Question

Prove that B (n) =${\mathbf{Z}}_{\mathbf{2}}\mathbf{\times }{\mathbf{Z}}_{\mathbf{2}}\mathbf{\times }{\mathbf{Z}}_{\mathbf{2}}\mathbf{\times }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\times }{\mathbf{Z}}_{\mathbf{2}}$(n factors) with coordinate wise addition is an abelian group of order${\mathbf{2}}^{\mathbf{n}}$.

Short answer

The value $\mathrm{B}\left(\mathrm{n}\right)$ with coordinate wise addition is an abelian group of order $2^{\mathrm{n}}$ .

Step-by-step solution

Conceptual Introduction

Coding theory is the study of a code's characteristics and how well-suited it is to various applications. Data compression, cryptography, error detection, and correction, as well as data transport and storage, all depend on codes.

Introduction

The value C is referred to as a (n, k) code, a linear code, or simply a code, and its constituents are referred to as code words. Only code words are sent, but any element of B(n) can be received as a word.

Where, $\mathrm{B}\left(\mathrm{n}\right)={\mathrm{Z}}_2\times {\mathrm{Z}}_2\times {\mathrm{Z}}_2\times ....\times {\mathrm{Z}}_2\mathrm{upto}\mathrm{n}\mathrm{copies}$.

The objective is to show that B (n) =$Z_2\times Z_2\times Z_2\times ....\times Z_2$is an abelian group of order $2^n$with coordinate wise addition.

Use encoding algorithm

By right multiplication, a typical generator matrix may be used as an encoding method to turn the elements of B(k) into code words (elements s of B(n)).

Each $\mathrm{u}\in \mathrm{B}\left(\mathrm{k}\right)$is considered as a row vector of length k.

Take uG n-dimensional row vector

The matrix product uG is therefore an n-dimensional row vector that is an element of B(n). The first k coordinates of the code word uG produce the equivalent message word $\mathrm{u}\in \mathrm{B}\left(\mathrm{k}\right)$because the first k columns to G constitute the identity matrix data-custom-editor="chemistry" ${\mathrm{I}}_{\mathrm{K}}$.

B(n) is an abelian group of order $2^{\mathrm{n}}$because coordinate wise addition satisfies commutative rule.

Hence, proved.