I am going through the lecture note How to get character tables of symmetric groups.

On page 2, it computes the character table of $S_4$. The procedure starts with building the table of the characters of permutation representations for $S_4$. The table is given below.

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The note says the following about the table.

The top row lists the different conjugacy classes by way of listing partitions; the left-hand edge lists the different representations we get for different shapes of tabloid. The top line is the trivial representation, while the bottom one is the regular representation.

I am trying to understand what is going on here. It is clear that the columns are the corresponding partitions, hence, the conjugacy classes. $\sigma_\lambda$ for a given shaped $\lambda$ is defined as follows.

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So, for example, how $\sigma_{3, 1}$ gives $2$ for the partition $2, 1, 1$?

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    $\begingroup$ Think about a specific element, say $(1,2)$. How does it act in the four $3,1$ partitions? How many does it fix? $\endgroup$ – Steve D May 21 '16 at 0:54
  • $\begingroup$ @SteveD, the four 3,1 partitions are {[2, 1, 1], [2, 2, 0], [3, 1, 0], [4, 0, 0]}, Right? $\endgroup$ – Omar Shehab May 21 '16 at 15:19
  • $\begingroup$ maybe we are talking past each other, but I think the shape $3,1$ corresponds to a partition of $\{1,2,3,4\}$ into a set of size $3$ and a set of size $1$. $\endgroup$ – Steve D May 21 '16 at 22:13
  • $\begingroup$ @SteveD, so the partitions are $\left\{\{1, 2, 3\}, \{4\}\right\}$, $\left\{\{1, 2, 4\}, \{3\}\right\}$, $\left\{\{1, 3, 2\}, \{4\}\right\}$, $\left\{\{1, 4, 2\}, \{3\}\right\}$, $\left\{\{2, 3, 4\}, \{1\}\right\}$, and $\left\{\{2, 4, 3\}, \{1\}\right\}$. So, there are six $3, 1$ partitions. Right? $\endgroup$ – Omar Shehab May 22 '16 at 7:47
  • $\begingroup$ There are only four... I think you're injecting some kind of order? $\endgroup$ – Steve D May 22 '16 at 7:47

Let's go over how these representations are defined:

We have a certain partition of the set $\{1,2,\ldots,n\}$; this partition is into an ordered collection of unordered sets. Such a partition might look like $[\{1,3\}, \{2,4\}]$; note this is exactly the same partition as $[\{3,1\}, \{2,4\}]$, but not the same as $[\{2,4\}, \{1,3\}]$. The problem with that last one is the first element of the array is not the same. All of these partitions have the same shape: $2,2$ means an array, with a set of size $2$ in the first spot, and another set of size $2$ in the second spot.

For each such shape, we can define a representation (over $\mathbb{C}$), and the character is what you are calling $\sigma_{\text{shape}}$. That character can be determined for each "permutation type" by simply asking how many partitions (of that shape) it fixes.

Here's a concrete example: for the "permutation type" $2,1,1$, and the shape $2,2$, we can choose any permutation in $S_4$ of that type: let's use $(1,2)$. The question is then: how many partitions of shape $2,2$ does $(1,2)$ fix? The answer is two: $[\{1,2\}, \{3,4\}]$ and $[\{3,4\}, \{1,2\}]$. Thus the entry in your table, under column $2,1,1$ and row $\sigma_{2,2}$, is $2$.

  • $\begingroup$ Looks like any permutation of type $2, 1, 1$ is going to fix $2$ partitions of shape $2, 2$. $\endgroup$ – Omar Shehab May 24 '16 at 20:38
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    $\begingroup$ @omarshebab: yes because they all have the same character values! $\endgroup$ – Steve D May 24 '16 at 23:42

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