# Why do partitions correspond to irreps in $S_n$?

As stated for example in these notes (Link to pdf), top of page 8, irreps of the symmetric group $$S_n$$ correspond to partitions of $$n$$. This is justified with the following statement:

Irreps of $$S_n$$ correspond to partitions of $$n$$. We've seen that conjugacy classes of $$S_n$$ are defined by cycle type, and cycle types correspond to partitions. Therefore partitions correspond to conjugacy classes, which correspond to irreps.

I understand the equivalence between partitions, cycle types, and conjugacy classes, but I do not fully get the connection with irreps:

1. I can associate to a partition $$\lambda\vdash n$$ the conjugacy class of permutations of the form $$\pi=(a_1,...,a_{\lambda_1})(b_1,...,b_{\lambda_2})\cdots (c_1,...,c_{\lambda_k}).$$

2. The fact that conjugacy classes are defined by cycle types comes from the fact that $$\sigma\pi\sigma^{-1}$$ has the same cycle type structure as $$\pi$$.

However, in what sense do conjugacy classes correspond to irreps? I can understand this if we restrict to one-dimensional representations, as then $$\rho(\pi)=\rho(\sigma\pi\sigma^{-1})$$ for all $$\sigma$$, but this is not the case for higher dimensional representations I think, being $$S_n$$ non-abelian.

• It is in fact a somewhat non-trivial fact that a finite group has the same number of irreducible complex representations and conjugacy classes. But it is covered in pretty much any introductory treatment of the topic. Jul 1 '20 at 17:38

## 1 Answer

This is a standard result in representation theory. The number of (isomorphism types of) irreducible complex representations of a finite group is the same as the number of conjugacy classes of that group. In short, this follows from the fact that the irreducible characters (and, in this setup, two representations are isomorphic iff they have the same character) form a basis of the space of class functions (complex-valued functions on the group that are constant on conjugacy classes). Clearly, the dimension of the space of class functions is the number of conjugacy classes. The space of class functions comes with a natural inner product and the linear independence of the irreducible characters follows from the Schur orthogonality relations. That they generate the space follows from examining the orthogonal complement of their span. Setting up the necessary preliminaries and giving these arguments in full amounts to the first couple chapters in a standard text on representation theory, so I shall not do as such.

In general, there is no natural bijection between conjugacy classes and irreducible representations. A nice, albeit a bit more advanced, discussion of this can be found here on MO. In the particular case of $$S_n$$, however, there is a natural choice of bijection between irreducible representations and conjugacy classes. Establishing this bijection will be part of the notes you are reading.

• This being said, it turns out that for $S_n$ the bijection is canonical, and is given by the whole theory of Young tableaux and such. Jul 1 '20 at 17:41
• Thanks for the remark. I've edited to contextualize this. Jul 1 '20 at 17:52