A Note on Linear Complementary Pairs of Group Codes

TL;DR

This paper provides a simple proof that in linear complementary pairs of group codes, one code uniquely determines the other and their duals are permutation equivalent, simplifying previous complex polynomial-based proofs.

Contribution

It introduces an elementary proof for the relationship between complementary group codes and their duals, extending earlier results with a more straightforward approach.

Findings

D code is uniquely determined by C in a complementary pair

D^ot is permutation equivalent to C

Simplifies previous polynomial-based proofs

Abstract

We give a short and elementary proof of the fact that for a linear complementary pair $(C,D)$, where $C$ and $D$ are $2$-sided ideals in a group algebra, $D$ is uniquely determined by $C$ and the dual code $D^\perp$ is permutation equivalent to $C$. This includes earlier results of Carlet et al. and G\"uneri et al. on nD cyclic codes which have been proved by subtle and lengthy calculations in the space of polynomials.

Equations

$$R_n=K[x_1,\dots ,x_n]/⟨x^{m_1}-1,\dots ,x^{m_n}-1⟩,$$

$$KG=\{a=\sum _{g\in G}{a}_{g}g\mid a_g\in K\}$$

$$ab=\sum _{g\in G}(\sum _{h\in G}{a}_h{b}_{h^{-1}g})g.$$

$$\langle g,h\rangle=\left\{\begin{array}[]{cl}1&\mbox{if }g=h\\ 0&\mbox{otherwise.}\end{array}\right.$$

$$⟨c,c^{\perp }g⟩=⟨c{g}^{-1},c^{\perp }⟩=0.$$

$$m(\alpha g)=(m{g}^{-1})\alpha$$

$$⟨ea,(1-\hat{e})b⟩=⟨e^{2}a,(1-\hat{e})b⟩=⟨ea,\hat{e}(1-\hat{e})b⟩=0$$

$$\begin{array}[]{rcl}\dim\,(1-\hat{e})KG&=&|G|-\dim\,\hat{e}KG\\ &=&|G|-\dim\,eKG^{*}\qquad\text{(by Lemma A)}\\ &=&|G|-\dim\,eKG\\ &=&\dim\,D^{\perp}\end{array}$$

$$e=e^2\in Z(KG)=\{a\mid ab=ba\text{for all}b\in KG\}$$

$$G=⟨a,b\mid a^7=1=b^2,a^b=a^{-1}⟩$$

$$e=1+a+a^2+a^4+b+a^{2}b+a^{5}b+a^{6}b,$$

$$\text{mbox}\mathrm{d}(D^{\perp })=\text{mbox}\mathrm{d}((1-\hat{e})KG)=\text{mbox}\mathrm{d}(KG(1-e))=3,$$

$$⟨\sum _{g\in G}{a}_{g}g,\sum _{h\in G}{b}_{h}h⟩=\sum _{g\in G}{a}_g{b}_g^q.$$

$$D^{\perp }=(1-\widehat{e^{(q)}})KG$$

$$\text{mbox}\mathrm{d}(D^{\perp })=\text{mbox}\mathrm{d}((1-e)KG{)}^{(q)})=\text{mbox}\mathrm{d}((1-e)KG)=\text{mbox}\mathrm{d}(C).$$

Full text

Keywords. Group code, linear complementary pair (LCP)

MSC classification. 94B05, 94B99, 20C05

Throughout this note let $K$ be a finite field. A pair $(C,D)$ of linear codes over $K$ of length $n$ is called a l**inear complementary pair (LCP) if $C\cap D=\{0\}$ and $C+D=K^{n}$, or equivalently if $C\oplus D=K^{n}$. In the special case that $D=C^{\perp}$ where the dual is taken with respect to the Euclidean inner product the code $C$ is referred to a l**inear complementary dual (LCD) code. LCD codes have first been considered by Massey in [7]. The nowadays interest of LCP codes aroused from the fact that they can be used in protection against side channel and fault injection attacks [1], [2], [4]. In this context the security of a linear complementary pair $(C,D)$ can be measured by the security parameter $\min\{\mbox{\rm d}(C),\mbox{\rm d}(D^{\perp})\}$. Clearly, if $D=C^{\perp}$, then the security parameter for $(C,D)$ is $\mbox{\rm d}(C)$.

Just recently, it has been shown in [3] that for linear complementary pairs $(C,D)$ the codes $C$ and $D^{\perp}$ are equivalent if $C$ and $D$ are both cyclic or $2$D cyclic codes under the assumption that the characteristic of $K$ does not divide the length. In [6], this result has been extended to the case that both $C$ and $D$ are $n$D cyclic for $n\in\mathbb{N}$. In both papers the proof is rather complicated and formulated in the world of polynomials.

Recall that an $n$D cyclic code is an ideal in the algebra

[TABLE]

and that $R_{n}$ is isomorphic to the group algebra $KG$ where $G=C_{m_{1}}\times\cdots\times C_{m_{n}}$ with cyclic groups $C_{m_{i}}$ of order $m_{i}$. Thus the above mentioned results are results on ideals in abelian group algebras.

A linear code $C$ is called a group code, or $G$-code, if $C$ is a right ideal in a group algebra

[TABLE]

where $G$ is a finite group. The vector space $KG\cong K^{|G|}$ with basis $g\in G$ serves as the ambient space and the weight function is defined by $\mbox{\rm wt}(a)=|\{g\in G\mid a_{g}\not=0\}|$ (which corresponds to the classical weight function via the isomorphism $KG\cong K^{|G|}$). Note that $KG$ carries a $K$-algebra structure via the multiplication in $G$. More precisely, if $a=\sum_{g\in G}a_{g}g$ and $b=\sum_{g\in G}b_{g}g$ are given, then

[TABLE]

In this sense $n$D cyclic codes are group codes for abelian groups $G$ and vice versa since a finite abelian group is the direct product of cyclic groups.

There is a natural $K$-linear anti-algebra automorphism $\hat{}:KG\longrightarrow KG$ which is given by $g\mapsto g^{-1}$ for $g\in G$ (in the isomorphism $KG\cong K^{|G|}$, the automorphism  $\hat{}$  corresponds to a permutation of the coordinates). Thus we may associate to each $a=\sum_{g\in G}a_{g}g\in KG$ the adjoint $\hat{a}=\sum_{g\in G}a_{g}g^{-1}$ and call $a$ self-adjoint if $a=\hat{a}$.

In addition, the group algebra $KG$ carries a symmetric non-degenerate $G$-invariant bilinear form $\langle.\,,.\rangle$ which is defined by

[TABLE]

Here $G$-invariance means that $\langle ag,bg\rangle=\langle a,b\rangle$ for all $a,b\in KG$ and all $g\in G$. Via the isomorphism $KG\cong K^{|G|}$, the above form corresponds to the usual Euclidean inner product. With respect to this form we may define the dual code $C^{\perp}$ of a group code $C\leq KG$ as usual and say that $C$ is self-dual if $C=C^{\perp}$. Note that for a group code $C$ the dual $C^{\perp}$ is a right ideal since for all $c\in C,c^{\perp}\in C^{\perp}$ and $g\in G$ we have

[TABLE]

Thus with $C$ the dual $C^{\perp}$ is a group code as well.

In [9] we classified completely group algebras which contain self-dual ideals. More precisely, a self-dual $G$-code exists over the field $K$ if and only if $|G|$ and the characteristic of $K$ are even. In [5] we investigated LCD group codes and characterized them via self-adjoint idempotents $e^{2}=e=\hat{e}$ in the group algebra $KG$.

In this short note we prove the following theorem which includes the above mentioned results of [3] and [6]. Observe that we require no assumption on the characteristic of the field $K$.

Theorem. Let $G$ be a finite group. If $C\oplus D=KG$ where $C$ and $D$ are $2$-sided ideals in $KG$, then $D$ is uniquely determined by $C$ and $D^{\perp}$ is permutation equivalent to $C$. In particular $\mbox{\rm d}(D^{\perp})=\mbox{\rm d}(C)$.

In order to prove the Theorem we state some elementary facts from representation theory.

Definition. If $M$ is a right $KG$-module, then the dual vector space $M^{*}=\mbox{\rm Hom}_{K}(M,K)$ becomes a right $KG$-module via

[TABLE]

where $m\in M,\alpha\in M^{*}$ and $g\in G$. With this action $M^{*}$ is called the dual module of $M$. Clearly, $\dim\,M^{*}=\dim\,M.$

Lemma A. (Okuyama-Tsushima, [8]) If $e=e^{2}\in KG$, then $\hat{e}KG\cong eKG^{*}$. In particular, $\dim\,\hat{e}KG=\dim\,eKG$.

Proof: A short proof is given in Lemma 2.3 of [5]. $\Box$

Lemma B. If $D=eKG$ with $e^{2}=e\in KG$, then $D^{\perp}=(1-\hat{e})KG$.

Proof: First observe that $\hat{e}^{2}=\hat{e}$ and that $\langle ab,c\rangle=\langle b,\hat{a}c\rangle$ for all $a,b,c\in KG$. Thus

[TABLE]

for all $a,b\in KG$. Hence, $(1-\hat{e})KG\subseteq D^{\perp}$. As

[TABLE]

we obtain $(1-\hat{e})KG=D^{\perp}$. $\Box$

Proof of the Theorem: Since $C\oplus D=KG$ we may write $D=eKG$ and $C=(1-e)KG$ for a suitable central idempotent

[TABLE]

(see [5]). Lemma B says that $D^{\perp}=(1-\hat{e})KG$. Via the map $\hat{}:KG\longrightarrow KG$ the $K$-linear code $(1-\hat{e})KG$ is permutation equivalent to the $K$-linear code $KG(1-e)$. But $KG(1-e)=(1-e)KG=C$ since $e$ is central, which completes the proof. $\Box$

If $C$ and $D$ are only right ideals, then $D$ is uniquely determined by $C$, but $D^{\perp}$, in general, is not necessarily permutation equivalent to $C$. It even may happen that $\mbox{\rm d}(D^{\perp})\not=\mbox{\rm d}(C)$ as the next example shows.

Example. Let $K=\mathbb{F}_{2}$ and let

[TABLE]

be a dihedral group of order 14. If we put

[TABLE]

then $e=e^{2}$. With Magma one easily computes $\mbox{\rm d}((1-e)KG)=2$ and $\mbox{\rm d}(KG(1-e))=3.$

Now let $C=(1-e)KG$ and $D=eKG$. By Lemma B, we have $D^{\perp}=(1-\hat{e})KG$. Thus

[TABLE]

but $\mbox{\rm d}(C)=\mbox{\rm d}((1-e)KG)=2.$ We like to mention here that $C$ and $D$ are quasi-cyclic codes. $\Box$

Remark. Let $|K|=q^{2}$. In this case we may consider the Hermitian inner product on $KG$ which is defined by

[TABLE]

For $a=\sum_{g\in G}a_{g}g$ we put $a^{(q)}=\sum_{g\in G}a_{g}^{q}g$. With this notation we have

[TABLE]

in Lemma B. Applying the anti-automorphism $\hat{}:KG\longrightarrow KG$ we see that $D^{\perp}$ is permutation equivalent to $KG(1-{e^{(q)}})$. If in addition $e$ is central, then $e^{(q)}$ is central. Thus $D^{\perp}$ is permutation equivalent to $(1-e^{(q)})KG=((1-e)KG)^{(q)}$.

It follow that

[TABLE]

Thus, in the Hermitian case a linear complementary pair $(C,D)$ of $2$-sided group codes $C$ and $D$ also has security parameter $\mbox{\rm d}(C)$.