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## Exhibit a group homomorphism from Z/(8) to Z/(4)

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 3.1 Exercise 3.1.10

Let $\varphi : \mathbb{Z}/(8) \rightarrow \mathbb{Z}/(4)$ be defined by $\overline{a} \mapsto \overline{a}$. (Note that $\overline{a}$ means two different things here.) Show that $\varphi$ is a well-defined surjective homomorphism and describe its kernel and fibers explicitly.

Solution:

Well-defined: Suppose $\overline{a} = \overline{b}$ in $\mathbb{Z}/(8)$. Then $a-b = 8k$ for some $k$. Then $a-b = 4(2k)$, and we have $$\varphi(\overline{a}) = \overline{a} = \overline{b} = \varphi(\overline{b}).$$ Homomorphism: We have \begin{align*}\varphi(\overline{a} + \overline{b}) =&\ \varphi(\overline{a+b}) \\=&\ \overline{a+b} = \overline{a} + \overline{b}\\ = &\ \varphi(\overline{a}) + \varphi(\overline{b}).\end{align*} Surjective: If $\overline{a} \in \mathbb{Z}/(4)$, then $\varphi(\overline{a}) = \overline{a}$.

Fix some $\overline{c} \in \mathbb{Z}/(4)$, and suppose $\varphi(\overline{a}) = \overline{c}$. Then we have $a-c = 4k$ for some $k$.

If $k=2\ell$ is even, we have $a-c = 8\ell$, so that $\overline{a} = \overline{c}$ in $\mathbb{Z}/(8)$.

If $k=2\ell + 1$ is odd, we have $a-c-4 = 8\ell$, so that $\overline{a} = \overline{c+4}$ in $\mathbb{Z}/(8)$.

Hence, the preimage of (i.e. fiber over) $\overline{c}$ is $\{ \overline{c}, \overline{c+4} \}$. In particular, the kernel of $\varphi$ is $\{ \overline{0}, \overline{4} \}$.