Question

Show that$\mathit{C}\mathbf{=}\left\{0000, 0101, 1010,1111 \right\}$ is a $\left(4,2\right)$code

Short answer

Since C is a subgroup of $B\left(4\right)$and$\left|C\right|=2^2=4$ ,so from the definition, it is a code $\left(4,2\right)$.

Step-by-step solution

Theorem 7.12 and Definition of Binary linear code

Let H be a nonempty finite subset of a group G . If H is closed under the operationin G, then H is a subgroup of G.

Definition

If$0<k<n$ , then an$\left(n, k\right)$ binary linear code consists of a subgroup C of$B\left(n\right)$

oforder$2^n$ .

Showing that  C is a 4,2code

Given in the question.

$C=\left\{0000, 0101, 1010,1111 \right\}$

We can see, that C consists of 4 elements.

We need to show that C is a subgroup of$B\left(4\right)$of order$2^2=4$$2^2=4$,

So, to prove that C is a subgroup of $B\left(4\right)$,according to Theorem 7.12 we need to show that C is closed under the binary addition.

We know that code is the identity element and it is closed under that addition.

Now checking the other elements

$$\begin{gathered} 0101+1010=1111\in C \\ 0101+1111=1010\in C \\ 1010+1111=0101\in C \end{gathered}$$

From above we can see C is closed under the binary addition.

Therefore, it is a subgroup of $B\left(4\right)$and $\left|C\right|=2^2=4$so from the definition, it is a $\left(4,2\right)$code.