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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.

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  • $\begingroup$ $21=1+1+1+9+9$.. $\endgroup$
    – user87543
    Commented Feb 17, 2015 at 16:33
  • $\begingroup$ but $21=1+1+9+4+4+1+1$? $\endgroup$
    – Trajan
    Commented Feb 17, 2015 at 16:36
  • $\begingroup$ why does that matter?? do you know how many irreducible representations you should have if you know no of conjugacy classes?? $\endgroup$
    – user87543
    Commented Feb 17, 2015 at 16:36
  • $\begingroup$ possibly havent got there yet in my course $\endgroup$
    – Trajan
    Commented Feb 17, 2015 at 16:37
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    $\begingroup$ there is some typo.. you have written 1 dimensional representations 4 times.. it should be only 3 $\endgroup$
    – user87543
    Commented Feb 17, 2015 at 16:45

1 Answer 1

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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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