= Solution
Order the <conjugacy classes> of the <symmetric group> $S_4$ by representatives $1,(12),(12)(34),(123),(1234)$; their sizes are $1,6,3,8,6$. For a three-dimensional <rotation> through an <angle> $\theta$, the <eigenvalues> are $1,e^{i\theta},e^{-i\theta}$, so the <trace of a three-dimensional rotation> is $1+2\cos\theta$.
Under the identification with the <rotational symmetry group of a cube>, elements of order two give half-turns, elements of order three give turns through $\pm2\pi/3$, and elements of order four give turns through $\pm\pi/2$. A half-turn can have an axis through opposite edges or opposite faces, but both kinds have <trace> $-1$. Thus
$$
\boxed{\begin{array}{c|rrrrr}
\text{representative}&1&(12)&(12)(34)&(123)&(1234)\\\hline
\text{class size}&1&6&3&8&6\\
\chi_C&3&-1&-1&0&1
\end{array}.}
$$
Its <character inner product> with itself is
$$
\langle\chi_C,\chi_C\rangle_{S_4}
=\frac{9+6+3+0+6}{24}=1.
$$
By <character orthogonality>, a nonzero <complex representation> has <character inner product> one with itself exactly when it is an <irreducible representation>. Therefore $C$ is an <irreducible representation> of <dimension> three.
For comparison useful in part (e), the <standard representation of the symmetric group> has <character> equal to the number of fixed letters minus one, namely $(3,1,-1,0,-1)$ in this order. Multiplication by the <sign representation> changes this <character> to $(3,-1,-1,0,1)$. Thus $C$ is the <tensor product> of the <standard representation of the symmetric group> with the <sign representation>.
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