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.

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 $b_2$.