for equivalence between abelian group codes are given in [3]. However, there are instances where despite two groups $G_1$ and $G_2$ are non-isomorphic, equivalence can be established between group ring codes in $F_2G_1$ and $F_2G_2$ as in [10], where $G_1=D_{2n}$ and $G_2=C_{2n}$ are the dihedral and cyclic groups of order 2n respectively. This led to the conjecture “Every zero-divisor code in $F_2D_{2n}$ is equivalent to some zero-divisor codes in $F_2C_{2n}$” [10].

To answer this conjecture, a comprehensive study on equivalence between zero-divisor codes over $F_2$ was done by Ong and Ang in 2019 [7]. In the same paper [7], the equivalence of zero-divisor codes having idempotents as generators were studied in details. This was done by using a new classification of idempotents in $F_2G$ which was introduced in [8, 9]. As a result, the conjecture “Every zero-divisor code in $F_2D_{2n}$ is equivalent to some in $F_2C_{2n}$” was affirmed in the cases where the generators are those classified idempotents [7].

An element of a ring $u\in R$ is called a nilpotent if there exists $n\in {\mathbb {Z}}^+$ such that $u^n=0$, where the smallest such n is termed the nilpotency degree of u. Consider a group algebra $F_2G$, where G is a finite group. In this paper, we are particularly interested with those $u\in F_2G$ with $u^2=0$, namely the nilpotents of nilpotency degree 2. The set of all nilpotents of nilpotency degree 2 together with 0 is called the 2-nilradical of $F_2G$, denoted by $Nil_2(F_2G)$.

As nilpotents are also zero-divisors, this paper is mainly devoted to study the equivalence between dihedral and cyclic zero-divisor codes having generators from their respective $Nil_2(F_2G)$. Section 2 comprises some relevant results on equivalent zero-divisor codes in [7], followed by some basic properties of nilpotents. In Sect. 3 and 4, we study $Nil_2(F_2G)$ for both commutative and non-commutative $F_2G$ then follow up with the specific cases when G is cyclic and dihedral. The results are then applied in Sect. 5 to establish the equivalence between some dihedral and cyclic zero-divisor codes having generators from their respective 2-nilradicals. In particular, given a dihedral zero-divisor code with generator from its $Nil_2(F_2G)$, we study the sufficient conditions that ensure the code being equivalent to a cyclic zero-divisor code having generator from its respective $Nil_2(F_2G)$.

2 Preliminaries

In 1962, MacWilliams had affirmed that two linear codes of the same length are equivalent if and only if there is a weight-preserving isomorphism between them [2]. Here, we give a definition of equivalent zero-divisor codes by slightly modifying MacWilliams’s into the following analogous version.

Definition 2.1

Let $G_1$ and $G_2$ be two finite groups with $\vert G_1 \vert = \vert G_2 \vert$. Let $C_1$ and $C_2$ be zero-divisor codes in $F_2G_1$ and $F_2G_2$ respectively. Then, $C_1$ and $C_2$ are said to be equivalent if there exists an $F_2$-linear isomorphism $\varphi : C_1 \rightarrow C_2$ such that for every codeword $c\in C_1$, $wt(c)=wt((c)\varphi )$.

The property $wt(c)=wt((c)\varphi )$ for every codeword $c\in C_1$ is called weight-preserving. Recall that for every $F_2$-submodule W of $F_2G$, Wu is an $F_2$-submodule of the corresponding group code $F_2Gu$. Then, the following proposition clearly holds.

Proposition 2.2

Let $\varphi : F_2G_1u_1 \rightarrow F_2G_2u_2$ be a weight-preserving $F_2$-linear isomorphism. Restrict the domain of $\varphi$ to $C_1u$, an $F_2$-submodule of $F_2G_1u_1$ by defining $\varphi _\downarrow :C_1u\rightarrow F_2G_2u_2$ in such a way that $(c)\varphi _\downarrow =(c)\varphi$ for every $c\in C_1u.$ Then, $C_1u$ is equivalent to $Im(\varphi _\downarrow )$, an $F_2$-submodule of $F_2G_2$.

Proposition 2.2 ensures that if $F_2G_1u_1$ and $F_2G_2u_2$ are equivalent, essentially each of the $F_2$-submodule of $F_2G_1u_1$ has an equivalent form in $F_2G_2u_2$. Sufficiently, the study of equivalence in this paper is done on group codes $F_2G_1u_1$ and $F_2G_2u_2$ in this paper.

For distinct $F_2G_1u_1$ and $F_2G_2u_2$, the problem of determining their equivalence has been structured into finding a weight-preserving $F_2$-linear isomorphism $\varphi :F_2G_1u_1\rightarrow F_2G_2u_2$. We concentrated on $\varphi$ which are expressible in terms of some bijective $\chi :G_1\rightarrow G_2$, that is $(\sum \limits _{g\in G_1}{a_gg})\varphi =\sum \limits _{g\in G_1}{a_g\chi (g)}$. Then, it is proven that the exhibition of $\varphi$ being weight-preserving $F_2$-linear isomorphic is sufficiently dependent on the properties of $\chi$ illustrated as follows.

Definition 2.3

Let $u_1$ and $u_2$ be zero-divisors in $F_2G_1$ and $F_2G_2$ respectively. Let $\chi :A\rightarrow B$ be a bijective function for $A\subseteq G_1$ and $B\subseteq G_2$. Then:

• 1. $\chi$ is denoted as $\chi _{u_1,u_2}$ if $\chi (supp(u_1))=supp(u_2)$.

• 2. $\chi$ is termed a $u_1$-homomorphism if $\chi (gh)=\chi (g)\chi (h)$ for every $g\in A$ and $h\in supp(u_1)$.

The above property of bijective map $\chi$ was first introduced in [7] and shown to be sufficient for resulting in equivalence between group codes as follows.

Theorem 2.4

Let $u_1$ and $u_2$ be zero-divisors in $F_2G_1$ and $F_2G_2$ respectively. If there exists $\chi _{u_1,u_2}:G_1\rightarrow G_2$ which is a $u_1$-homomorphism, then $F_2G_1u_1$ and $F_2G_2u_2$ are equivalent codes.

Proof Refer to Theorem 4.4 in [7].

In the same paper [7], Theorem 4.4 was utilized to study the equivalence of zero-divisor codes having idempotents as generators. This was done by using a new classification of idempotents in $F_2G$ which was introduced by us in [8, 9]. As a result, we are able to show that each dihedral zero-divisor codes in $F_2D_{2n}$ with our classified idempotent generators has an equivalent cyclic form in $F_2C_{2n}$ [7].

The remaining of this section is devoted to discuss some properties of nilpotents. Let R be a commutative ring. The set of all nilpotents in R is called the nilradical and is denoted by Nil(R). It is well known that Nil(R) is an ideal when R is commutative. In this paper, we use the notion of Nil(R) to represent the set of all nilpotents in R, for general R. Let I be an ideal of R. If $I\subseteq Nil(R)$, then I is called a nil-ideal of R.

Let G be a finite group, then $g\in G$ is called an involution if $ord(g)=2$. We denote the set of all involutions in G by Inv(G). Let $u=1+g\in F_2G$ where $g\in Inv(G)$, then clearly $u^2=(1+g)^2=0$, that is $u\in Nil(F_2G)$. If $u\in Nil(F_2G)\setminus \{0\}$ with $u^2=0$, then u is a nilpotent of nilpotency degree 2. Recall that the set of all nilpotents of nilpotency degree 2 together with 0 is called the 2-nilradical of $F_2G$ and is denoted by $Nil_2(F_2G)$. Clearly, $\lbrace 1+g\,\vert \,g\in Inv(G)\rbrace \subseteq Nil_2(F_2G)$. In addition, any ideal $I\subseteq Nil_2(F_2G)$ is called a 2-nil ideal.

3 $Nil_2(F_2G)$ for commutative $F_2G$

Throughout this section, G is fixed to be finite abelian and thus $F_2G$ is a commutative ring. First, we show that $Nil_2(F_2G)$ possesses ideal structure.

Proposition 3.1

Let $F_2G$ be a commutative ring. Then, $Nil_2(F_2G)$ is an ideal of $F_2G$.

Proof

Let $u_1,u_2\in Nil_2(F_2G)$, then $u_1^2=0$ and $u_2^2=0$. Note that $(u_1+u_2)^2=u_1^2+u_1u_2+u_2u_1+u_2^2=0+2u_1u_2+0=0$ and $(u_1u_2)^2=u_1u_2u_1u_2=u_1^2u_2^2=0$. Also, for $v\in F_2G$, $vu_1\in Nil_2(F_2G)$ since $(vu_1)^2=vu_1vu_1=v^2u_1^2=v^2(0)=0$. $\square$

Let $u=\sum \limits _{i=1}^kg_i\in F_2G$ for some $k\in {\mathbb {Z}}^+$. As $char(F_2)=2$, it can be easily verified that for $F_2G$, $u^2=(\sum \limits _{i=1}^kg_i)^2= \sum \limits _{i=1}^kg_i^2$. Hence $u^2=0$ if and only if $\sum \limits _{i=1}^kg_i^2=0$ and k must be even. This leads to Proposition 3.2.

Proposition 3.2

Let $u\in F_2G$ for some commutative $F_2G$. Then, $u=\sum \limits _{i=1}^kg_i\in Nil_2(F_2G)\setminus \lbrace 0\rbrace$ if and only if supp(u) can be partitioned into $\frac{k}{2}$ pairs of $g_s\ne g_t$, with $s,t\in \lbrace 1,2,\dots ,k\rbrace$ such that $g_s^2=g_t^2$ or equivalently $g_s+g_t\in Nil_2(F_2G)\setminus \lbrace 0\rbrace$.

Now, our task is to identify $Nil_2(F_2G)$. As a starting point, we show that when $\vert G\vert$ is odd, $Nil_2(F_2G)$ is trivial.

Proposition 3.3

Let G be an abelian group of odd order. Then, $Nil_2(F_2G)=\lbrace 0\rbrace$.

Proof

Write $G=\mathop {\times }\limits _{i=1}^k C_{n_i}$ where each $C_{n_i}=\langle x_i\rangle$ has odd order $n_i$. Let $u=\sum \limits _{i=1}^lg_i\in Nil_2(F_2G)$. Suppose that $Nil_2(F_2G)$ is non-trivial, by Proposition 3.2, we can always partition supp(u) into pairs of $g_s\ne g_t$ with $g_s^2=g_t^2$. Let $g_s=\prod \limits _{i=1}^kx_i^{s_i}$ and $g_t=\prod \limits _{i=1}^kx_i^{t_i}$. Then, $g_s^2=g_t^2$ implies that for every $i\in \lbrace 1,2,\dots ,k\rbrace$, $2s_i=2t_i(\mathrm {mod}\ n_i)$. Since for each i, $n_i$ is odd, this results in $s_i=t_i(\mathrm {mod}\ n_i)$, that is $g_s=g_t$. Therefore, $Nil_2(F_2G)=\lbrace 0\rbrace$. $\square$

Next, for the case when G is of even order, note that Inv(G) is always non-empty. The following theorem identifies $Nil_2(F_2G)$ using elements in Inv(G). $\square$

Theorem 3.4

Let G be an abelian group of even order. Then, $Nil_2(F_2G)=F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$.

Proof

Let $G=H\times K$ where $\vert H\vert$ is odd and $\vert K\vert =2^l$ for some $l\in {\mathbb {Z}}^+$. Let $K=\mathop {\times }\limits _{i=1}^k C_{n_i}$ where each $C_{n_i}=\langle x_i\rangle$. Let $u\in Nil_2(F_2G)\setminus \lbrace 0\rbrace$. It follows from Proposition 3.2 that supp(u) can be partitioned into pairs of $g_s\ne g_t$, with $s,t\in \lbrace 1,2,\dots ,k\rbrace$ such that $g_s^2=g_t^2$. Let $g_s=h_s\prod \limits _{i=1}^kx_i^{s_i}$ and $g_t=h_t\prod \limits _{i=1}^kx_i^{t_i}$ for some $h_s,h_t\in H$. Then, $g_s^2=g_t^2$ implies that $h_s^2\prod \limits _{i=1}^kx_i^{2s_i}=h_t^2\prod \limits _{i=1}^kx_i^{2t_i}$. As $ord(h_s)$ and $ord(h_t)$ are both odd, as illustrated in the proof of Proposition 3.3, we have $h_s=h_t$. Now, from $h_s^2\prod \limits _{i=1}^kx_i^{2s_i}=h_s^2\prod \limits _{i=1}^kx_i^{2t_i}$, we have $2s_i\equiv 2t_i (\mathrm {mod} \, n_i)$ for every $i\in \lbrace 1,2,\dots ,k\rbrace$. Since $n_i$ is even, this results in $s_i\equiv t_i (\mathrm {mod} \, \frac{n_i}{2})$. Hence, for each $i\in \lbrace 1,2,\dots ,k\rbrace$, either $s_i\equiv t_i (\mathrm {mod} \, n_i)$ or $s_i\equiv t_i +\frac{n_i}{2} (\mathrm {mod} \, n_i)$. Then, re-enumerate $\prod \limits _{i=1}^kx_i^{s_i}$ and $\prod \limits _{i=1}^kx_i^{t_i}$ in the form of $\prod \limits _{i=1}^mx_i^{s_i}\prod \limits _{i=m+1}^kx_i^{s_i}$ and $\prod \limits _{i=1}^mx_i^{t_i}\prod \limits _{i=m+1}^{k}x_i^{t_i}$ respectively, where $s_i\equiv t_i (\mathrm {mod} \, n_i)$ for every $i\in \lbrace 1,2,\dots ,m\rbrace$ and $s_i\equiv t_i +\frac{n_i}{2} (\mathrm {mod} \, n_i)$ for every $i\in \lbrace m+1,m+2,\dots ,k\rbrace$. This leads to $\prod \limits _{i=1}^mx_i^{t_i}=\prod \limits _{i=1}^mx_i^{s_i}$ and $\prod \limits _{i=m+1}^{k}x_i^{t_i}=\prod \limits _{i=m+1}^{k}x_i^{s_i+\frac{n_i}{2}}$. Hence:

$$\begin{aligned} g_s+g_t= & {} h_s\prod \limits _{i=1}^mx_i^{s_i}\prod \limits _{i=m+1}^kx_i^{s_i}+h_t\prod \limits _{i=1}^mx_i^{t_i}\prod \limits _{i=m+1}^{k}x_i^{t_i} \\= & {} h_s\prod \limits _{i=1}^mx_i^{s_i}\left( \prod \limits _{i=m+1}^kx_i^{s_i}+\prod \limits _{i=m+1}^{k}x_i^{s_i+\frac{n_i}{2}}\right) \\= & {} h_s\prod \limits _{i=1}^mx_i^{s_i}\prod \limits _{i=m+1}^kx_i^{s_i}\left( 1+\prod \limits _{i=m+1}^{k}x_i^{\frac{n_i}{2}}\right) \in F_2G\left( 1+\prod \limits _{i=m+1}^{k}x_i^{\frac{n_i}{2}}\right) \end{aligned}$$

Thus, $g_s+g_t\in F_2G(1+\prod \limits _{i=m+1}^{k}x_i^{\frac{n_i}{2}})\subseteq F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$ as $\prod \limits _{i=m+1}^{k}x_i^{\frac{n_i}{2}}\in Inv(G)$. Then, it follows from the closure property of addition that $u\in F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$ and $Nil_2(F_2G)\subseteq F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$. Conversely, let $u\in F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$. Express $u=\sum \limits _{g\in Inv(G)}u_g(1+g)$ where each $u_g\in F_2G$. Then, $u^2=(\sum \limits _{g\in Inv(G)}u_g(1+g))^2=\sum \limits _{g\in Inv(G)}u_g^2(1+g)^2=\sum \limits _{g\in Inv(G)}u_g^2(0)=0$. Hence, $u\in F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace \subseteq Nil_2(F_2G)$. This concludes that $Nil_2(F_2G)= F_2G\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$. $\square$

We conclude this section with the case when G is cyclic. Consider $C_n=\langle x\rangle$ for some even $n\in {\mathbb {Z}}^+$, since $Inv(C_n)=\lbrace x^{\frac{n}{2}}\rbrace$, Corollary 3.5 follows directly from the theorem above.

Corollary 3.5

Let $C_n=\langle x\rangle$ with n being even. The 2-nilradical of $F_2C_n$ is $F_2C_n(1+x^{\frac{n}{2}})$.

4 $Nil_2(F_2G)$ for non-commutative $F_2G$

Throughout this section, G is fixed to be finite non-abelian and thus $F_2G$ is a non-commutative ring, unless otherwise stated. For $u_1,u_2\in F_2G$, their commutator is defined as $[u_1,u_2]= u_1u_2+u_2u_1$. Note that $u_1u_2=u_2u_1$ if and only if $[u_1,u_2]=0$.

For commutative $F_2G$, Theorem 3.4 implies that for every $g\in Inv(G)$ and $h\in G$, we have $h(1+g)\in Nil_2(F_2G)$. However, for non-commutative $F_2G$, this desirable condition need not hold for every $h\in G$. In fact, the details are given by the following lemma.

Lemma 4.1

Let $g\in Inv(G)$. Then, for each $h\in G$, $h(1+g)\in Nil_2(F_2G)$ if and only if $[g,h]=0$.

Let us clarify some notations. The centralizer of a set $S\subseteq G$ is denoted by $C_{G}(S)$. For convenience, if $S=\lbrace s\rbrace$, we write $C_{G}(\lbrace s\rbrace )$ as $C_{G}(s)$. Also, Z(G) denotes the centre of G. Then, the next result follows from Lemma 4.1.

Proposition 4.2

Let $g\in Inv(G)\cap Z(G)$. Then, for every $h\in G$, $h(1+g)\in Nil_2(F_2G)$. Moreover, $F_2G(1+g)$ is a 2-nil ideal.

Recall that for commutative $F_2G$, $\lbrace 1+g \, \vert \, g\in Inv(G)\rbrace$ serves as a generating set of $Nil_2(F_2G)$. Unfortunately, this property fails to hold for non-commutative $F_2G$ in two different ways.

First, we claim that there exists $u\in Nil_2(G)$ which is not expressible as a summation of $h(1+g)\in Nil_2(F_2G)$ for $g\in Inv(G)$ and $h\in G$. Before proving this, the following proposition is necessary. Note that for a subgroup H of G, the notation $\bar{H}$ is used to denote $\sum \limits _{h\in H}h\in F_2G$.

Proposition 4.3

Let H be an even order subgroup of G.Then, $\bar{H}\in Nil_2(F_2G)$.

Proof Since $\vert H\vert$ is even, this results in $\bar{H}^2=\bar{H}\bar{H}=\vert H\vert \bar{H}=0$. $\square$

Consider the dihedral group $D_6=\langle a,b\,\vert \, a^3=b^2=1,ab=ba^{-1}\rangle$. Proposition 4.3 guarantees that $\bar{D_{6}}\in Nil_2(F_2D_6)$. It can be easily verified that $Inv(D_6)=\lbrace a^ib\,\vert \, i\in \lbrace 0,1,2\rbrace \rbrace$ where for each $a^ib$, $C_{G}(a^ib)=\lbrace 1, a^ib\rbrace$. Note that $a^ib(1+a^ib)=1(1+a^ib)=1+a^ib$. This concludes that $1+b,1+ab$ and $1+a^2b$ are the only possible choices of $h(1+g)\in Nil_2(F_2D_{2n})$. Clearly, summation between those elements would never yield $\bar{D_{6}}$.

Second, note that $Nil_2(F_2G)$ need not be closed under addition, as shown by the counterexample $1+b,1+ab\in F_2D_6$. To understand the algebraic structure of $Nil_2(F_2G)$, the conditions that are necessary or sufficient for $u_1+u_2\in Nil_2(F_2G)$ for distinct $u_1,u_2\in Nil_2(F_2G)$ is studied.

Proposition 4.4

Let H and K be even order subgroups of G where H is normal. Then, $\bar{H}+\bar{K}\in Nil_2(F_2G)$.

Proof Note that $(\bar{H}+\bar{K})^2=\bar{H}^2+\bar{H}\bar{K}+\bar{K}\bar{H}+\bar{K}^2$. By Proposition 4.3, we have $\bar{H}^2=\bar{K}^2=0$ and this yields $(\bar{H}+\bar{K})^2=\bar{H}\bar{K}+\bar{K}\bar{H}=\sum \limits _{k\in K}\bar{H}k+\sum \limits _{k\in K}k\bar{H}=\sum \limits _{k\in K}k\bar{H}+\sum \limits _{k\in K}k\bar{H}=0$.

$\square$

The remaining of this section is devoted to discuss the case when G is dihedral, where the results are significantly used to prove our main result later. In this paper, $D_{2n}=\langle a,b\,\vert \, a^n=b^2=1, ab=ba^{-1}\rangle$ denotes a dihedral group of order 2n where $n>2$. Note that $Inv(D_{2n})=\lbrace a^ib\,\vert \, 0\le i\le n-1\rbrace$ if n is odd and $Inv(D_{2n})=\lbrace a^ib\,\vert \, 0\le i\le n-1\rbrace \cup \lbrace a^{\frac{n}{2}}\rbrace$ if n is even.

Starting from the case of odd n, for each $g\in Inv(G)$, we first identify those corresponding $h\in G$ which results in $h(1+g)\in Nil_2(F_2D_{2n})$, then followed up by a study of the closure of addition between each pair of $h(1+g)\in Nil_2(F_2D_{2n})$. Note that $D_{2n}$ for odd n is centerless, that is $Z(D_{2n})=\lbrace 1 \rbrace$.

Proposition 4.5

Let $g\in Inv(D_{2n})$ for odd $n>2$. Then, for every $h\in D_{2n}$, either $h(1+g)=1+g$ or $h(1+g)\not \in Nil_2(F_2D_{2n})$.

Proof For each $g=a^ib\in Inv(D_{2n})$ with $i\in \lbrace 0,1,2,\cdots ,n-1\rbrace$, note that $C_{D_{2n}}(a^ib)=\lbrace 1,a^ib\rbrace$ and $a^ib(1+a^ib)=1(1+a^ib)=1+a^ib$. Hence, for every $h\in C_{D_{2n}}(a^ib)$, $h(1+g)=1+g$. Suppose that $h\not \in C_{D_{2n}}(a^ib)$, then $[h,g]\ne 0$. It follows from Lemma 4.1 that $h(1+g)\not \in Nil_2(F_2D_{2n})$. $\square$

Now, for distinct $g_1,g_2\in Inv(D_{2n})$ for odd $n>2$, we show that the sum of $(1+g_1)$ and $(1+g_2)$ can never be in $Nil_2(F_2D_{2n})$.

Proposition 4.6

For distinct $g_1,g_2\in Inv(D_{2n})$ for odd $n>2$, $(1+g_1)+(1+g_2)\not \in Nil_2(F_2D_{2n})$.

Proof Let $g_1=a^ib$ and $g_2=a^jb$ for some distinct $i,j\in \lbrace 0,1,2,\cdots ,n-1\rbrace$. Note that $(1+a^ib)+(1+a^jb)=a^ib+a^jb$. Then, $a^ib+a^jb\in Nil_2(F_2D_{2n})$ if and only if $[a^ib,a^jb]=0$.

Assume that $[a^ib,a^jb]=a^iba^jb+a^jba^ib=a^{i-j}+a^{j-i}=0$. This results in $i-j\equiv j-i(\mathrm {mod}\,n)$ or equivalently $2i\equiv 2j(\mathrm {mod}\,n)$. Since n is odd, we have $i\equiv j(\mathrm {mod}\,n)$, a contradiction. Hence, $[a^ib,a^jb]\ne 0$. $\square$

Now, for even $n>2$, recall that $Inv(D_{2n})=\lbrace a^ib\,\vert \, 0\le i\le n-1\rbrace \cup \lbrace a^{\frac{n}{2}}\rbrace$. Note that $a^{\frac{n}{2}}\in Z(D_{2n})\cap Inv(D_{2n})$, then by Proposition 4.2, the following result holds trivially.

Proposition 4.7

Let $n>2$ be even. Then, $F_2D_{2n}(1+a^\frac{n}{2})$ is a 2-nil ideal.

If $g\in Inv(D_{2n})\setminus \lbrace a^{\frac{n}{2}}\rbrace$, note that we have $C_{D_{2n}}(g)=\lbrace 1,a^{\frac{n}{2}},g,ga^{\frac{n}{2}}\rbrace$. Then, the proposition below follows from Lemma 4.1.

Proposition 4.8

Let $g\in Inv(D_{2n})\setminus \lbrace a^{\frac{n}{2}}\rbrace$ for even $n>2$. Then, for $h\in D_{2n}$, $h(1+g)\in Nil_2(F_2D_{2n})$ if and only if $h\in \lbrace 1,a^{\frac{n}{2}},g,ga^{\frac{n}{2}}\rbrace$.

It is instructive to note that for every $g\in Inv(D_{2n})\setminus \lbrace a^{\frac{n}{2}}\rbrace$, $g(1+g)=1+g$ and $ga^\frac{n}{2}(1+g)=a^\frac{n}{2}(1+g)$. Hence, for every $g\in Inv(D_{2n})\setminus \lbrace a^{\frac{n}{2}}\rbrace$, we have $1+g,a^\frac{n}{2}(1+g)\in Nil_2(F_2D_{2n})$. Then, this leads to the result below.

Proposition 4.9

Consider distinct $a^ib,a^jb\in Inv(D_{2n})$ for even $n>2$ and let $k_1,k_2\in \lbrace 0,1\rbrace$. Then, $(a^\frac{n}{2})^{k_1}(1+a^ib)+(a^\frac{n}{2})^{k_2}(1+a^jb)\in Nil_2(F_2D_{2n})$ if and only if $i\equiv j+\frac{n}{2} (\mathrm {mod}\, n)$.

Proof Note that $(a^\frac{n}{2})^{k_1}(1+a^ib)+(a^\frac{n}{2})^{k_2}(1+a^jb)\in Nil_2(F_2D_{2n})$ if and only if $[(a^\frac{n}{2})^{k_1}(1+a^ib),(a^\frac{n}{2})^{k_2}(1+a^jb)]=0$ or equivalently $[1+a^ib,1+a^jb]=0$ since $(a^\frac{n}{2})^{k_1},(a^\frac{n}{2})^{k_2}\in Z(G)$. Clearly, $[1+a^ib,1+a^jb]=0$ if and only if $[a^ib,a^jb]=0$. Then, it follows from the proof of Proposition 4.6 that $[a^ib,a^jb]=0$ if and only if $2i\equiv 2j (\mathrm {mod}\, n)$. As n is even, this leads to $i\equiv j (\mathrm {mod}\, \frac{n}{2})$. $\square$

In addition, each of the summations obtained from Proposition 4.9 is in fact an element in $F_2D_{2n}(1+a^\frac{n}{2})$.

Proposition 4.10

Consider distinct $a^ib,a^jb\in Inv(D_{2n})$ for even $n>2$ and let $k_1,k_2\in \lbrace 0,1\rbrace$. If $i\equiv j+\frac{n}{2} (\mathrm {mod}\, n)$, then $(a^\frac{n}{2})^{k_1}(1+a^ib)+(a^\frac{n}{2})^{k_2}(1+a^jb)\in F_2D_{2n}(1+a^\frac{n}{2})$.

Proof Suppose that $k_1=k_2$, if $i\equiv j+\frac{n}{2} (\mathrm {mod}\, n)$, then:

$$\begin{aligned} (a^\frac{n}{2})^{k_1}(1+a^{j+\frac{n}{2}}b)+(a^\frac{n}{2})^{k_1}(1+a^jb)&= (a^\frac{n}{2})^{k_1}(a^jb+a^{j+\frac{n}{2}}b)\\&= (a^\frac{n}{2})^{k_1}a^jb(1+a^\frac{n}{2})\in F_2D_{2n}(1+a^\frac{n}{2}) \end{aligned}$$

On the other hand, suppose that $k_1\ne k_2$, if $i\equiv j+\frac{n}{2} (\mathrm {mod}\, n)$, then:

$$\begin{aligned} (a^\frac{n}{2})^{k_1}(1+a^{j+\frac{n}{2}}b)+(a^\frac{n}{2})^{k_2}(1+a^jb)&= (a^\frac{n}{2})^{k_1}(1+a^{j+\frac{n}{2}}b+a^\frac{n}{2}(1+a^jb))\\&= (a^\frac{n}{2})^{k_1}(1+a^\frac{n}{2})\in F_2D_{2n}(1+a^\frac{n}{2}). \end{aligned}$$

$\square$

5 Main results

This section is devoted to study the equivalence of dihedral and cyclic zero-divisor codes having generators from their respective 2-nilradicals. Recall that Corollary 3.5 suggests that the 2-nilradical of $F_2C_{2n}$ is $F_2C_{2n}(1+x^n)$ whereas Proposition 4.7 indicates that $F_2D_{2n}(1+a^\frac{n}{2})$ is a 2-nil ideal for even $n>2$.

As a beginning, we start with a more general result on the equivalence between zero-divisor codes $F_2G_1(1+g_1)$ and $F_2G_2(1+g_2)$ for some $g_1\in Inv(G_1)$ and $g_2\in Inv(G_2)$. We show that the condition $\vert G_1\vert =\vert G_2\vert$ is sufficient to ensure that equivalence between $F_2G_1(1+g_1)$ and $F_2G_2(1+g_2)$ holds, regardless of the algebraic structures of $G_1$ and $G_2$.

Theorem 5.1

Let $G_1$ and $G_2$ be finite groups with $\vert G_1\vert =\vert G_2\vert$ being even. If $g_1\in Inv(G_1)$ and $g_2\in Inv(G_2)$, then $F_2G_1(1+g_1)$ and $F_2G_2(1+g_2)$ are equivalent codes.

Proof Let $\vert G_1\vert =\vert G_2\vert =2n$. As $\lbrace 1,g_1\rbrace$ and $\lbrace 1,g_2\rbrace$ are order 2 subgroups of $G_1$ and $G_2$ respectively, let us consider a complete listing of the cosets of $\lbrace 1,g_1\rbrace$ and $\lbrace 1,g_2\rbrace$ in $G_1$ and $G_2$ respectively as follows:

$$\begin{aligned} H_1=h_1\lbrace 1,g_1\rbrace \,\,\,&\,\,\, H'_1=h'_1\lbrace 1,g_2\rbrace \\ H_2=h_2\lbrace 1,g_1\rbrace \,\,\,&\,\,\, H'_2=h'_2\lbrace 1,g_2\rbrace \\&\vdots \\ H_n=h_n\lbrace 1,g_1\rbrace \,\,\,&\,\,\, H'_n=h'_n\lbrace 1,g_2\rbrace \end{aligned}$$

The mutually disjoint property of the cosets results in $\lbrace \bar{H_i}\,\vert \, i\in \lbrace 1,2,\dots ,n\rbrace \rbrace$ and $\lbrace \bar{H'_i}\,\vert \, i\in \lbrace 1,2,\dots ,n\rbrace \rbrace$ are bases for $F_2G_1(1+g_1)$ and $F_2G_2(1+g_2)$ respectively. Define $\varphi :F_2G_1(1+g_1)\rightarrow F_2G_2(1+g_2)$ as a $F_2$-linear extension in such a way that $\varphi : \bar{H_i}\mapsto \bar{H'_i}$ for each $i\in \lbrace 1,2,\dots ,n\rbrace$. Clearly, $\varphi$ is a $F_2$-linear isomorphism. In addition, the fact that $\varphi$ is weight-preserving follows directly from the mutually disjoint property of cosets. $\square$

Despite Theorem 5.1 affirms the equivalence between $F_2C_{2n}(1+x^n)$ and $F_2D_{2n}(1+a^\frac{n}{2})$, we give an alternative proof below by constructing a weight-preserving $F_2$-linear isomorphism $\varphi :F_2D_{2n}(1+a^\frac{n}{2})\rightarrow F_2C_{2n}(1+x^n)$ in the form of $(\sum \limits _{g\in G_1}{a_gg})\varphi =\sum \limits _{g\in G_1}{a_g\chi (g)}$ for some bijective $\chi :D_{2n}\rightarrow C_{2n}$, which is essential for further studies.

Consider the following bijection $\chi :D_{2n}\rightarrow C_{2n}$ introduced in Theorem 5.2 from [7]:

$$\begin{aligned} \chi (a^ib^j)= \left\{ \begin{array}{cc} x^{-2i-1} &{} j=1 \\ x^{2i} &{} j=0 \\ \end{array} \right. \end{aligned}$$

Note that $\chi (\lbrace 1,a^\frac{n}{2}\rbrace )=\lbrace 1,x^n\rbrace$. Let $g=a^ib^j\in D_{2n}$ for $i\in \lbrace 0,1,\cdots ,n-1\rbrace$ and $j\in \lbrace 0,1\rbrace$, clearly $\chi (g(1))=\chi (g)\chi (1)$. We claim that $\chi (ga^\frac{n}{2})=\chi (g)\chi (a^\frac{n}{2})$. Suppose that $j=0$, that is $g=a^i$, then:

$$\begin{aligned} \chi (a^ia^\frac{n}{2})&= \chi (a^{i+\frac{n}{2}})\\&= x^{2i+n}\\&=x^{2i}x^n\\&=\chi (a^i)\chi (a^\frac{n}{2}). \end{aligned}$$

Suppose that $j=1$, that is $g=a^ib$, then:

$$\begin{aligned} \chi (a^iba^\frac{n}{2})&= \chi (a^{i+\frac{n}{2}}b)\\&= x^{-2i-n-1}\\&=x^{-2i-1}x^{-n}\\&=\chi (a^ib)\chi (a^\frac{n}{2}). \end{aligned}$$

As a whole, $\chi =\chi _{1+a^\frac{n}{2},1+x^n}$ and $\chi$ is a $(1+a^\frac{n}{2})$-homomorphism. Hence by Theorem 2.4, $F_2D_{2n}(1+a^\frac{n}{2})$ and $F_2C_{2n}(1+x^n)$ are equivalent group codes.

Theorem 5.2

Let $n>2$ be even. Then, $F_2D_{2n}(1+a^\frac{n}{2})$ and $F_2C_{2n}(1+x^n)$ are equivalent codes.

For every $u\in F_2D_{2n}(1+a^\frac{n}{2})$, note that $F_2D_{2n}u$ is a subideal of $F_2D_{2n}(1+a^\frac{n}{2})$. Restrict the domain of $\varphi$ to $F_2D_{2n}u$ by defining $\varphi _\downarrow :F_2D_{2n}u\rightarrow F_2C_{2n}(1+x^n)$ in such a way that $(c)\varphi _\downarrow =(c)\varphi$ for every $c\in F_2D_{2n}u$. By Proposition 2.2, we have $F_2D_{2n}u$ is equivalent to $Im(\varphi _\downarrow )$ with respect to the weight-preserving $F_2$ linear isomorphism $\varphi _\downarrow$.

While identifying $Im(\varphi _\downarrow )$, one might conjecture that $Im(\varphi _\downarrow ) = F_2C_{2n}(u)\varphi _\downarrow$ is a subideal of $F_2C_{2n}(1+x^n)$. Here, we construct a counterexample for this conjecture by first showing that although $\chi (supp(u))=supp((u)\varphi _\downarrow )$, $\chi$ need not be a u-homomorphism.

Let $u=\sum \limits _{i=1}^kv_i(1+a^\frac{n}{2})$ for some $k\in {\mathbb {Z}}^+$ and each $v_i\in D_{2n}$. Note that $(u)\varphi _\downarrow =\sum \limits _{i=1}^k\chi (v_i)(1+x^n)$. For each codeword $c\in F_2D_{2n}u$, write $c=wu=(\sum \limits _{j=1}^lw_j)u$ for some $l\in {\mathbb {Z}}^+$ and each $w_i\in D_{2n}$. This leads to $c=(\sum \limits _{j=1}^lw_j)\sum \limits _{i=1}^kv_i(1+a^\frac{n}{2})=\sum \limits _{j=1}^l\sum \limits _{i=1}^kw_jv_i(1+a^\frac{n}{2})$. Denote each $w_jv_i=u_{ij}$, then $c=\sum \limits _{j=1}^l\sum \limits _{i=1}^ku_{ij}(1+a^\frac{n}{2})$ and the $(1+a^\frac{n}{2})$-homomorphic property ensures that for every $u_{ij}$, $\chi (u_{ij}h)=\chi (u_{ij})\chi (h)$ for each $h\in \lbrace 1,a^\frac{n}{2}\rbrace$.

To further show that $\chi$ is not necessarily a u-homomorphism, it is sufficient to prove that not all $\chi (u_{ij})$ are expressible as $\chi (w_j)\chi (v_i)$. We divide the proof into four cases:

Case 1 When $w_j=a^s$ and $v_i=a^t$ for some $s,t\in \lbrace 0,1,2, \dots n-1\rbrace$, then it can be easily shown that $\chi (w_jv_i)=\chi (w_j)\chi (v_i)$.

Case 2 When $w_j=a^s$ and $v_i=a^tb$ for some $s,t\in \lbrace 0,1,2,\dots n-1\rbrace$, then:

$$\begin{aligned} \chi (w_jv_i)=\chi (a^sa^tb)&= \chi (a^{s+t}b)\\&= x^{-2(s+t)-1}\\&=x^{-2s}x^{-2t-1}\\&=\chi (a^{-s})\chi (a^tb). \end{aligned}$$

Note that $\chi (a^{-s})\chi (a^tb)\ne \chi (a^{s})\chi (a^tb)$ whenever $s\not \in \lbrace 0,\frac{n}{2}\rbrace$.

Case 3 When $w_j=a^sb$ and $v_i=a^t$ for some $s,t\in \lbrace 0,1,2,\dots n-1\rbrace$, then it can be easily shown that $\chi (w_jv_i)=\chi (w_j)\chi (v_i)$.

Case 4 When $w_j=a^sb$ and $v_i=a^tb$ for some $s,t\in \lbrace 0,1,2,\dots n-1\rbrace$, then:

$$\begin{aligned} \chi (w_jv_i)=\chi (a^sba^tb)&= \chi (a^{s-t})\\&= x^{2(s-t)}\\&=x^{-2(-s-1)-1)}x^{-2t-1}\\&=\chi (a^{-s-1}b)\chi (a^tb). \end{aligned}$$

Note that $\chi (w_jv_i)=\chi (a^{-s-1}b)\chi (a^tb)\ne \chi (a^{s}b)\chi (a^tb)$.

Now, suppose that $u=\sum \limits _{i=1}^kv_i(1+a^\frac{n}{2})$ for some $k\in {\mathbb {Z}}^+$ and each $v_i=a^t\in D_{2n}$ for some $t\in \lbrace 0,1,2,\dots ,n-1\rbrace$, we have $\chi (w_jv_i)=\chi (w_j)\chi (v_i)$, then:

$$\begin{aligned} (c)\varphi _\downarrow&= \left( \sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }u_{ij}h\right) \varphi _\downarrow \\&= \left( \sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }w_jv_ih\right) \varphi _\downarrow \\&=\sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (w_jv_ih)\\&=\sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (w_j)\chi (v_i)\chi (h)\\&=\left( \sum \limits _{j=1}^l\chi (w_j)\right) \left( \sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (v_i)\chi (h)\right) \\&=(w)\varphi _\downarrow (u)\varphi _\downarrow \in F_2C_n(u)\varphi _\downarrow . \end{aligned}$$

Hence, by Theorem 2.4, the u-homomorphic property of $\chi$ with $\chi =\chi _{u,(u)\varphi _\downarrow }$ leads to the following proposition.

Proposition 5.3

Let $n>2$ be even and $u\in F_2D_{2n}(1+a^\frac{n}{2})$ such that $a^ib\not \in supp(u)$ for every $i\in \lbrace 0,1,\dots ,n-1\rbrace$. Then, $F_2D_{2n}u$ and $F_2C_{2n}(u)\varphi _\downarrow$ are equivalent codes.

Next, suppose that $u=\sum \limits _{i=1}^kv_i(1+a^\frac{n}{2})$ for some $k\in {\mathbb {Z}}^+$ and each $v_i=a^tb\in D_{2n}$ for some $t\in \lbrace 0,1,2,\dots ,n-1\rbrace$, we have $\chi (w_jv_i)=\chi (w_j^{-1})\chi (v_i)$ when $w_j=a^s$ for some $s\in \lbrace 0,1,2,\dots ,n-1\rbrace$ and $\chi (w_jv_i)=\chi (w_j^*)\chi (v_i)$ when $w_j=a^sb$ for some $s\in \lbrace 0,1,2,\dots ,n-1\rbrace$, where $w_j^*=a^{-s-1}b$. Without loss of generality, we write $w=\sum \limits _{j=1}^\alpha w_j+\sum \limits _{j=\alpha +1}^l w_j$ for some $0 \le \alpha \le l$ such that $w_j=a^s$ when $0 \le j\le \alpha$ and $w_j=a^sb$ when $\alpha +1 \le j\le l$. Then:

$$\begin{aligned} (c)\varphi _\downarrow&= \left( \sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }u_{ij}h\right) \varphi _\downarrow \\&= \left( \sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }w_jv_ih\right) \varphi _\downarrow \\&=\sum \limits _{j=1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (w_jv_ih)\\&=\sum \limits _{j=1}^\alpha \sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (w_j^{-1})\chi (v_i)\chi (h)+\sum \limits _{j=\alpha +1}^l\sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (w_j^*)\chi (v_i)\chi (h)\\&=\left( \sum \limits _{j=1}^\alpha \chi (w_j^{-1})+\sum \limits _{j=\alpha +1}^l\chi (w_j^*)\right) \left( \sum \limits _{i=1}^k\sum \limits _{h\in \lbrace 1,a^\frac{n}{2}\rbrace }\chi (v_i)\chi (h)\right) \\&=\left( \sum \limits _{j=1}^\alpha \chi (w_j^{-1})+\sum \limits _{j=\alpha +1}^l\chi (w_j^*)\right) (u)\varphi _\downarrow \in F_2C_n(u)\varphi _\downarrow . \end{aligned}$$

Define a map $f:D_{2n}\rightarrow D_{2n}$ in such a way that $f: a^s\mapsto a^{-s}$ and $f: a^sb\mapsto a^{-s-1}b$ for each $s\in \lbrace 0,1,\dots ,n-1\rbrace$. Note that f is bijective, this implies that $Im(\varphi _\downarrow )=F_2C_{2n}((u)\varphi _\downarrow )$.

Proposition 5.4

Let $n>2$ be even and $u\in F_2D_{2n}(1+a^\frac{n}{2})$ such that $a^i\not \in supp(u)$ for every $i\in \lbrace 0,1,\dots ,n-1\rbrace$. Then, $F_2D_{2n}u$ and $F_2C_{2n}(u)\varphi _\downarrow$ are equivalent codes.

The following example shows that $Im(\varphi _\downarrow )$ need not possess ideal structure in general, thus need not be $F_2C_{2n}(u)\varphi _\downarrow$.

Example 5.5

Consider the case when $n=4$ and let $u=1+a^2+b+a^2b=1(1+a^2)+b(1+a^2)\in F_2D_8(1+a^2)$. Note that $(u)\varphi _\downarrow =1+x^3+x^4+x^7=1(1+x^4)+x^3(1+x^4)\in F_2C_8(1+x^4)$. Note that $F_2D_8u$ and $F_2C_8(u)\varphi _\downarrow$ are not equivalent since it can be verified that $F_2D_8u=L_{F_2}(u,au)$ and $F_2D_8u=L_{F_2}((u)\varphi _\downarrow ,x(u)\varphi _\downarrow ,x^2(u)\varphi _\downarrow )$, having dimension 2 and 3 respectively. In fact, $F_2D_8u=L_{F_2}(u,au)$ is equivalent to $W_2(u)\varphi _\downarrow =L_{F_2}((u)\varphi _\downarrow ,(au)\varphi _\downarrow )$. Note that $W_2(u)\varphi _\downarrow \subset F_2C_8(u)\varphi _\downarrow$.

The main results of this section are summarized as follows. We proved that for $u\in \{1+g\vert g\in Inv(D_{2n})\}\subseteq Nil_2(F_2D_{2n})$, $F_2D_{2n}u$ has an equivalent form of cyclic zero-divisor code, that is $F_2C_{2n}(1+x^n)$ by Proposition 5.1. In addition, for even $n>2$, partition $D_{2n}$ into $A_1=\{1,a,a^2,\dots ,a^{n-1}\}$ and $A_2=\{b,ab,a^2b,\dots ,a^{n-1}b\}$. Let $u\in F_2D_{2n}(1+a^\frac{n}{2})\subseteq Nil_2(F_2D_{2n})$. If either $supp(u)\subseteq A_1$ or $supp(u)\subseteq A_2$, then Proposition 5.3 and Proposition 5.4 ensure that $F_2D_{2n}u$ has an equivalent form of cyclic zero-divisor code. The remaining studies on whether $F_2D_{2n}u$ has an equivalent form of cyclic zero-divisor code for the case when $supp(u)\not \subseteq A_1$ and $supp(u)\not \subseteq A_2$ are left as a future direction.