Is there an example of a finite group $A$ that acts on a finite group $C$ irreducibly (that is, $C$ has no proper nontrivial $A$-invariant subgroup)
such that there exists an epimorphism $$\tau \colon A \ltimes C \to A$$ which does not split?



A finite group that splits and does not split
I don't believe this is possible. Let $G$ be a minimal counterexample. Then $G = AC$ with $C \unlhd G$ and $A$ a complement of $C$, and there exists $D \unlhd G$ with $G/D \cong A$, but $D$ has no complement if $G$. Note that $G/D \cong G/C \Rightarrow |C|=|D|$. Since $A$ acts irreducibly on $C$, we have $C \cap D = 1$. 


We prove first that $D \le A$. Since $C \cap D = 1$, we have $CD \cap A \cong D$, and $CD \cap A$ is a subgroup of $C \times D$ of the form $\{(\phi(d),d) : d \in D \}$, where $\phi:D \to C$ is a homomorphism. 


But $CD \cap A \unlhd A$, and for $a \in A$ (with notation $g^h=h^{-1}gh$) we have $(\phi(d),d)^a = (\phi(d)^a, d^a)$, but this must also equal $(\phi(d^a),d^a)$, so $\phi(d^a) = \phi(d)^a$ and hence ${\rm Im}(\phi)$ is $A$-invariant so, by assumption, it must be $C$ or $1$.


If ${\rm Im}(\phi) = C$ then $\ker(\phi)=1$, so $D \cap A = 1$. But then $A$ is a complement of $D$ in $G$, contrary to assumption. So $ {\rm Im}(\phi) = 1$ and $D \le A$ as claimed.


So now $G/D$ is a split extension of $C$ by $A/D$ and, since $D$ centralizes $C$, $A/D$ must act irreducibly on $C$.


If $D$ had a complement $B$ in $A$, then $CB$ would be a complement of $D$ in $G$, contrary to assumption. So $G/C \cong A$ is a nonsplit extension of $D$ by $A/D$. But we are assuming that $G/C \cong G/D$, so $G/C$ is a smaller counterexample than $G$.


