# The Unapologetic Mathematician

## Mathematics for the interested outsider

We’ve been talking a lot about the general theory of finite group representations. But our goal is to talk about symmetric groups in particular. Now, we’ve seen that the character table of a finite group is square, meaning there are as many irreducible representations of a group $G$ as there are conjugacy classes $K\subseteq G$. But we’ve also noted that there’s no reason to believe that these have any sort of natural correspondence.

But for the symmetric group $S_n$, there’s something we can say. We know that conjugacy classes in symmetric groups correspond to cycle types. Cycles correspond to integer partitions of $n$. And from a partition we will build a representation.

For a first step, let $n=\lambda_1+\lambda_2+\dots+\lambda_k$ be a partition, with $\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k$. We can use this to come up with a subgroup $S_\lambda\subseteq S_n$. Given a set $X$ we will write $S_X$ for the group of permutations of that set. For example $S_{\{1,\dots,\lambda_1\}}$ permutes the first $\lambda_1$ positive integers, and $S_{\{\lambda_1+1,\dots,\lambda_2\}}$ permutes the next $\lambda_2$ of them. We can put a bunch of these groups together to build

$\displaystyle S_\lambda=S_{\{1,\dots,\lambda_1\}}\times S_{\{\lambda_1+1,\dots,\lambda_2\}}\times\dots\times S_{\{\lambda_{k-1}+1,\dots,\lambda_k\}}$

Elements of $S_\lambda$ permute the same set as $S_n$, and so $S_\lambda\subseteq S_n$, but only in certain discrete chunks. Numbers in each block can be shuffled arbitrarily among each other, but the different blocks are never mixed. Really, all that matters is that the chunks have sizes $\lambda_1$ through $\lambda_k$, but choosing them like this is a nicely concrete way to do it.

So, now we can define the $S_n$-module $M^\lambda=1\!\!\uparrow_{S_\lambda}^{S_n}$ by inducing the trivial representation from the subgroup $S_\lambda$ to all of $S_n$. Now, the $M^\lambda$ are not all irreducible, but we will see how to identify a particular irreducible submodule $S^\lambda\subseteq M^\lambda$ of each one, and the $S^\lambda$ will all be distinct. Since they correspond to partitions $\lambda$, there are exactly as many of them as there are conjugacy classes in $S_n$, and so they must be all the irreducible $S_n$-modules, up to isomorphism.

December 7, 2010 -

## 1 Comment »

1. […] It doesn’t really matter which we pick; any two tableaux in the same orbit — and they’re all in the same single orbit — have isomorphic stabilizers. But like we mentioned last time the usual choice lists the numbers from to on the first row, from to on the second row, and so on. We write for the stabilizer of this choice, and this is the subgroup of we will use. Notice that this is exactly the same subgroup we described earlier. […]

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