Showing that there is no Left splitting in the following exact sequence of groups.

Question

Here is the exact sequence of groups I am asking about:

$$0 \to A_3 \xrightarrow{f} S_3 \xrightarrow{g} \mathbb Z/ 2 \mathbb Z \to 0$$

I want to show that there is no $\psi: S_3 \to A_3 $ such that $\psi \circ f = 1_{A_3}.$

My thoughts are:

Any map to $3$ elements that is not zero is onto (I do not remember how to prove this statement or exactly from where I learned it but I am sure it is correct, so if anyone can help me in proving it, I would appreciate this) so we have a kernel. But the kernel of $S_3$ has $6$ elements while the Kernel of $A_3$ has 2 elelments. But then how can I reach that there is no such map? could someone help me please?

EDIT

I studied this theorem but I still can not see how it leads to a counter example:

Theorem

For groups $G,H,K,$ 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 $H \cap K = \{1\}.$

Can anyone clarify this to me please?

Answer 1

Any homomorphism $S_3\to A_3$ maps all transpositions to the identity, since the only element in $A_3$ whose square is $1$ is $1$ itself. Now, every element of $S_3$ is equal to a product of transpositions, so...