Exercise 3.1.10 (Homomorphism $\varphi : \mathbb{Z}/8\mathbb{Z} \to \mathbb{Z}/ 4 \mathbb{Z}$)

Proof.

Consider the map

• $\phi$ is well defined: If $a\equiv b(\mathit{mod}8)$, then $8\mid b-a$, thus $4\mid b-a$, so $a\equiv b(\mathit{mod}4)$. This shows that $\phi ({[a]}_8)$ doesn’t depend of the choice of the representative $a$ of ${[a]}_8$.

• $\phi$ is a homomorphism: If $u={[a]}_8$ and $v={[b]}_8$ are elements of $\mathbb{Z}∕8\mathbb{Z}$, then

  $$\phi (u+v)=\phi ({[a]}_8+{[b]}_8)=\phi ({[a+b]}_8)={[a+b]}_4={[a]}_4+{[b]}_4=\phi ({[a]}_8)+\phi ({[b]}_8)=\phi (u)+\phi (v).$$

• Let $u={[a]}_8$ in $\mathbb{Z}∕8\mathbb{Z}$, where $0\le a<8$. Since

  $$u\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )⟺{[a]}_4=0⟺4\mid a⟺a=0\text{ or }a=4⟺{[a]}_8={[0]}_8\text{ or }[a_8]={[4]}_8,$$

  we obtain

  $$\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )=\{{[0]}_8,{[4]}_8\}.$$

• Note that if $w={[b]}_4\in \mathbb{Z}∕4\mathbb{Z}$, then $w=\phi ({[b]}_8)$, therefore, if $u={[a]}_8\in \mathbb{Z}∕8\mathbb{Z}$,

  $$\begin{array}{llll}\hfill u\in {\phi }^{-1}(w)& ⟺\phi (u)=w& \hfill & \\ \hfill & ⟺\phi (u)=\phi ({[b]}_8)& \hfill & \\ \hfill & ⟺\phi (u-{[b]}_8)={[0]}_8& \hfill & \\ \hfill & ⟺u-{[b]}_8\in \ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& \hfill & \\ \hfill & ⟺u\in {[b]}_8+\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )& \hfill & \\ \hfill & ⟺u\in \{{[b]}_8,{[b]}_8+{[4]}_8\},& \hfill & \end{array}$$

  so the fiber ${\phi }^{-1}(w)$ is the translate of $\ker <!--\; FUNCTION\; APPLICATION\; -->(\phi )$ by ${[b]}_8$:

  $${\phi }^{-1}({[b]}_4)=\{{[b]}_8,{[b]}_8+{[4]}_8\}.$$

  Therefore the fibers of $\phi$ are

  $$\begin{array}{llll}\hfill {\phi }^{-1}({[0]}_4)& =\{{[0]}_8,{[4]}_8\},& \hfill & \\ \hfill {\phi }^{-1}({[1]}_4)& =\{{[1]}_8,{[5]}_8\},& \hfill & \\ \hfill {\phi }^{-1}({[2]}_4)& =\{{[2]}_8,{[6]}_8\},& \hfill & \\ \hfill {\phi }^{-1}({[3]}_4)& =\{{[3]}_8,{[7]}_8\}.& \hfill & \\ \hfill & & \hfill \end{array}$$