# 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.

• $21=1+1+1+9+9$.. – user87543 Feb 17 '15 at 16:33
• but $21=1+1+9+4+4+1+1$? – Permian Feb 17 '15 at 16:36
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.