= Solution
The <conjugacy classes> are specified by <cycle type>, with sizes $1,6,3,8,6$ for $1,(12),(12)(34),(123),(1234)$. The trivial and <sign representations> give the first two rows. The <permutation representation> on four letters contains a trivial line; its <orthogonal complement> has <character> equal to the number of fixed letters minus one. Tensoring that <representation> by the sign gives another row. Finally the quotient $S_4/V_4\cong S_3$ supplies the two-dimensional standard <representation>. The resulting table is
$$
\begin{array}{c|rrrrr}
&1&(12)&(12)(34)&(123)&(1234)\\\hline
\text{class size}&1&6&3&8&6\\
\mathbf1&1&1&1&1&1\\
\mathrm{sgn}&1&-1&1&1&-1\\
V&3&1&-1&0&-1\\
V\otimes\mathrm{sgn}&3&-1&-1&0&1\\
W&2&0&2&-1&0
\end{array}
$$
With the class-size-weighted <inner product>, these <characters> have <norm> one and are pairwise orthogonal, so all are irreducible and inequivalent. Their squared dimensions sum to $1+1+9+9+4=24=|S_4|$, proving completeness.
Back to article page
