Proving a criterion for recognizing when a group $G$ is a direct product of 2 groups

Question

Here is the question I want to answer:

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 thoughts are:

I know that a short exact sequence is split if it both left- and right-split, so 2 implies 3 trivially.

I 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. $$rf = id_K$$ Where $f: K \rightarrow G$ and it is injective.

And, I also know that a short exact sequence $1 \rightarrow K \rightarrow G \rightarrow H \rightarrow 1$ is right-split if there is a section $s: H \rightarrow G$ s.t. $$gs = id_H$$ Where $g: G \rightarrow H$ and it is onto.

Now I am stuck in proving $1 \implies 2$ (I do not know if this is the smartest way of proving the equivalences, I received some hints of proving the equivalences in the following order $1 \implies 4 \implies 2 \implies 3 \implies 1 $ but I have no clue if that is better or easier than proving $1 \implies 2 \implies 3 \implies 4$), I am guessing that this problem here Decomposing a group as a direct product of its kernel and image may help but I do not know how, could anyone help me in proving that please?

Answer 1

The statement $(2) \Rightarrow (1)$ is false. You need some extra assumptions, because the existence of of a split short exact sequence is not enough.

You have a short exact sequence $0\to A_n\to S_n \to \{\pm 1\} \to 0$, where the first morphism is inclusion and the second one is the signature map.

The sequence is right-split by $s: \{\pm 1\}\to S_n$, which sends $-1$ to a fixed transposition $\tau$.

However, you don't have $S_n\simeq A_n\times \{\pm 1\}$, as $S_n$ has no normal subgroup of order $2$. If you don't like this argument, just take $n=3$: $S_3$ is not abelian, but $A_3$ and $\{\pm 1\}$ are, and so is their direct product.

In fact , the existence of a split short exact sequence is equivalent to have $G\simeq K\rtimes H$.