TITLE: Field of definition of representations of symmetric groups
QUESTION [15 upvotes]: 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.

REPLY [4 votes]: 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 $S_n$.