Here is my trial for the proof after receiving a lot of help here:
For groups $G,H,K,$ show that the following conditions are equivalent.
$G \cong K \times H.$
There exists a split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$
There exists a left-split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$
$H \triangleleft G, K \triangleleft G, G = HK $ and $H \cap K = \{1\}.$
My question is:
Could anyone show me please how can I prove $3 \implies 4$ here? or $3 \implies 1$ (although I received a counterexample here Proving a criterion for recognizing when a group $G$ is a direct product of 2 groups showing that I can not show that $3 \implies 1$ at all, but I do not fully understand the example. So if anyone can re-explained it to me I would appreciate that)?
For $3 \Rightarrow 4$ I'd say it is better to prove $3 \Rightarrow 1$ because $1 \Rightarrow 4$ is clear I think.
So for $3 \Rightarrow 1$ : you have the exact sequence with $i:K \to G$ and $p:G \to H$. By assumption you have $r:G \to K$ such that $r \circ i= \mathrm{Id}_K$. You can define the map $\varphi : G \to K \times H$ by $\varphi(g)=(r(g),p(g))$. Then show that $\varphi$ is an isomorphism.