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Let $G$ be a finite group of order $20$, and consider the representation of $G$ over $\Bbb C$. I know $G$ has representative of conjugacy class $e$,$g_1$,$g_2$,$g_3$,$g_4$. Number of conjugacy class of $g_1$,$g_2$,$g_3$,$g_4$ is $1,4,5,5,5$.

And character table is like the following

$$ \begin{array}{|c|c|c|c|c|} \hline & e & g_1 & g_2 & g_3 & g_4 \\ \hline \chi_0 & 1 &1 & 1& 1&1 \\ \hline \chi_1 & 1 & \ 1 & \ i & -1 & -i \\\hline \chi_2 & 1 & 1 & -1 & 1 & -1 \\ \hline \chi_4 & 1&a_2 &a_3 &a_4 &a_5 \\ \hline \chi_5 & 4 & b_2 & b_3 & b_4 & b_5 \\ \hline \end{array} $$

I want to figure out what is $a_i,b_i(2≦i,j≦5)$.

Calcuraing product of characters,

$\left<\chi_1,\chi_4\right>=0$ deduces $1+4a_2+5a_4+5a_5=0$・・・①. In the same way, from $\left<\chi_2,\chi_4\right>=0$ deduces $1+4a_2-5ia_3-5a_4+5ia_5=0・・・②$. From $\left<\chi_3,\chi_4\right>=0$, $1+4a_2-5a_3+5a_4-5a_5=0・・・③$. From $\left<\chi_4,\chi_4\right>=0$, $1+4|a_2|^2+5|a_3|^2+5|a_4|^2+5|a_5|^2=20・・・④$.

I just need to solve simultaneous equation ①②③④. But complicated and I may mistook or going wrong way. Where I mistook the way ? Thank you for your help.

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  • $\begingroup$ The dot product is Hermitean? $\endgroup$
    – markvs
    Commented Nov 27, 2021 at 6:25
  • $\begingroup$ It looks like the group is isomorphic to the semidirect product $C_5\rtimes C_4$ (=the holomorph of $C_5$). Why didn't you say so? You know, there are a few non-isomorphic groups of order $20$ :-) $\endgroup$ Commented Nov 27, 2021 at 7:02
  • $\begingroup$ Anyway, why is $\chi_1\chi_2$ the character of an irreducible representation? How does that help? $\endgroup$ Commented Nov 27, 2021 at 7:02
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    $\begingroup$ No. In a Hermitean dot product you use complex conjugates of coordinates of one of the vectors otherwise $<u,u>$ may not be real. $\endgroup$
    – markvs
    Commented Nov 27, 2021 at 7:05
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    $\begingroup$ Mind you, there is a complex conjugation in the inner product so $$\langle\chi_4,\chi_4\rangle=1+4|a_1|^2+5|a_2|^2+5|a_3|^2+5|a_4|^2.$$ $\endgroup$ Commented Nov 27, 2021 at 7:05

1 Answer 1

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Hint the product of two linear characters is obviously again a linear character, so $\chi_4=\chi_1\chi_2$. This gives you the $a_i$'s. From this point you can use the orthogonality relations to find the $b_i$'s. Or, note that the product of a linear character with an irreducible character is again an irreducible character. So, since $\chi_5$ is the unique character of degree $4$, $\chi_1\chi_5=\chi_2\chi_5=\chi_5$. This immediately gives $b_3=b_4=b_5=0$. I leave it to you to calculate $b_2$.

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  • $\begingroup$ 1.I'm beginner of representation theory, sorry to ask elementary thing, but could you tell me the definition of product of character (or references) ? 2. Is it impossible to solve just $4+4b_2+5b_3+5b_4+5b_5=0$ and $4+4b_2−5ib_3-5b_4+5ib_5=0$ and $4+4b_2-5b_3+5b_4-5b_5=0$ and $16+4|b_2|^2+5|b_3|^2+5|b_4|^2+5|b_5|^2=20$. The answer seems to be $b_2=-1, b_2=b_4=b_5=0$ and from orthogonality relation, $a_2=1, a_4=-1, a_5=1, a_3=-1$. Is it impossible to solve the simultaneously equation directly ? $\endgroup$ Commented Nov 27, 2021 at 12:59
  • $\begingroup$ If $\lambda$ and $\mu$ are linear representations, they are just homomorphisms from $G$ into $\mathbb{C}^*$. Their pointwise product $\lambda \mu$ is again a homomorphism of $G$ into $\mathbb{C}^*$. $\endgroup$ Commented Nov 27, 2021 at 13:04
  • $\begingroup$ Could you tell me why ' product of linear character is linear character' induces $χ_4=χ_1χ_2$? $\endgroup$ Commented Nov 27, 2021 at 13:26
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    $\begingroup$ @Catsoup He just did. It completely straightforward from the definitions that this is the case (and since the product $\chi_1\chi_2$ must be one of the linear characters and it is clearly not $\chi_0$, $\chi_1$ or $\chi_2$, it must be $\chi_4$ (not sure why there is no $\chi_3$) $\endgroup$ Commented Nov 27, 2021 at 13:29
  • $\begingroup$ Could you tell me why χ1χ5 is 4 dimmensional ? $\endgroup$ Commented Nov 30, 2021 at 16:46

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