Field of definition of representations of symmetric groups

Question

Can one show in an elementary way, without recourse to Young tableaux etc., that the complex representations of symmetric groups are realisable over $\mathbb{R}$? It is easy to show that they are all self-dual, since the conjugacy classes of symmetric groups are self-dual, so one just has to exclude the possibility of quaternionic representations. Surely, there must be a similarly elementary argument? If there is, it is escaping me at the moment.

Answer 1

Some thoughts. Let $r_2(g)$ denote the number of square roots of $g$. Then $\langle r_2, \chi \rangle$ is the Frobenius-Schur indicator of $\chi$, and we want to show that this is equal to $1$ for all irreducibles $\chi$. This would follow if we could directly construct a representation of $S_n$ with character $r_2$, since then we would immediately have $\langle r_2, \chi \rangle \ge 0$.

The representation associated to this character would have dimension $r_2(e)$, or the number of involutions of $S_n$. And, indeed, there is a natural permutation representation of $S_n$ on the set of involutions (by conjugation), but it has the wrong character. In fact $r_2$ cannot in general be the character of a permutation representation: since it contains the trivial representation only once, it must be transitive, but its degree does not divide $n!$ in general.

But it still might be possible to construct this representation in a reasonably elementary way without going through the full construction of the irreducible representations of $S_n$.