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I am trying to understand the character table of $S_4$. I have obtained the trivial, signature and standard representations. The fourth one is the product of signature and standard.

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Now for the last representation, I have tried two things.

  1. I took the standard basis for $C$, defined $\pi(\sigma)e_i = e_{\sigma i}$. I took the trace of the matrices as characters, but there it turned out that $\pi$ is just direct sum of trivial and standard representations.

  2. I have read that I should take the normal subgroup corresponding to (12)(34). I don't know how to go about it. Can somebody explain? Any other method to determine the last representation is also welcome.

Thanks in advance.

user519535
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    Hint: There exists an epimorphism from $S_4$ to $S_3$ allowing you to pull back irreducible reps of $S_3$ to $S_4$. – Jyrki Lahtonen Feb 21 '24 at 20:52
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    Also, when only one irreducible is missing from a character table, you can use orthogonality relations to figure it out. – Jyrki Lahtonen Feb 21 '24 at 20:55
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    If all you want is the last row of the character table, you can conclude the $e$ column has to be 2 since the sum of squares has to reach $|S_4| = 24$; and then the representation of $S_4$ acting on $\bigoplus_{\sigma \in S_4} \mathbb{C}$ (with total character 24,0,0,0,0) has to decompose into 1 copy each of the first and second, 3 copies each of the third and fourth, and 2 copies of the fifth. – Daniel Schepler Feb 21 '24 at 20:58

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