No group has commutator isomorphic to S_4

Question

I'm wondering about an alternative solution to the following question from Dummit and Foote:

Show that no group has commutator subgroup isomorphic to $S_4$.

The hint says to use the previous question: $G^{(1)}/G^{(2)}$ and $G^{(2)}/G^{(3)}$ cyclic $\,\Rightarrow\,$ $G^{(2)}=1$. The application of this fact is easy enough. The proof of this fact is rather tedious.

My first thought was:

Suppose not. Let $G$ be such that $[G,G]\cong S_4$. Note, certainly $|G|=: n \geq 5$. By Cayley's theorem, $\exists H\leq S_n$ with $[H,H]=S_4\leq [S_n,S_n]=A_n$. Now, I would like to apply the fact that $A_n$ is simple for $n\geq 5$, but of course $[H,H]$ is not necessarily normal in $[G,G]$ whenever $H$ is a subgroup of $G$.

Any ideas on whether or not and why an argument along these lines is bound to fail? or on an alternative solution?

Thanks.

Answer 1

I see that you want to know how to deal with this kind of problems . How about splitting your question into two paths? if S4 is normal then you know what to do , if not then you have to use this fact "S4 is not normal in A(n)" in order to complete your answer .

As you know the answer to the question already, you can't say that "an argument along these lines is bound to fail"