Exercise 14.3.11

Proof. (a) Let $H$ be a subgroup of ${𝔽}_p^n$. If $v=({\alpha }_1,\dots ,{\alpha }_n)\in H$, we show by induction on $k\in \mathbb{N}$ that ${[k]}_{p}v\in H$. First ${[0]}_{p}v=(0,\dots ,0)\in H$. If ${[k]}_{p}v\in H$, then ${[k+1]}_{p}v={[k]}_{p}v+v\in H$. Therefore

Since for every $\lambda \in {𝔽}_p$, there is some $k\in \mathbb{N}$ such that $\lambda ={[k]}_p$, we can conclude that, for all $v\in H$, for all $\lambda \in {𝔽}_p$, $\mathit{\lambda v}\in H$. Thus $H$ is a subspace of ${𝔽}_p^n$. (b) Let $\gamma :{𝔽}_p^n\to {𝔽}_p^n$ a group homomorphism. if $v\in {𝔽}_p^n$, We show by induction that $\gamma ({[k]}_{p}v)={[k]}_p\gamma (v)$ for $k\in \mathbb{N}$.

For $k=0$, $\gamma ({[0]}_{p}v)=0={[0]}_p\gamma (v)$. If $\gamma ({[k]}_{p}v)={[k]}_p\gamma (v)$ for some $k\in \mathbb{N}$, then, since $\gamma$ is a group homomorphism,

Therefore $\gamma ({[k]}_{p}v)={[k]}_p\gamma (v)$ for all $k\in \mathbb{N}$. This proves that $\gamma (\mathit{\lambda v})=\mathit{\lambda f}(v)$ for all $v\in {𝔽}_p^n$, for all $\lambda \in {𝔽}_p$, thus $\gamma$ is linear. □