Prove that there exists the exact sequence (over Z) 0 → Z 2 →Z 4 → Z 4 → Z 2 → 0 .

Question

$$ 0 \to \mathbb Z /2 \to \mathbb Z /4 \to \mathbb Z /4 \to \mathbb Z / 2 \to 0$$

To show that it is exact, I need $f : \mathbb Z / 2 \to \mathbb Z / 4$ monic, and $g : \mathbb Z /4 \to \mathbb Z /2$ epic.

homomorphism $f$ by $f([a])=[2a]$ , $[a]$ is in $\mathbb Z /2$ and $[2a]$ is in $\mathbb Z /4$ and homomorphism $g$ by $g([b]2)=[b]4$ .

when I take like that by finding all homomorphism between them, image of $f$ is not equal to kernel of $h : \mathbb Z /4 \to \mathbb Z /4$. Also image of $h$ is not equal to kernel of $g$.

Could someone help me out?

Answer 1

Homomorphism $h: \mathbb{Z}/4 \to \mathbb{Z}/4$ can be defined as following:

Let $h(\bar{0})=h(\bar{2})=\bar{0}$, and $h(\bar{1})=h(\bar{3})=\bar{2}$.

It can be verified that $h$ is a homomorphism on $\mathbb{Z}/4$, which is a $\mathbb{Z}$ module.

And it's clear that Ker$(h)$=Im$(h)$=$\left\{\bar{0},\bar{2}\right\}$.