Write the complete character table for the symmetric group \(S_4\), giving the construction and justification for each irreducible character.
Solution
The conjugacy classes of \(S_4\) are indexed by cycle type: \[ 1^4,\quad 2\,1^2,\quad 2^2,\quad 3\,1,\quad 4, \] with class sizes \(1,6,3,8,6\).
There are therefore five irreducible complex characters. Two are the trivial and sign characters. The permutation representation on \(\mathbb C^4\) has character equal to the number of fixed points; subtracting the trivial representation gives the standard character \[ \chi_{\mathrm{std}}=(3,1,-1,0,-1). \] Tensoring with sign gives \[ \chi_{\mathrm{std}}\otimes\mathrm{sgn}=(3,-1,-1,0,1). \] Finally \(V_4\trianglelefteq S_4\) and \(S_4/V_4\cong S_3\). Pulling back the \(2\)-dimensional irreducible character of \(S_3\) gives \[ \chi_2=(2,0,2,-1,0). \] Thus the complete table is \[ \begin{array}{c|ccccc} &1^4&2\,1^2&2^2&3\,1&4\\ \hline \mathbf 1&1&1&1&1&1\\ \mathrm{sgn}&1&-1&1&1&-1\\ \chi_2&2&0&2&-1&0\\ \chi_{\mathrm{std}}&3&1&-1&0&-1\\ \chi_{\mathrm{std}}\otimes\mathrm{sgn}&3&-1&-1&0&1 \end{array} \] The squared dimensions sum to \(1+1+4+9+9=24=|S_4|\), and the displayed rows are pairwise orthonormal with respect to the class-function inner product, so these are all irreducibles.