How Do Characters of Group Actions Define Values on Conjugacy Classes?

[Kreizhn]

Homework Statement

Consider the symmetric group on n-letters denoted [itex] S_n [/itex]. For [itex] \lambda [/itex] a positive integer partition of n (under the usual cycle type notation) define [itex] \sigma_\lambda[/itex] to be the character of the permutation representation of [itex] S_n [/itex] acting on the set of all ways to divide n into sets of size [itex] \lambda[/itex].

Show that these characters have the following values on conjugacy classes of [itex] S_4 [/itex]

[tex] \begin{array}{l|rrrrr}
& [1 1 1 1] & [2 1 1] & [2 2] & [3 1] & [4] \\ \hline
\sigma_4 & 1 & 1 & 1 & 1 & 1 \\
\sigma_{3,1} & 4 & 2 & 0 & 1 & 0 \\
\sigma_{2,2} & 6 & 2 & 2 & 0 & 0 \\
\sigma_{2,1,1} & 12 & 2 & 0 & 0 & 0 \\
\sigma_{1,1,1,1} & 24 & 0 & 0 & 0 & 0
\end{array}
[/tex]

The Attempt at a Solution

Even just trying to figure out what "the character of the permutation representation of [itex] S_n [/itex] acting on the set of all ways to divide n into sets of size [itex] \lambda[/itex]" means is hurting my head, though I think I might get it. In particular, let [itex] \mu [/itex] be another partition of n and take [itex] S_\mu [/itex] to be the set of all [itex] \mu [/itex] partitions of [itex] \{ 1, 2, \cdots, n \} [/itex]. Then [itex] (\lambda,\mu) [/itex] element of the table is the number of fixed points of [itex] S_\mu [/itex] under permutations of cycle type [itex] \lambda [/itex].

First Question: Have I interpreted this correctly? Some scratch calculations imply the numbers in the table may actually be the number of orbits (for any fixed element) of the action.

Second Question: Is this well defined? Namely, by choosing an arbitrary [itex] \lambda [/itex] cycle, are we always ensured the number of fixed points of [itex] S_\mu [/itex] are the same? I think this is true (and must be for the table to make sense), and works because [itex] S_\mu [/itex] contains ALL the possible partitions of n by [itex] \mu[/itex].

Third Question: How does one calculate these table values in a reasonable way. I can kind of see it, but it's not precise in my head.

 

[kurt.physics]

Thanks in advance for any help.A:The character is well defined, because if you have two permutations of the same cycle type, they conjugate each other, and so the number of fixed points of $S_\mu$ under them must be the same.To calculate the values in the table, you can use the Orbit-Stabilizer theorem. The idea is that you consider the action of $S_4$ on $S_\mu$, and look at the orbit of a single element of $S_\mu$ under this action. If you take a permutation $\sigma \in S_4$ of cycle type $\lambda$, then the number of elements of the orbit of this single element of $S_\mu$ is $\frac{|S_4|}{|\text{Stab}(\mu)|}$, where $\text{Stab}(\mu)$ is the stabilizer of that element of $S_\mu$. Since the size of the orbit does not depend on the particular element of $S_\mu$ chosen, the same formula holds for all elements of $S_\mu$.The size of the stabilizer of an element of $S_\mu$ can be calculated using the fact that it is the product of the stabilizers of each point under the action of the symmetric group. We can calculate the stabilizer of a point $p$ by considering how it is permuted by elements of $S_4$. If $p$ is in a part of size $k$, then the stabilizer of $p$ will be $S_{k-1}$ (it is the set of permutations that fix $p$). So if the partition is $\mu = (\mu_1, \mu_2, \cdots, \mu_r)$, then $$|\text{Stab}(\mu)| = |S_{\mu_1 - 1}||S_{\mu_2 - 1}| \cdots |S_{\mu_r - 1}|.$$You can then use the result of the Orbit-Stabilizer theorem to calculate the size of the orbit of an element of $S_\mu$, and so the character value.