= Solution
For a finite <group> over $\mathbb C$, the number of inequivalent <irreducible representations> is its number of <conjugacy classes>, and the <sum of squares of irreducible representation dimensions> equals the <group order>. There are therefore five irreducibles for $S_4$, with squared dimensions summing to $24$.
First establish that part (c) found every <one-dimensional representation>. In such a representation, conjugate elements have the same scalar image. All <transpositions> are conjugate and square to the identity, so they all have one common image $s$ with $s^2=1$. <Transpositions> generate $S_4$, hence $s=1$ gives exactly the <trivial representation>, while $s=-1$ gives exactly the <sign representation>. There are precisely two degree-one irreducibles.
The other three dimensions $d_1,d_2,d_3$ are integers at least two and satisfy
$$
d_1^2+d_2^2+d_3^2=24-1-1=22.
$$
No one of them can be at least four, since $4^2+2^2+2^2=24>22$. They are therefore all two or three. Starting from three twos gives sum of squares $12$; each replacement of a two by a three adds $5$. To reach $22$ requires exactly two replacements. Thus the complete list is
$$
\boxed{1,\ 1,\ 2,\ 3,\ 3}.
$$
The count of <conjugacy classes> guarantees that these are all the inequivalent complex <irreducible representations>, not just dimensions of a proposed incomplete list.
Back to article page
