Spring 2016 - Problem 8

Solution.

Recall that $S_4$ has $5$ conjugacy classes. Thus, by Artin-Wedderburn and algebraic closure of $\C$, we have

with $n_1 \leq \cdots \leq n_4$ and $24 = \abs{S_4} = n_1^2 + \cdots + n_5^2$. If $n_i \leq 2$ for each $i$, then $n_1^2 + \cdots + n_5^2 \leq 20 < 24$, so at least one $n_i$ is larger than $2$, say $n_5$. Note that $n_5 \leq 4$, since $5^2 > 24$, so there are two possibilities for $n_5$: if $n_5 = 4$, then we have $n_1^2 + \cdots + n_4^2 = 24 - 16 = 8$. At least one of these must be larger than $1$, so $n_4 = 2$, but then we'd have $n_1^2 + n_2^2 + n_3^2 = 4$ so this case isn't possible.

If $n_5 = 3$, then we have $n_1^2 + \cdots + n_4^2 = 24 - 9 = 15$. By quickly checking the possibilities, we see $n_4 = 2$ isn't possible, so $n_4 = 3$ also. From here, we see that we have $n_1 = n_2 = 1$, $n_3 = 2$, $n_4 = n_5 = 3$. Thus, our character table is of the form

where $\chi_i$ is a character (with $\chi_1$ the trivial character).

Recall the sign homomorphism $S_4 \to \set{\pm 1}$, which gives a $1$-dimensional character. This gives the row

We also have a (reducible) representation $\rho$ of $S_4$ in $\C^4$ by sending each permutation to its corresponding permutation matrix. This representation is reducible: let $V = \span\set{e_1 + e_2 + e_3 + e_4}$ and decompose $C^4 \simeq V \oplus W$ where $W$ is the orthogonal complement to $V$. Note that $\restr{\rho}{V}$ is the trivial representation and that $W = \span\set{e_1 - e_2, e_1 - e_3, e_1 - e_4}$.

Let $\chi_4 = \restr{\rho}{W}$. Then

and

so $\chi_4$ is actually an irreducible $3$-dimensional representation. This also gives $\chi_5$, since $\chi_2\chi_4$ is another character. We now have the table

Finally, to complete the table, we just need to use the orthogonality relations. Columns $1$ and $2$ give $2a = 0$, so $a = 0$; columns $1$ and $3$ give $2 + 2b - 6 = 0$, so $b = 2$; columns $1$ and $4$ give $2 + 2c = 0$, so $c = -1$; and columns $1$ and $5$ give $2d = 0$, so $d = 0$. Thus, the character table is