Here is the sequence that I want to show that it is exact and non-split:
Let $K = (a)$, $O(a) = 2$ like $\mathbb{Z}/2\mathbb{Z}$ and Let $L = (b)$, $o(b) = 4$ like $\mathbb{Z}/4\mathbb{Z}.$ Consider the following exact sequence of abelian groups ($\mathbb{Z}$-modules) $$0 \to K \xrightarrow{i} L \xrightarrow{p} K \to 0 $$ where the map $i$ is defined by $a \mapsto 2b$ and the map $p$ is defined by $p(b) = a$.
My try:
Showing that it is exact:
I know that I should show that $i$ is injection, $p$ is onto and $\ker(p) = \operatorname{Im}(i).$ First $i$ is injection because it is the inclusion map. Second $p$ is onto because it is the projection map. Third. To show that $\ker(p) = \operatorname{Im}(i)$, I will calculate $\ker(p)$ and $\operatorname{Im}(i)$. Since $\ker(p) = \{b \in L \mid p(b) = 0\} = \{b \in L \mid a = 0\}$ and since $\operatorname{Im}(i) = \{i(a) \mid a \in K \}$, then I do not know how to conclude that they are equal.
For the non-split part:
I do not know exactly how should I prove it. 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:
There exists $\psi \colon M \to M'$ with $\psi \circ f = 1_{M'}$.
$M = \ker(g) \oplus \ker(\psi)$.
There exists $\phi \colon M'' \to M$ with $g \circ \phi = 1_{M''}$.
$M = \operatorname{Im}(f) \oplus \operatorname{Im}(\phi)$.
But I can not conclude/guess which condition will not be fulfilled in our sequence. Could anyone help me please?
Best Answer
We have that $K=\{0,a\}$, and that $L=\{0,b,2b,3b\}$. Hence, all the values of $i$ and $p$ are $$ i(0)=0, \quad i(a)=2b, \\ p(0)=0, \quad p(b)=a, \quad p(2b)=2a=0, \quad p(3b)=3a=a. $$ Thus, $$ \ker i = \{x \in K : i(x)=0\} = \{0\}, \\ \operatorname{im} i = \{i(x) : x \in K\} = \{0,2b\}, \\ \ker p = \{y \in L : p(y)=0\} = \{0,2b\}, \\ \operatorname{im} p = \{p(y) : y \in L\} = \{0,a\} = K. $$ That is, the sequence is exact.
Finally, if there were a map $r \colon L \to K$ such that $r \circ i = 1_K$, then $$ a=r(i(a)) = r(2b)=2r(b) = 0, $$ (in $K$, every element $x$ satisfy that $2x=0$) which is absurd.
Hence, the sequence doesn’t split.