Let’s try to calculate the characters of the Young tabloid modules we’ve constructed. Since these come from actions of on various sets, we have our usual shortcut to calculate their characters: count fixed points.
So, let’s write for the character of the representation corresponding to the partition . For a permutation , the character value is the number of Young tabloids such that . This might be a little difficult to count on its face, but let’s analyze it a little more closely.
First of all, pick a canonical Young tableau . The easiest one just lists the numbers from to in order from left to right on rows from top to bottom of the tableau, like
but it really doesn’t matter which one we choose. The important thing is that any other tableau has the form for some unique . Now our fixed-point condition reads , or . But as runs over , the conjugate runs over the conjugacy class of . What’s more, it runs evenly over the conjugacy class — exactly values of give each element in . So what we need to count is how many elements give a tableau that is row-equivalent to . We multiply this by to get , right?
Well, no, because now we’ve overcounted. We’ve counted the number of tableaux with . But we want the number of tabloids with this property. For example, let’s try to count : there’s only one element in , and it leaves fixed. Our rule above would have us multiply this by to get , but there are not always tabloids of shape !
The story is evidently more complicated than we might have hoped. Instead of letting above range over all of , we could try letting it only range over a transversal for the subgroup of that preserves the rows of . But then there’s no obvious reason to assume that the conjugates of should be evenly distributed over , which complicates our counting. We’ll have to come back to this later.