Show $S_4$ can't be derived group 
Prove that there is no group $G$ such that $G' = S_4$.

We know that every group of order $24$ is not simple. and $S_4$ is an non-abelian group.
I tried to show that group of order $24$ can not be normal in any group (and because $G'$ is normal in $G, G'$ can not be $S_4$), but I can't prove that group of order $24$ can not be normal.
 A: Write $C=C_G(G')$, then $C \cap G'=Z(S_4)=1$. By the N/C theorem $G/C$ injects homomorphically in ${\rm Aut}(S_4)=S_4$. Hence $(G/C)' =G'C/C \cong G'/(G' \cap C) \cong G'$ injects homomorphically in $(S_4)'=A_4$, which is impossible.
A: There is a nice generalization, that appeared in a paper entitled Which groups are commutator subgroups? in the Liber Amicorum dedicated to Rob van der Waall on the occasion of his retirement. 
Proposition (René J. Schoof, 2006) Let $n \gt 2$, then $S_n$ never appears as the commutator subgroup of some group.
Proof For $n \gt 2$ and $n \neq 6$, the argument of my previous answer works, since in those cases $Z(S_n)=1$ and Aut$(S_n) \cong S_n$. So we only have to deal with $n=6$. It is a famous fact that in that case actually $S_6$ allows for an outer automorphism and if we identify $S_6$ with $\rm{Inn}(S_6)$, then $|\rm{Aut}(S_6):S_6|=2$. Now $A_6 \text{ char } S_6 \lhd \rm{Aut}(S_6)$, so $A_6 \lhd \text{Aut}(S_6)$ and $|\rm{Aut}(S_6):A_6|=4$. This means that $\rm{Aut}(S_6)/A_6$ is abelian, thus $\rm{Aut}(S_6)' \subset A_6$. Now $Z(S_6)=1$, so we can apply again the argument in my other answer and that would imply $S_6 \subset \rm{Aut}(S_6)' \subset A_6$, which is a contradiction. $\square$
