Construct the character table of \(S_5\).
Solution
The conjugacy classes are indexed by partitions of \(5\). In the order \[ 1^5,\quad 2\,1^3,\quad 2^2 1,\quad 3\,1^2,\quad 3\,2,\quad 4\,1,\quad 5, \] their sizes are \[ 1,\ 10,\ 15,\ 20,\ 20,\ 30,\ 24. \] The irreducible characters are indexed by the same partitions. The full table is
\[ \begin{array}{c|rrrrrrr} &1^5&2\,1^3&2^2 1&3\,1^2&3\,2&4\,1&5\\ \hline [5] &1& 1& 1& 1& 1& 1& 1\\ [4,1] &4& 2& 0& 1&-1& 0&-1\\ [3,2] &5& 1& 1&-1& 1&-1& 0\\ [3,1,1] &6& 0&-2& 0& 0& 0& 1\\ [2,2,1] &5&-1& 1&-1&-1& 1& 0\\ [2,1,1,1] &4&-2& 0& 1& 1& 0&-1\\ [1^5] &1&-1& 1& 1&-1&-1& 1 \end{array} \]
Here is a construction of the rows.
Let \(V\) be the \(4\)-dimensional standard representation, obtained from the permutation representation on five letters by removing the trivial summand. Its character is \[ \chi_V(g)=\#\operatorname{Fix}(g)-1, \] which gives the row \([4,1]\). Tensoring by the sign character gives the row \([2,1,1,1]\).
For any representation with character \(\chi\), \[ \chi_{\operatorname{Sym}^2 V}(g) =\frac{\chi(g)^2+\chi(g^2)}2, \qquad \chi_{\wedge^2 V}(g) =\frac{\chi(g)^2-\chi(g^2)}2. \] For the standard representation of \(S_5\), \[ \operatorname{Sym}^2V\cong [5]\oplus[4,1]\oplus[3,2], \] so subtracting the trivial and standard rows yields the character \([3,2]\). Also \[ \wedge^2V\cong[3,1,1], \] which yields the \(6\)-dimensional row. Finally, tensoring \([3,2]\) by sign gives \([2,2,1]\).
The weighted inner products using the displayed class sizes are \[ \langle\chi_\lambda,\chi_\mu\rangle =\frac1{120}\sum_C |C|\chi_\lambda(C)\chi_\mu(C) =\delta_{\lambda\mu}, \] so these seven characters are pairwise distinct irreducibles. Their squared dimensions sum to \[ 1^2+4^2+5^2+6^2+5^2+4^2+1^2=120=|S_5|, \] so the table is complete.