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My question comes from a section of Representations and Characters of Groups, by James & Liebeck. They discuss the following example in the chapter $15$ on The Number of Irreducible Characters

$15.7$ Example
We shall see in Section $18.4$ that there is a certain group $G$ of order $12$ which has exactly six conjugacy classes with representatives $g_1, \ldots, g_6$ (where $g_1=1$), and six irreducible characters $\chi_1, \ldots, \chi_6$ given as follows:

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Suppose we are given characters $\chi$ and $\psi$ of $G$ as follows:

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Then it is easy to spot that $$\chi = \chi_2+\chi_6, \quad \psi=\chi_1+\chi_2+\chi_3+\chi_4.$$ For example, the second entry in the row vector for $\chi$ is equal to minus the first entry. Inspecting the values of the irreducible characters $\chi_i$, we see that $\chi$ must be a combination of $\chi_2, \chi_4$ and $\chi_6$. The correct answer now comes quickly to mind.

I don't see what's so "easy to spot" and how the answer "comes quickly to mind." I understand the answer that they got, but I must be missing some trick that makes this example so "easy."

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  • $\begingroup$ Just trial and error? There just aren't that many possibilities to try. $\endgroup$ Commented Mar 14, 2023 at 2:20
  • $\begingroup$ Okay, I was wondering if there were some trick I was missing. Thanks anyways. If you post your comment as an answer I'll be happy to accept it. $\endgroup$ Commented Mar 14, 2023 at 2:22

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This is definitely an overstatement; it's not immediately obvious that $\chi$ must be a combination of $\chi_2$, $\chi_4$, and $\chi_6$. Certainly any linear combination of these three irreducible characters would satisfy the desired condition that the entries in the first and second columns are negatives of each other, but so would $\chi_1 - \chi_3$, for example!

Edit Eric Wofsey makes a good point below: in order for $\chi$ to actually be a character, it must be a linear combination of $\chi_1, \dots, \chi_6$ with natural-number coefficients, which does force $\chi$ to be a combination of $\chi_2, \chi_4, \chi_6$.

As Eric says in the comments, this approach is just trial and error (and making lucky guesses whenever possible).

A different approach: write $\chi = \alpha_1 \chi_1 + \dots + \alpha_6 \chi_6$, and then use the values in your tables to write a system of $6$ linear equations that you can solve to find the values of the $\alpha_i$'s.

Or, if you've learned the orthogonality relations, you can use inner products to speed things up dramatically! If you haven't seen this yet, you will soon :)

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    $\begingroup$ The trial and error required is greatly reduced if you know that this is actually a character of a representation so the coefficients must be natural numbers. (This restriction also makes it actually true that you need to use only $\chi_2,\chi_4,$ and $\chi_6$.) $\endgroup$ Commented Mar 14, 2023 at 2:39
  • $\begingroup$ Good point, thank you $\endgroup$ Commented Mar 14, 2023 at 2:44

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