The importance of $G \cong K \times H$ is just to create the split short exact sequence.

Question

Here is the problem I am trying to prove:

For groups $G,H,K,$ show that the following conditions are equivalent.

1. $G \cong K \times H.$
2. There exists a split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$
3. There exists a left-split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$
4. $H \triangleleft G, K \triangleleft G, G = HK $ and $H \cap K = \{1\}.$

My questions are:

Here is the proof of $1 \implies 2.$

Assume that $G \cong K \times H.$ We want to show that there exists a split short exact sequence: $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1.$

But we know that a short exact sequence is split if it is both left- and right-split. Also, know that a short exact sequence $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1$ is left-split if there is a retraction $r: G \rightarrow K$ s.t. $$r\circ i = id_K \quad \quad (1) $$ Where $i: K \rightarrow G$ and it is injective. And also, we know that a short exact sequence $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1$ is a right-split if there is a section $s: H \rightarrow G$ s.t. $$p\circ s = id_H \quad \quad (2)$$ Where $p: G \rightarrow H$ and it is onto.

So, we need to find the functions $i, p, r$ and $s$ that satisfies $1.$ and $2.$ above.

So, since $G \cong K \times H,$ we can say that we have $1 \rightarrow K \rightarrow K \times H \rightarrow H \rightarrow 1,$ with $i: K \rightarrow K \times H$ the embedding $i(k) = (k,1)$ and $p: K \times H \rightarrow H$ the projection $p(k,h) = h.$

If we define $r: K \times H \rightarrow K $ with $r(k,h) = k$ then we have $(r\circ i) (k) = r((k,1)) = k$ as required.

Also, if we define $s: H \rightarrow K \times H $ with $s(h) = (1,h)$ then we have $(p\circ s)(h) = p ((1,h)) = h $ as required.

1- Is it correct? and is the importance of $G \cong K \times H$ is just to create the split short exact sequence?

2- Is the statement of my problem completely correct?

3- How can I prove $1 \implies 4$?

4- Is the right sequence of implications for proving the problem is proving $1 \implies 2 \implies 3 \implies 1$ and then proving $1 \Leftrightarrow 4$?

5- Is $2 \implies 3 \implies 1$ in my problem means $2 \implies 1$? or $2 \implies 1$ is a different thing

Answer 1

1. Your proof of $1\implies 2$ seems correct.

2. The problem statement is wrong. As stated in the comments, the equivalence between $1$ and $2$ is incorrect. We cannot get from $2$ to $1$, since $2$ is one way of defining a semidirect product $\bar{G}=K\rtimes_{\varphi}H$, where $\varphi:\bar{G}\to H$ is a homomorphism that is the identity on $H$ with kernel $K$. Not all semidirect products are direct products. See p.788 of Rotman's "Advanced Modern Algebra" (or simply look up the definition given in other editions).

3. This is standard. Here $4$ is the definition of the internal direct product. See, for instance, p.196 of Gallian's "Contemporary Abstract Algebra (Eighth Edition)".

4. There is no right sequence, since $2$ does not imply $1$.

5. Yes, that's what it would mean. Implication is transitive.