Let $\Sigma_n$ be the $n$th symmetric group. Then the complex representation ring $R\Sigma_n$ has the augmentation ideal $I_n$, with the $k$th power ideal $I_n^k$. Since $\Sigma_n\times\Sigma_m\leq \Sigma_{n+m}$ we have an induction map

$$ \operatorname{Ind}: R[\Sigma_n\times\Sigma_m]\longrightarrow R\Sigma_{n+m} $$

given by taking $M\in R\Sigma_n$ and $N\in R\Sigma_m$ and inducing up $M\otimes N$. Suppose $M\in I_n^k$ and $N\in I_m^l$. Then the induced module is isomorphic to a direct sum of copies of $M\otimes N$ with an induced action. Therefore the induced module still has virtual dimension 0. My question is whether $\operatorname{Ind}(M\otimes N)\in I_{n+m}^{k+l}$.