# Representations of symmetric groups of order $2n$ and $n$

Background: Denote by $S_n$ the symmetric group of order $n$. There are many ways to embed $S_n$ as a subgroup into $S_{2n}$. Given a symmetric group, we can use Young diagrams to classify all irreducible representations of it. A standard reference is Fulton & Harris's book.

Question : Given a irreducible representation of $S_{2n}$, which corresponds to certain young diagram, when we view this irreducible representation as a representation of $S_n$, can we decompose it into the direct sum of irr. reps. of $S_n$? What is the corresponding Young diagrams of $S_n$? Until now, I have no good ideas about this question. So I'm looking forward to your answer. Thanks a lot!

To my knowledge, the "usual" way to embed $S_{n-1}$ into $S_n$ is to let $S_{n-1}$ act on $\{1,\ldots,n-1\}$ and fix $n.$ We can iterate this construction to get an embedding of $S_n$ into $S_{2n}.$ These are the embeddings we use in the following.
In the semisimple case (i.e. the characteristic of the coefficient field is greater than $n$), we have the "branching rule" to decompose an irreducible representation of $S_n$ into irreducible representations of $S_{n-1}.$ Namely, for $\lambda \vdash n,$ the corresponding irreducible representation of $S_n$ is the Specht module $S^\lambda.$ The restriction to $S_{n-1}$ is given by $$S^\lambda\downarrow_{S_{n-1}} = \bigoplus_{\mu \in M}S^\mu,$$ where $M$ is the set of all $\mu \vdash n-1$ obtained from $\lambda$ by removing one square.
We can iterate this construction to decompose an irreducible representation of $S_{2n}$ into irreducible representations of $S_n.$ However, I fear that the calculations get a bit "bulky".