Solvability and nonnilpotence of small symmetric groups

Both are abelian, and every abelian group is solvable.

The chain \[ S_3\trianglerighteq A_3\trianglerighteq\{e\} \] is a normal series. Its factors are \[ S_3/A_3\cong\ZZ_2, \qquad A_3\cong\ZZ_3, \] which are abelian. Hence \(S_3\) is solvable.

Let \[ V=\{e,(12)(34),(13)(24),(14)(23)\}. \] Conjugation in \(S_4\) preserves cycle type, so it permutes the three nonidentity elements of \(V\); hence \(V\trianglelefteq S_4\). Also \(V\le A_4\). Thus \[ S_4\trianglerighteq A_4\trianglerighteq V\trianglerighteq\{e\} \] is a normal series, and its factors have orders \(2\), \(3\), and \(4\) respectively. More precisely, \[ S_4/A_4\cong\ZZ_2, \qquad A_4/V\cong\ZZ_3, \qquad V\cong\ZZ_2^2. \] All factors are abelian, so \(S_4\) is solvable.

A central element must commute with \((12)\). Of the six elements of \(S_3\), the centralizer of \((12)\) is \(\{e,(12)\}\). But \((12)\) does not commute with \((23)\), so the only central element is \(e\).

Let \(\sigma\in Z(S_4)\). Since \(\sigma\) commutes with every transposition, \[ \sigma(ij)\sigma^{-1}=(ij) \] for every \(i\ne j\). But conjugation relabels a transposition: \[ \sigma(ij)\sigma^{-1}=(\sigma(i)\ \sigma(j)). \] Hence \(\{\sigma(i),\sigma(j)\}=\{i,j\}\) for every pair \(i\ne j\). Fixing \(i\) and choosing two distinct \(j,k\ne i\) forces \(\sigma(i)=i\). Thus every point is fixed and \(\sigma=e\).

Every nontrivial nilpotent group has nontrivial center. Indeed, the upper central series is defined by \(Z_0=\{e\}\) and \(Z_{i+1}/Z_i=Z(G/Z_i)\); if \(Z(G)=Z_1\) is trivial, induction gives \(Z_i=\{e\}\) for all \(i\), so the series never reaches a nontrivial \(G\). By 4 and 5, both \(S_3\) and \(S_4\) are nontrivial with trivial center. Therefore neither is nilpotent.