Proving that another sequence is exact and non-split (Correcting the function $p$ to make it exact)

Question

In connection to my previous question here Proving that a sequence is exact and non-split I want to solve the question below:

Let $K = (a)$, $O(a) = 2$ like $\mathbb{Z}/2\mathbb{Z}$ and Let $L = (b)$, $o(b) = 4$ like $\mathbb{Z}/4\mathbb{Z}.$

Let $N$ be any $\mathbb Z$-module. Show that the sequence $$0 \to K \xrightarrow{i} L \oplus N \xrightarrow{p} K \oplus N \to 0 $$ with the maps $i$ defined by $a \mapsto (2b, 0)$ and $p$ defined by $(b, n) = (p(b), n)$ is exact and non-split.

My thoughts

I believe it is not exact as I have that $\operatorname{ker}(p) = \{(0,0)\}$ while $\operatorname{Im}(i) = \{(0,0), (2b, 0)\}.$ am I correct?

And then, according to the following criteria it is non-split because it is not exact. right?

The splitting criteria I know is:

Let $0 \to M' \xrightarrow{f} M \xrightarrow{g} M'' \to 0$ be an exact sequence of $A$-modules. The following are equivalent:

1. There exists $\psi \colon M \to M'$ with $\psi \circ f = 1_{M'}$.

2. $M = \ker(g) \oplus \ker(\psi)$.

3. There exists $\phi \colon M'' \to M$ with $g \circ \phi = 1_{M''}$.

4. $M = \operatorname{Im}(f) \oplus \operatorname{Im}(\phi)$.

Now, to correct it to be exact, I believe we should change the definition of $p$ to $(b, n) = (p(b), n),$ right?

But then, for non-splitting, I applied the same idea of the referred question above but with the following modifications:

if there were a map $r \colon L \oplus N \to K$ such that $r \circ i = 1_K$, then $$ a=r(i(a)) = r((2b,0))=2r((b,0)) = 0, $$ (in $K$, every element $x$ satisfy that $2x=0$) which is absurd. Hence, the given sequence is non-split.\

am I correct?

Could anyone help me please?

Answer 1

Assuming that $p$ is defined by $$\forall n \in N, \quad p((x,n)) = \begin{cases} (0,n) & \textrm{if } x=0 \\ (a,n) & \textrm{if } x=b \\ (0,n) & \textrm{if } x=2b \\ (a,n) & \textrm{if } x=3b \\ \end{cases} $$ the sequence is exact: Since $p((2b,0))=(0,0)$, we have that $\ker p$ contains $\{(0,0),(2b,0)\}$. On the other hand, if $(x,n) \in \ker p$, the equality $p((x,n)) = (0,0)$ implies that $x$ is equal to $0$ or $2b$ and that $n=0$. Hence $\ker p = \{(0,0),(2b,0)\} = \operatorname{im} i$.

The argument of that doesn’t exists a map $r \colon L \oplus N \to K$ such that $r \circ i = 1_K$ is correct. Hence, the sequence is exact and non-split.