Difference between revisions of "Aufgaben:Problem 7"

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Exercise

Compute the character table of \(S_4\)

Solution

Let \(G\) be a finite group

Let \(g \in G\)

Let \(\psi \text{ and } \phi \in \mathbb{C}^G\)

There is an inner product defined as \( (\psi , \phi)_G = \frac{1}{\left|G\right|} \sum_{y \in G} \psi (y) \phi (y)^{*} \)

""We know that \( \psi_r , \psi_s \) are orthonormal if \( (\psi_r , \psi_s)_G = \delta_{rs} \)""

Step 0:

We know that \( S_4 \) is the symmetric group, that means the group of all permutations of a set with 4 elements, and has \( n! = 4! = 24 \) elements which belong to 5 conjugacy classes which are represented by "e = identity", "(12) = permutation of two elements = \(C_2\)", "(123) = permutation of 3 elements = \(C_3\)", "(12)(34) = permutation of 2 times 2 elements = \(C_{2,2}\)" and "(1234) = permutation of 4 elements = \(C_4\)".

We can also count or calculate the number of elements in each conjugacy class and get the following: \( c[e]=1, c[C_2]=6, c[C_3]=8, c[C_4]=6, c[C_{2,2}]=3 \)

The rows of the character table are the irreducible representations of the group and the columns are the conjugacy classes. Because the number of the irreducible representations is equal to the number of conjugacy classes, the table has to be quadratic. Now we can draw it like that:

Step 1: \( \chi^{triv} \) is the trivial map from \( S_4 \to S^1 \text{ where } \chi^{triv} \equiv 1\) Therefore the table now looks like this:

Step 2: We let act the the conjugacy class on \( (1,2,3,4) \in S_4\) and calculate the signum (for example by counting the inversion number). We use this as the second irreducible representation.

Charactertable looks like:

References

https://unapologetic.wordpress.com/2010/11/08/the-character-table-of-s4/

https://www.itp.uni-hannover.de/~flohr/lectures/symm/handout2.pdf