May 112018
 

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{equation} \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} \end{equation}

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!

 Posted by at 11:43 am

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