Orbits of pairs of multi-indices under the diagonal action of the symmetric group This question concerns a statement made on page 168 of
Dipper, R. and Donkin, S., 1991. Quantum GLn. Proceedings of the London Mathematical Society, 3(1), pp.165-211.
I have tried to include all necessary definitions, terminology, and notation below. If I have missed some definition or something isn't clear I'm happy to clarify.

Given any pair of multi-indices $(i,j) \in I^{2}(n,r)$, let $(i',j') \in I^{2}(n,r)^{+}$ be its corresponding initial double index. Then there exists a unique element $\pi \in \mathfrak{S}_r$ of minimal length such that $(i,j) = (i',j') \cdot \pi$. Let $\alpha = \mathsf{c}(i)$, $\beta = \mathsf{c}(j)$ be the contents of $i$, $j$ respectively.
Then the statement is that there exists $u \in \mathfrak{S}_{\alpha} \cap \mathscr{D}_{\beta}$ and $v \in \mathscr{D}_{\alpha}$ such that $\pi = uv$.

I'm having trouble seeing this result. I suspect that this is clear if you look at it in the correct way but I can't quite seem to get the correct perspective. Could anyone perhaps help elucidate this result? 

Background: 
For two positive integers $n,r$, denote by $I(n,r)$ the set of functions $\{1,\dots,r\} \to \{1,\dots,n\}$, which we can think of as multi-indices. This set is ordered lexicographically. The set $I^{2}(n,r) = I(n,r) \times I(n,r)$ is ordered with the induced order from each $I(n,r)$, that is to say that $(i,j) <(k,\ell)$ if either $i < k$, or $i = k$ and $j < \ell$. 
For a multi-index $i \in I(n,r)$, denote by $\mathsf{c}(i)$ its content, that is to say the vector in $\mathbb{N}^{n}$ whose $j^{\text{th}}$ entry is the number of occurrences of $j$ in $i$. The set of possible contents is the set of compositions of $r$ into at most $n$ non-zero parts (that is the collection of elements of $\mathbb{N}^{n}$ (where $\mathbb{N}$ includes zero) whose weight, i.e the sum of entries, is $r$). We shall denote this set by $C(n,r)$.
Then the symmetric group $\mathfrak{S}_r$ acts on $I(n,r)$ from the right by place permutation, and the content of the elements of $I(n,r)$ is an invariant of this action and thus parametrises the orbits. We thus identify elements of $C(n,r)$ with its corresponding orbit in $I(n,r)$. Then given $\alpha \in C(n,r)$ there is a unique lexicographically minimal multi-index in $\alpha$, which we shall denote by $i_{\alpha}$ and is called the initial index corresponding to $\alpha$ . We shall then denote by $\mathfrak{S}_{\alpha}$ the $\mathfrak{S}_r$-stabiliser of $i_{\alpha}$. These subgroups $\mathfrak{S}_{\alpha}$ are Young's subgroups. 
Now each right coset $\mathfrak{S}_{\alpha} \cdot \sigma$ of right coset space $\mathfrak{S}_{\alpha} \backslash \mathfrak{S}_r$ contains a unique element of minimal length, where here by length we mean the traditional length of a permutation (i.e its number of inversions). We shall call this element the distinguished coset representative of $\sigma$ with respect to $\alpha$, and we shall denote by $\mathscr{D}_{\alpha}$ the collection of distinguished coset representatives with respect to $\alpha$. $\mathscr{D}_{\alpha}$ is, by definition, a transversal for the right coset space $\mathfrak{S}_{\alpha} \backslash \mathfrak{S}_r$. In particular for every $i \in I(n,r)$ with content $\alpha$ say, there exists a unique $\sigma \in \mathscr{D}_{\alpha}$ such that $i = i_{\alpha} \cdot \sigma$.
Now $\mathfrak{S}_r$ acts on $I^{2}(n,r)$ diagonally, and for each pair $(i,j) \in I^{2}(n,r)$ there exists a unique lexicographically minimal pair $(i',j') \in I^{2}(n,r)$ in the $\mathfrak{S}_r$-orbit of $(i,j)$ in $I^{2}(n,r)$. We refer to $(i',j')$ as the initial double index corresponding to $(i,j)$, and we shall denote by $I^{2}(n,r)^+$ the collection of initial double indices in $I^{2}(n,r)$ which is definitionally a transversal for the $\mathfrak{S}_r$-orbits in $I^{2}(n,r)$. Notice that by definition of the ordering on $I^{2}(n,r)$ we always have $i' = i_{\alpha}$ where $\alpha = \mathsf{c}(i)$, but the same may not be true for $j'$.

I should say that in the referenced material the letter $G$ is used in place of $\mathfrak{S}$, and $\lambda$, $\mu$ in place of $\alpha, \beta$ respectively, and $k$ in place of $j$.
 A: Edit: As discussed in the comments below, while this answer provides a useful perspective on the problem, it doesn’t actually prove the statement, which turns out to be false.

This is perhaps best viewed in terms of stable sorting. Among multi-indices with the same content, the lexicographically minimal one is the one sorted in ascending order. The permutation that performs this sorting with the least number of inversions is the one that results from a stable sort, that is, which doesn’t change the order of elements with the same value. A pair of multi-indices $((i_1,\ldots,i_r),(j_1,\ldots,j_r))$ can instead be viewed as a multi-index of pairs $((i_1,j_1),\ldots,(i_r,j_r))$. The lexicographically minimal one with the same content is obtained by sorting these pairs in ascending lexicographical order, and the permutation $\pi$ that achieves this with the least number of inversions is the one resulting from a stable sort of the pairs, that is, one that doesn’t change the order of pairs with the same values. I think in this perspective the statement becomes more transparent: This stable sort of the pairs is achieved by first stably sorting the first elements and then stably sorting each group of pairs with the same first element; the first part is $v$ and the second part is $u$.
