I have the following homework question:

A group of order 18 has the following partial character table, where $y=-\frac{1}{2} + xi$:

\begin{array}{c | c c c c c} \hline\hline & g_1 & g_2 & g_3 & g_4 & \dots \\ \hline \chi_1 & 1 & 1 & 1 & 1 & \dots \\ \chi_2 & 1 & y & 1 & 1 & \dots \\ \chi_3 & 1 & 1 & 1 & -1 & \dots \\ \chi_4 & 2 & 2 & -1 & 0 & \dots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \\ \hline \end{array}

We're first asked to calculate what y is. I know it must be an m-th root of 1, so $x = \pm \frac{\sqrt{3}}{2}$, but I can't see how to discern which it is. We're then asked to calculate five new irreducible characters to form a complete set of irreducible characters, which I have done ($\chi_5 = \chi_2 \chi_3, \; \chi_6 = \chi_2 \chi_2, \; \chi_7 = \chi_2 \chi_5, \; \chi_8 = \chi_2 \chi_4, \; \chi_9 = \chi_6 \chi_4$) and so there are 9 conjugacy classes in $G$. Finally, we're asked to complete the character table, given that the character values for $\chi_4$ are real. I've filled out the first four columns of the next five rows. I've then argued that since y is complex, $g_2 ^{-1} \not\in g_2 ^G$, and so $g_2 ^{-1} \in g_5 ^G$, say, with $\chi_i(g_5) = \overline{\chi_i(g_2)}$. I think I could now proceed using column relations, but I feel that I can probably glean more information before having to brute force it

Any hints/suggestions please?

  • $\begingroup$ It appears to be the [0.5 ex] and [1 ex] that was preventing the table from working. I also changed tabular to array, because I know that works. I don't know what they were supposed to do, but it renders now. Please take a look and see if you like it. $\endgroup$ Apr 21, 2013 at 20:38
  • 2
    $\begingroup$ Re-check your computation of $x$. Your $x=\pm\sqrt2$ does not make $y=-\frac12+xi$ a root of unity. $\endgroup$ Apr 21, 2013 at 20:51
  • $\begingroup$ Sorry, I meant $\pm\frac{\sqrt{3}}{2}$. That's what I'd written down during my working anyway! And thanks a lot Ross! $\endgroup$
    – lokodiz
    Apr 21, 2013 at 21:00
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    $\begingroup$ The character table is invariant under conjugation, so you're not going to be able to determine the sign of $x = \pm\frac{\sqrt{3}}{2}$ from the given information (since all the entries you're given are real). $\endgroup$
    – Ted
    Apr 21, 2013 at 21:07
  • $\begingroup$ Is anyone able to give any hints then please? I managed to make no more progress after another attempt. $\endgroup$
    – lokodiz
    Apr 22, 2013 at 10:16

1 Answer 1


You have deduced that all characters are of the form $\chi_2^i \chi_j$ for $j \in \{1,3,4\}$. One similarly has an expression for the conjugacy classes: $g_2^i g_j$ for $j \in \{1,3,4\}$. Since $\chi_j(g_2) = \chi_j(1)$, we have a nice multiplicative relation $(\chi_2^i\chi_j)( g_2^{i'} g_{j'} ) = \chi_2(g_2)^{ii'} \chi_j(g_{j'})$.

In other words, the square sub-table with indices $\{1,3,4\} \times \{1,3,4\}$ along with the table $\chi_2^{i}(g_2^{i'})$ is sufficient. The group is simply the direct product of the groups $\langle g_2 \rangle$ and $\langle g_3, g_4 \rangle^G$

You can directly check each of the $g_2^i g_j$ are non-conjugate, since they have distinct values for at least one character.


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