Theorem. There does not exist a group whose commutator subgroup is isomorphic to $$S_4$$.

The relevant facts are that $$S_4$$ is a complete group(no outer automorphisms, trivial center) which is not perfect(that is, the commutator subgroup of $$S_4$$ is not $$S_4$$ itself). Any group which has these properties is never a commutator subgroup of anything. Here’s why.

Lemma. If $$K$$ is a complete group and $$K\lhd G$$, then $$G$$ is the direct product $$K\times H$$ of $$K$$ by its centralizer $$H=C_G(K)$$.

In other words, a complete group can be a normal subgroup only in the most trivial fashion: the large group is just a direct product of the normal group by something.

Proof of the lemma. Let $$H=C_G(K)$$ be the centralizer of $$K$$ in $$G$$, namely the set of all elements which commute with all elements of $$K$$. $$H$$ is a normal subgroup of $$G$$, and $$H\cap K=Z(K)=1$$ since $$K$$ has trivial center. Any element $$g\in G$$ induces an automorphisms $$\phi_g$$ of $$K$$ by conjugation: $$\phi_g(k)=g^{-1}kg$$. But $$K$$ has no outer automorphisms, so $$\phi_g$$ must equal some inner automorphism of $$K$$, that is, for some $$k\in K$$, $$\phi_g=\phi_k$$. Now conjugation by $$gk^{-1}$$ does nothing to $$K$$, so $$gk^{-1}=h\in H$$. In other words $$g=kh$$: every element of $$G$$ is expressible as product of an element of $$K$$ and an element of $$H$$. Since $$H$$ and $$K$$ commute, $$G$$ is the direct product of $$H$$ and $$K$$.            $$\Box$$

Proof. Now suppose that $$G$$ is some group such that $$K=G^\prime=[G,G]$$, the commutator subgroup, is such that $$K$$ is complete and non-perfect. By the lamma, $$G=K\times A$$ where $$A\cong G/K$$ is an abelian group. So any element of $$G$$ can be writeen as $$ka$$ with $$k\in K$$ and $$a\in A$$, and moreover, $$ka=ak$$ for any $$k\in K$$, $$a\in A$$.

Consider a commutator $$c=xyx^{-1}y^{-1}$$ in $$G$$. Write $$x=ka$$ and $$y=lb$$. Since $$K$$ and $$A$$ commute, $$c=aba^{-1}b^{-1}xyx^{-1}y^{-1}$$. Since $$A$$ is abelian, the first part vanishes and $$c=xyx^{-1}y^{-1}$$. So any commutator of $$G$$ lies in the commutator sungroup of $$K$$, and it follows that $$G^\prime=K^\prime$$. Since $$K^\prime\ne K$$, $$G^\prime\ne K$$, as well.

It remains to show that $$S_4$$ is complete and non-perfect.

(i) $$S_4$$ has trivial center: this is obvious. No permutation commutes with all other permutations.

(ii) $$S_4$$ has no outer automorphisms. This is true for all $$S_n$$ except $$n=2$$, $$6$$. It’s a standard result.

(iii) $$S_4$$ is not perfect. This is also ovious: for any two permutations $$\sigma$$, $$\tau\in S_n$$, the commutator $$\sigma^{-1}\tau^{-1}\sigma\tau$$ is an even permutation, so the commutator subgroup is contained in the alternating group $$A_n$$. In fact the commutator group eauals $$A_n$$, but we don’t need that here.

This completes the proof the theorem.                                  $$\Box$$

Remark. a simple modification: $$K^\prime\subset G^\prime$$ is clear, and the other direction follows since $$G/K^\prime=K/{K^\prime\times A}$$ is abelian.

Here is a proof not using those well-known facts about $$S_4$$. (though it’s easy to derive them with it)

Proof. $$S_4$$ has exactly $$4$$ Sylow  $$3$$-subgroups

$P_1=\langle(234)\rangle, P_2=\langle(134)\rangle, P_3=\langle(124)\rangle, P_4=\langle(123)\rangle,$

where $$P_i$$ is the only Sylow $$3$$-subgroup of the stabilizer of $$i$$ in $$S_4$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. So $$\sigma P_i\sigma^{-1}=P_{\sigma(i)}$$ for all $$\sigma\in S_4$$ and $$i=1$$, $$2$$, $$3$$, $$4$$.

Assume $$G^\prime=S_4$$ and take $$g\in G$$. The conjugation with $$g$$ permutes $$P_1$$, $$P_2$$, $$P_3$$, $$P_4$$, so we find $$\rho\in G$$ with $$g P_ig^{-1}=P_{\rho(i)}=\rho P_i\rho^{-1}$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. So $$h\colon \rho^{-1}g\in G$$ satisfies $$h P_ih^{-1}=P_i$$ for $$i=1$$, $$2$$, $$3$$, $$4$$.

Then for all $$\sigma\in S_4$$ we have also $$h^{-1}\sigma h\in S_4$$ and

$$\begin{split}P_{h^{-1}\sigma h(i)}&=(h^{-1}\sigma h)P_i(h^{-1}\sigma h)^{-1}\\&=h^{-1}\sigma hP_ih^{-1}\sigma^{-1}h\\&=h^{-1}\sigma P_i\sigma^{-1}h\\&= h^{-1} P_{\sigma(i)}h=P_{\sigma(i)}.\end{split}$$

and therefore $$h^{-1}\sigma h(i)=\sigma(i)$$ for $$i=1$$, $$2$$, $$3$$, $$4$$.

This gives $$h\in C_G(S_4)$$ and $$g=\rho h\in S_4C_G(S_4)$$ for all $$g\in G$$. So $$G=S_4C_G(S_4)$$ and $$|G\colon A_4C_G(S_4)|\leqslant2$$. But then $$A_4C_G(S_4)\trianglelefteq G$$ with abelian factor, so $$S_4=G^\prime\leq A_4C_G(S_4)$$, and by Dedekind we get

$S_4=A_4C_G(S_4)\cap S_4=A_4(C_G(S_4)\cap S_4)=A_4Z(S_4)=A_4$

since $$Z(S_4)=1$$, as $$\sigma\in Z(S_4)$$ would give $$P_i=\sigma P_i\sigma^{-1}=P_{\sigma(i)}$$ and $$i=\sigma(i)$$ for $$i=1$$, $$2$$, $$3$$, $$4$$. Contradiction!

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