I think that these conjectures are essentially equivalent to results in the literature, but the translation is not straightforward. I know corresponding results for the Hecke algebras of the symmetric groups. When you work in this generality you have to change some of the statements a little in order to get them to work. Primarily this is because your column stabliser $C_T$ will not usually be a standard parabolic subgroup so there is no corresponding natural subalgebra of the Hecke algebra to work with. To get around this you take the corresponding standard parabolic, for which the Hecke algebra does have a subalgebra, and then in the Hecke algebra you conjugate by $T_d$ where $d$ is a minimal length coset representative which makes the two parabolics conjugate. Ultimately, this will prove the results that you want, but it gives an additional layer of complications, which add to the notational confusion.
I think that the easiest way to approach these questions is using seminormal forms. As you probably know, Okounkov and Vershik [1] have given a really nice approach to this subject.
The main fact we need is that over $\Bbb{Q}$ the Specht module $S^\lambda$ has a seminormal basis $\{v_T\}$ indexed by the standard $\lambda$-tableaux with the property that
$$ y_k v_T = c_k(T) v_T.\qquad\qquad(1)$$
Here $c_k(T)$ is the content of $k$ the tableau $T$. (This conflicts slightly with your notation, but it shouldn't cause any problems.) That is, $c_k(T)$ is equal to the column index of $k$ in $T$ minus the row index of $k$ in $T$. (If I have misread your conventions above then you will need to replace $c_k(T)$ with $-c_k(T)$.)
Each seminormal basis element $v_T$ is a simultaneous eigenvector for $y_1,\dots,y_n$. So $v_T$ is uniquely determined up to multiplication by a non-zero scalar. The exact choice of seminormal basis $\{v_T\}$ doesn't matter in what follows.
Now, for each standard tableau $S$ define
$$F_S = \prod_{k=1}^n\prod_{c\ne c_k(S)}\frac{y_k-c}{c_k(S)-c},$$
where in the second product $c\in\{c_m(V)\}$ where $1\le m\le n$ and $V$ ranges over all of the standard tableaux of size $n$ (so any shape). Acting on our seminormal basis using (1) shows that
$$F_S v_T = \delta_{ST}v_T.$$
To see this you only need to observe that if $S$ and $T$ are two standard tableaux, not necessarily of the same shape, then $S=T$ if and only if $c_m(S)=C_m(T)$ for
$1\le m\le n$. In turn, this is easily proved by induction on $n$. Letting $\lambda$ range over all partitions of $n$, and using (1) again, you can now prove that $\{F_S\mid S\text{ standard}\}$ is a complete set of pairwise orthogonal primitive idempotents in $\Bbb{Q}S_n$.
To connect with what you are saying let me set $a_\lambda=a_T$ and $b_\lambda=b_T$, where $T=T_\lambda$ is your initial tableau. Also let $T^\lambda$ be the final tableau, that is, the unique standard $\lambda$-tableau which has the numbers $1,2,\dots,n$ entered in order from top to bottom and then left to right down the columns of $\lambda$. (To add to the notational confusion, in my papers, I call these tableaux $t^\lambda$ and $t_\lambda$ respectively.)
The connection with your questions is that $a_\lambda b_\lambda$ is a scalar multiple of $F_{T^\lambda}$ and the multiple is exactly the product of the hook lengths. As far as I am aware, the first place that this appears in the literature is in Murphy's paper [3]. After translation, (I believe!) this follows from the displayed formula at the bottom of page 512. In Prop. 4.4 of my paper [2] you will find a statement more in keeping with the notation above, except that this paper considers the cyclotomic Hecke algebra case (so more notational complications).
Once you have this fact, then your conjecture 6 follows and conjecture 5 should follow from the definition of $F_s$. In the mathematical-physics literature, I think that your conjecture 5 is often referred to as a fusion system -- actually, this is not quite the same thing, but it is very similar.
Unfortunately, translating all of the notation back to your setting is slightly painful. Perhaps there are some (old?) papers in the representation theory of the symmetric groups that deal with seminormal forms which just give the result you are interested in, but I am not aware of any. I suspect that it might be easier to just re-prove these things in your setting using the seminormal idempotents $\{F_T\}$.
References
- A. Okounkov and A. Vershik, A new approach to representation theory of symmetric groups, Selecta Math. (N.S.), 2 (1996), 581–605.
- Mathas, Andrew Matrix units and generic degrees for the Ariki-Koike algebras. J. Algebra 281 (2004), no. 2, 695–730. arXiv:0108164
- G.E. Murphy, On the representation theory of the symmetric groups and associated Hecke algebras, J. Algebra 152 (1992) 492–513.
Best Answer
This can be shown using the following two facts:
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).