# Let $G$ be a finite group of order $20$, and consider the representation of $G$ over $\Bbb C$

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.

• The dot product is Hermitean? Nov 27, 2021 at 6:25
• 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 $20$ :-) Nov 27, 2021 at 7:02
• Anyway, why is $\chi_1\chi_2$ the character of an irreducible representation? How does that help? Nov 27, 2021 at 7:02
• No. In a Hermitean dot product you use complex conjugates of coordinates of one of the vectors otherwise $<u,u>$ may not be real. Nov 27, 2021 at 7:05
• Mind you, there is a complex conjugation in the inner product so $$\langle\chi_4,\chi_4\rangle=1+4|a_1|^2+5|a_2|^2+5|a_3|^2+5|a_4|^2.$$ Nov 27, 2021 at 7:05

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 $$b_2$$.

• 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 ?
– Pont
Nov 27, 2021 at 12:59
• 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 $\mathbb{C}^*$. Nov 27, 2021 at 13:04
• Could you tell me why ' product of linear character is linear character' induces $χ_4＝χ_1χ_2$?
– Pont
Nov 27, 2021 at 13:26
• @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 $\chi_3$) Nov 27, 2021 at 13:29
• Could you tell me why χ1χ5 is 4 dimmensional ?
– Pont
Nov 30, 2021 at 16:46