<s>[INST] Complete the character table of group of order $21$ You are given the incomplete character table of a group $G$ with order $21$ which has $5$ conjugacy classes, $C_1,\dots,C_5$, which have sizes $1,7,7,3,3$.
$$ \begin{array}{|c|c|c|c|c|}
\hline
& C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline
 &  & & & & \\ \hline
 &  &  && &  \\ \hline
\chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline
 &  &  & & &  \\ \hline
\chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline 
\end{array}
$$
Complete the character table.
Im guessing that $\chi_0$ has to be the trivial representation so we get that
$$ \begin{array}{|c|c|c|c|c|}
\hline
& C_1 & C_2 & C_3 & C_4 & C_5 \\ \hline
 \chi_0 & 1 &1 & 1& 1&1 \\ \hline
 & 1 &  && &  \\ \hline
\chi_2 & 1 & \zeta_3 & \zeta_3^2 & 1 & 1 \\ \hline
 & 3 &  & & &  \\ \hline
\chi_4 & 3 & 0 & 0 & \zeta_7+\zeta_7^2+\zeta_7^4 & \zeta_7^{-1}+\zeta_7^{-2}+\zeta_7^{-4} \\ \hline 
\end{array}
$$
and as $21=1+1+1+9+9$.
Im sure you have to obtain something from the fact we have 3rd and 7th roots of unity which correspond to the sizes of the conjugacy classes but I cannot see what I am meant to glimmer from this.
Hints only please. 
 [/INST] The remaining two characters are algebraic conjugates of $\chi_2$ and $\chi_4$. You get them by replacing $\zeta_3$ by $\zeta_3^2$, and $\zeta_7$ by $\zeta_7^3$, respectively.
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