Decomposition into irreducible of a representation of the wreath product $S_d \wr S_m$ (2) This is a question following Decomposition into irreducible of a representation of the wreath product $S_d\wr S_n$
Let $F$ be the trivial and $S$ be the standard representations of $S_d$ (of dimension $1$ and $d-1$).
Let:
$$ R_m= \bigl( F^{\widetilde{\otimes n-m}} \boxtimes S^{\widetilde{\otimes m}} \bigr)\bigl\uparrow_{S_d\wr S_{n-m} \times S_d\wr S_{m}}^{S_d\wr S_n} : $$
This is an irreducible representation of $S_d \wr S_n$.
Let $S^{(n-1,1)}$ and $S^{(n-2,1^2)}$ be irreducible representations (Specht modules) of the symmetric group $S_n$. $S_d \wr S_n$ naturally acts on it (just considering $S_n\subset S_d \wr S_n$).
Are $R_m\otimes S^{(n-1,1)}$ and $R_m\otimes S^{(n-2,1^2)}$ irreducibles?
If not, how difficult is this problem, would you have some reference to advice to takle this kind of problem?
 A: Contrary to a comment I've now deleted, the representation is reducible. It's important to be clear about the action of $S_n$: given a representation $\rho: S_n \rightarrow \mathrm{GL}(V)$, composing with the quotient homomorphism $S_d \wr S_n \rightarrow S_n$ gives a representation of $S_d \wr S_n$, denoted $\mathrm{Inf}_{S_n}^{S_d \wr S_n} V$.
Using this, I'll show more generally that if $U$ is any representation of $S_d \wr S_{n-m}$ and $V$ is any representation of $S_d \wr S_m$ where $1 \le m < n$, then
$$ \begin{split} (U \boxtimes V)\bigl\uparrow_{S_d \wr S_{n-m} \times S_d \wr S_m}^{S_d \wr S_n} & \otimes \; \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \\ &= \Bigl( (U \boxtimes V) \otimes \bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)}\bigl\downarrow_{S_d \wr S_{n-m} \times S_d \wr S_m} \bigr)\Bigr)\bigl\uparrow^{S_n} \end{split}$$
is reducible. Since restriction commutes with inflation,
$$\bigl( \mathrm{Inf}_{S_n}^{S_d \wr S_n} S^{(n-1,1)} \bigr) \bigl\downarrow_{S_d\wr S_{n-m} \times S_d \wr S_m}
\, \cong \mathrm{Inf}^{S_d \wr S_{n-m} \times S_d \wr S_m}_{S_{n-m} \times S_m} \bigl( S^{(n-1,1)} \bigl\downarrow_{S_{n-m} \times S_{m}} \bigr). $$
Since $S^{(n-1,1)}\!\!\downarrow_{S_{n-m} \times S_m}\, \cong \bigl( F \boxtimes F \bigr) \oplus \bigl( F \boxtimes S^{(m-1,1)} \bigr) \oplus \bigl( S^{(n-m-1,1)} \boxtimes F\bigr)$ is reducible (this follows from Young's rule using Frobenius reciprocity, or the more general Littlewood–Richardson rule, or could be proved using an explicit basis for $S^{(n-1,1)}$), so is the right-hand side immediately above, and hence so is the original module. A similar argument will work replacing $S^{(n-1,1)}$ with $S^{(n-2,1,1)}$.
