Question

If $\mathbf{k}\mathbf{<}\mathbf{n}$and $\mathbf{f}\mathbf{:}\mathbf{B}\left(k\right)\mathbf{\to }\mathbf{B}\left(n\right)$is a homomorphism of groups, is lm f a linear code? Is lm f an data-custom-editor="chemistry" $\left(n,k\right)$ linear code?

Short answer

lm f it is not a linear code and lm f is not a $\left(\mathrm{n},\mathrm{k}\right)$ linear code.

Step-by-step solution

Definition of binary linear code

If $\mathbf{0}\mathbf{<}\mathbf{k}\mathbf{<}\mathbf{n}$, then an data-custom-editor="chemistry" $\left(n,k\right)$ binary linear code consists of a subgroup C ofdata-custom-editor="chemistry" $\mathbf{B}\left(n\right)$of order data-custom-editor="chemistry" ${\mathbf{2}}^{\mathbf{n}}$.

Finding that is lm f a linear code

Let assume, $\mathrm{n}=1$since data-custom-editor="chemistry" $\mathrm{k}<\mathrm{n}$

So, $\mathrm{k}=1$

Therefore, homomorphism can be given as

$\mathrm{f}:\mathrm{B}\left(1\right)\to \mathrm{B}\left(2\right)$

Suppose the function defining homomorphism is

$$\begin{gathered} \mathrm{f}\left(0\right)=\left(0,0\right) \\ \mathrm{f}\left(1\right)=\left(0,0\right) \end{gathered}$$

Now, we must poove that this is a homomorphism

$$\begin{gathered} \mathrm{f}\left(0+0\right)=\mathrm{f}\left(0\right)=\left(0,0\right)=\left(0,0\right)+\left(0,0\right)=\mathrm{f}\left(0\right)+\mathrm{f}\left(0\right) \\ \mathrm{f}\left(0+1\right)=\mathrm{f}\left(1\right)=\left(0,0\right)=\left(0,0\right)+\left(0,0\right)=\mathrm{f}\left(0\right)+\mathrm{f}\left(1\right) \\ \mathrm{f}\left(1+1\right)=\mathrm{f}\left(1\right)=\left(0,0\right)=\left(0,0\right)+\left(0,0\right)=\mathrm{f}\left(0\right)+\mathrm{f}\left(1\right) \end{gathered}$$

From the above we can see, f it is not an injective homomorphism, since $\mathrm{f}=\left(0,0\right)\le \mathrm{B}\left(2\right)$ is not the order of $2^1=2$, therefore, according to the definition of linear code lm f it is not a linear code or lm f is not a data-custom-editor="chemistry" $\left(\mathrm{n},\mathrm{k}\right)$linear code.