I am trying to find a short exact sequence of the following form:
$$0 \to \mathbb Z_4 \xrightarrow{f} \mathbb Z_8 \oplus \mathbb Z_2 \xrightarrow{g} \mathbb Z_4 \to 0$$
I assumed that actually there is one and I am trying to put constraints on $f$ and $g$ that makes it exists. what I reached so far is that if $f$ is injective then the image of it will be one of the following $(\pm \bar{ 2}, \bar{0}),(\pm \bar{2}, \bar{1}).$ But then I want to use the surjectivity of $g$ to find the image of those elements and I hope that the image of one of them would be 0 and so it would be in the kernel of $g$ and so image g would be kernel f in this case and I would have find the sequence I want. The problem that faces me is that:
What are the values of $g$ on $(0,1)$ and $(1,0)$? so that I can express $(\pm \bar{ 2}, \bar{0}),(\pm \bar{2}, \bar{1})$ in terms of these two elements. could anyone help me in this please?
EDIT:
I also know that, If $\mathrm{ker}\ g=\mathrm{Im}\ f,$ then by the first isomorphism theorem applied to $g$, we would have $$ \left(\mathbb Z_8 \oplus \mathbb Z_2\right)\ /\ \mathrm{Im}(f) \cong \mathbb Z_4, $$ but how this will help me?
$f(1)\in\mathbb Z_8\oplus\mathbb Z_2$ has order exactly $4$, so the $1$st component is $\pm2$. Without loss we may assume $f(1)=(2,x)$ for $x\in\mathbb Z_2$.
If $f(1)=(2,0)$ then $\mathrm{im}(f)=\ker(g)=2\mathbb Z_8\oplus0$, so $\mathbb Z_8\oplus\mathbb Z_2/\ker(g)\cong\mathbb Z_2\oplus\mathbb Z_2$, which is not $\mathbb Z_4$, so this is impossible.
Thus $f(1)=(2,1)$, where $\mathbb Z_8\oplus\mathbb Z_2/\ker(g)=\mathbb Z_8\oplus\mathbb Z_2/\langle(2,1)\rangle\cong\mathbb Z_4$, since $(1,0)$ has order $4$.
Now one choice for $g$ is $(1,0)\mapsto 1$ and $(0,1)\mapsto 2$ (so that $g\circ f(1)=g(2,1)=0$.)