Prove that $\mathbb{Z} / \mathrm{Ker}$ $f$ is isomorphic to $\mathrm{f}(\mathbb{Z} )$

Question

Let $f\colon (\mathbb{Z},+) \rightarrow(\mathbb{Q}-\{0\}, \cdot)$ be defined by $$f(m)=\left\{\begin{array}{c}1, \text { if } m \text { is even } \\ -1, \text { if } m \text { is odd }\end{array}\right.$$ Show that $f$ is a homomorphism. Determine $\ker f$ and also prove that $\mathbb{Z} / \ker f$ is isomorphic to $f(\mathbb{Z})$ ( $\mathbb{Z} $ and $\mathbb{Q}$ are the set of integers and rational numbers respectively).

My approach:

Let $m,n\in \mathbb{Z}$. Suppose both $m$ and $n$ are even. Then $m+n$ is even and so $$f(m+n)=1=1\cdot 1=f(m)f(n)$$

If both $m$ and $n$ are odd. Then $m+n$ is even and so $$f(m+n)=1=(-1)\cdot (-1)=f(m)f(n)$$ If one of $m$ and $n$ is even and other is odd then $m+n$ is odd. In this case $f(m+n)=-1=f(m)f(n)$.
Hence $f$ is a homomorphism.

2nd part

$\ker(f)=\{x\in \mathbb{Z}\mid f(x)=1\}=\{x\in \mathbb{Z}\mid x \mbox{ is even}\}.$. I stuck here to show $\mathbb{Z} / \ker$ $f$ is isomorphic to $f(\mathbb{Z} )$. How can I show this?

Answer 1

Hint : If $\varphi : G \rightarrow G'$ is a group homomorphism, then there exists an injective map $$\overline{\varphi} : G/\mathrm{Ker}(\varphi) \rightarrow G'$$

such that $\mathrm{Im}(\overline{\varphi})= \mathrm{Im}(\varphi)$.

Answer 2

Your 1st part is OK.

For the 2nd part, note that $\operatorname{ker}f=\{m\in\Bbb Z\mid f(m)=1 \}=\{m\in\Bbb Z\mid m\text{ is even}\}=2\Bbb Z$; hence, $\Bbb Z/\operatorname{ker} f=\Bbb Z/2\Bbb Z=\{0,1\}^+$; show that the map $\{0,1\}^+\to \{1,-1\}^\times=f(\Bbb Z)$, defined by $0\mapsto 1, \space1\mapsto-1$ (there's no other candidate for an isomorphism) is indeed a bijective(!) homomorphism, namely an isomorphism.

Answer 3

It is easy to check $E=\mbox{ker}(f)$ and $O=\mathbb{Z}-E$ partitions $\mathbb{Z}$, each of them corresponds to the image $1, -1$. Then construct the set $\overline{Z}=\mathbb{Z}/E=\{E, O\}$ and define an operation $+$ on $\overline{Z}$ such that $E+E=E, E+O=O+E=O, O+O=O$. Then check that $(\overline{Z},+)$ is a group. Now think about a function $\phi:\overline{Z}\to \mbox{im}(f)$ such that $\phi(E)=1, \phi(O)=-1$. Can you proceed from here?

The point is (1) to define an operation $+$ to make $\overline{Z}=\mathbb{Z}/\mbox{ker}(f)$ into a group (2) to check there is a group isomorphism(a bijective homomorphism between groups) from $\overline{Z}$ to $\{-1, 1\}$.