Let $V_\lambda$ be the irreducible $S_n$-representation (a left $kS_n$-module) over a field $k$ of characteristic $0$ associated to the partition $\lambda\vdash n$. By abuse of notation let $S_a$ and $S_{n-a}$ be the subgroups of $S_n$ permuting $\{1,\dots,a\}$ and $\{a+1,\dots,n\}$, respectively. Let $W= k\otimes_{kS_{n-a}}V_\lambda$, which is just the direct sum of the trivial constituents of $V_\lambda$ viewed as an $S_{n-a}$-representation. One can easily prove that $W$ is then an $S_a$-representation via $\sigma \cdot (1\otimes v) = 1\otimes \sigma v$ because $S_a$ and $S_{n-a}$ commute.
My question is: Can you prove that $W$ is the direct sum of all irreducible $S_a$-representations $V_\mu$ associated to $\mu\vdash a$ that are obtained from $\lambda$ by removing $n-a$ boxes, but at most one from each column?
The context of this question is http://arxiv.org/abs/1204.4533v2, where in the proof of Lemma 2.40 this argument is used.
We will use Pieri's formula which is a special case of the Littlewood-Richardson-rule. It states:
Let us introduce the notation $\alpha\prec \lambda$ if $\lambda$ is obtained from $\alpha$ by adding boxes but at most one per column. Then we can reformulate the statement to saying $$ \langle V_{\lambda},\mathrm{Ind}_{S_a\times S_b}^{S_n} V_\alpha \otimes V_{(b)}\rangle = \begin{cases}1 &\text{if $ \alpha \prec \lambda$}\\ 0&\text{otherwise.}\end{cases} $$
Considering $W$ as an $S_a\times S_b$-representation, we know from the assumptions we have $$ \langle W, V_\alpha \otimes V_\beta \rangle = \begin{cases} \langle \mathrm{Res}_{S_a\times S_b}^{S_n} V_\lambda, V_\alpha \otimes V_\beta \rangle &\text{if $\beta= (b)$}\\0&\text{otherwise.} \end{cases}$$ Now Frobenius' Reciprocity formula and $$ \langle \mathrm{Res}_{S_a}^{S_a\times S_b} W, V_\alpha \rangle = \langle W, V_\alpha \otimes V_{(b)}\rangle $$ proves the assertion.