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Consider symmetric group $S_n$.

Let $\lambda=(\lambda_1,\lambda_2,\cdots,\lambda_k)$ be a partition of $n$ with $\lambda_1\ge\lambda_2\cdots$ (so $\lambda_1+\lambda_2+\cdots + \lambda_k=n$).

Let $\mu=(\mu_1,\mu_2,\cdots,\mu_r)$ be another partition of $n$.

In the representation theory of symmetric group, one important relation on partitions comes is the dominance relation: $$\lambda \trianglerighteq \mu \mbox{ if } (\lambda_1 + \lambda_2 + \lambda_i) \ge (\mu_1 + \mu_2 + \cdots + \mu_i) \mbox{ for all } i\ge 1.$$

Let $\lambda,\mu$ be two partitions of $n$ as above. Then the partition of $\lambda$ determines a conjugacy class of subgroups of $S_n$, isomorphic to $$S_{\lambda_1}\times S_{\lambda_2}\times \cdots \times S_{\lambda_k} $$ where

  • $S_{\lambda_1}$ is symmetric group on first $\lambda_1$ symbols $1,2,\cdots,\lambda_1$,

  • $S_{\lambda_2}$ is the permutation group on next $\lambda_2$ symbols, and so on.

Question. If $\lambda\trianglerighteq \mu$, then is it true that some conjugate of $S_{\mu_1}\times S_{\mu_2}\times \cdots \times S_{\mu_r}$ is contained in some conjugate of $S_{\lambda_1}\times S_{\lambda_2}\times \cdots \times S_{\lambda_k}$? (conjugates taken by may be different permutations in $S_n$ for these two subgroups.)

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    $\begingroup$ No: $(4,2)$ dominates $(3,3)$ but the order of $S_3 \times S_3$ is divisible by $3^2$ while that of $S_4 \times S_2$ is not. $\endgroup$ – Stephen Aug 20 '18 at 18:16
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No: $(4,2)$ dominates $(3,3)$ but the order of $S_3\times S_3$ is divisible by $3^2$ while that of $S_4×S_2$ is not. – Stephen yesterday

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