The Jucys-Murphy elements of the group algebra of a finite symmetric group (here's the [definition](http://en.wikipedia.org/wiki/Jucys%E2%80%93Murphy_element) in Wikipedia) are known to correspond to operators diagonal in the Young basis of an irreducible representaion of this group. As one can see from the Wikipedia entry, all of the elements of such a diagonal matrix (in other words the operator's eigenvalues) are integers.


I'm looking for a simple way of explaining this fact (that the eigenvalues are wholes). By simple I mean without going into more or less advanced representation theory of the symmetric group (tabloids, Specht modules etc.), so trying to prove the specific formula given in Wikipedia is not an option. (I'm considering the Young basis as the Gelfand-Tsetlin basis of the representation for the inductive chain of groups $S\_1\subset S\_2\subset \ldots\subset S\_n$, which is uniquely defined thanks to this chain's spectrum's simplicity, not as a set of vectors in correspondence with the standard tableaux.)


In fact, I'm trying to prove the first statement ($a\_i\in \mathbb{Z}$) of proposition 4.1 in this [article](http://www.stat.berkeley.edu/users/evans/Vershik_Okounkov_05.pdf).



Why are Jucys-Murphy elements' eigenvalues whole numbers?
This can be shown using the following two facts:


1. $X\_n=(1,n)+(2,n)+\ldots+(n-1,n)$ commutes with any element of $\mathbb Z S\_{n-1}$
2. Any irreducible $\mathbb Q S\_n$-module $V$ restricts to a multiplicity-free $\mathbb Q S\_{n-1}$-module (this follows from the classical branching rule; of course you said you didn't want to use tableaux's etc., so I'm not entirely sure whether this is compatible with your idea of "elementary").


The first one implies that $$X\_n\in End\_{\mathbb Q S\_{n-1}}(Res^n\_{n-1}V)$$ for any
irreducible $\mathbb Q S\_n$-module $V$. But, due to the second point and the fact that $\mathbb Q$ is a splitting field for $S\_{n-1}$, there is an isomorphism of algebras
$$
End\_{\mathbb Q S\_{n-1}}(Res^n\_{n-1}V) \cong \mathbb Q \oplus \ldots \oplus \mathbb Q
$$
This shows that $X\_n$ acts semisimply on $V$ with eigenvalues in $\mathbb Q$. That the eigenvalues lie in $\mathbb Z$ then actually follows from the fact that $\mathbb Z S\_n$ is a $\mathbb Z$-order (elements of $\mathbb Z$-orders have integral characteristic polynomial and therefore their eigenvalues will be integral over $\mathbb Z$ no matter on what module we let them act). 


The assertion for all JM elements reduces to this (hope this is clear).


