Complete the character table of a group with 4 conjugacy classes containing 1, 3, 4, 4 elements, respectively

Question

Consider a group $G$ with exactly 4 conjugacy classes $C_1,\;C_2,\;C_3\;C_4$ containing exactly 1, 3, 4, 4 elements, respectively. Let $\zeta$ be a primitive third root of 1 and the following table its character table. Complete it.

My attempt: The group has 4 irreducible representations. Let $\chi_0, \chi_1, \chi_2, \chi_3$ the corresponding irreducible characters. Using the character of the regular representation I found that $\chi_0(1)=\chi_1(1)=\chi_2(1)=1$ and $\chi_3(1)=3$.

Now, since each element of $C_2$ has order 2, then $\chi_0(g)=\chi_1(g)=\chi_2(g)=\chi_3(g)=1$, where $g$ is an element of $C_2$.

Also using that $1=<\chi_2,\chi_1>$, I found that $\chi_2(g)=2-\zeta$, where g is an element of $C_4$.

Therefore, so far the table goes like this.

Problem: I don't know how to find the other four values. I have tried with the orthogonality rules but I have not been able to eliminate any of the 4 unknowns.

Answer 1

(a) At the outset we have (if we add the trivial character) $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\\ \\ \\ \end{matrix}. $$

(b) Recall that the complex conjugate of an irreducible character is also an irreducible character and get $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\\ 1 & &\zeta^2\\ \\ \end{matrix}. $$

(c) Use the fact that $|G|=1+3+4+4$ to get $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\\ 1 & &\zeta^2\\ 3& \\ \end{matrix}. $$

(d) Use the fact that $|C_3|=4$ to see that the length of the 3rd column is $12/4=3$ and so get $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\\ 1 & &\zeta^2\\ 3& &0 &\\ \end{matrix}. $$

(e) Note that as the second character is not real on $C_3$ the elements of $C_3$ are not conjugate to their inverses, so their inverses form a distinct class, say $C_4$, and each character on $C_4$ is the complex conjugate of the values on $C_3$, since the eigenvalues of the representation on $C_4$ are the inverses of the eigenvalues of the representation on $C_3$. So we have $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\zeta^2\\ 1 & &\zeta^2&\zeta\\ 3& &0 &0\\ \end{matrix}. $$

(f) Use the fact that the second column is orthoogonal to each of the other columns, to see first that it is orthogonal to the sum of the other three columns, and so get $$ \begin{matrix} 1 & 1 &1 &1\\ 1& & \zeta&\zeta^2\\ 1 & &\zeta^2&\zeta\\ 3&-1 &0 &0\\ \end{matrix}. $$

(g) Use whatever orthogonality relations you like to complete the table and get $$ \begin{matrix} 1 & 1 &1 &1\\ 1& 1& \zeta&\zeta^2\\ 1 &1 &\zeta^2&\zeta\\ 3&-1 &0 &0\\ \end{matrix}. $$

Note. I think it is possible to do this without the hypotheses that the four classes have orders $1,3,4,4$ but it's a bit harder.

Answer 2

Here is a proof that does not use the sizes of the classes, just that it has four classes. Let $x_i$ be an element from class $C_i$. We start by inserting the trivial character, the action of $\chi_2$ on $x_3^{-1}$ (which, since $\chi_2$ acts as $\zeta\not\in\mathbb R$, must not be conjugate to $x_3$), then the dual of $\chi_2$, to obtain the following partial table.

$$\begin{matrix} &x_1&x_2&x_3&x_3^{-1}\\ \chi_1&1&1&1&1 \\\chi_2&1&&\zeta&\zeta^2 \\\chi_3&1&&\zeta^2&\zeta \\\chi_4&a \end{matrix}$$

Now let's compute the order, which is $1+1+1+a^2$. Since the order of $x_3$ is a multiple of $3$ (as the image of $x_3$ under the homomorphism $\chi_3$ has order $3$), $3\mid |G|=3+a^2$. Thus $a$ is a multiple of $3$, and in particular, $|G/G'|=3$. Thus $G'\neq 1$, and some non-trivial class must be in it. This is $x_2$.

$$\begin{matrix} &x_1&x_2&x_3&x_3^{-1}\\ \chi_1&1&1&1&1 \\\chi_2&1&1&\zeta&\zeta^2 \\\chi_3&1&1&\zeta^2&\zeta \\\chi_4&a&b \end{matrix}$$

Now by column orthogonality, $ab=-3$. But $3\mid a$, and $b$ is an algebraic integer. Thus $a=3$, $b=-1$. Finally, $\chi_4$ is clearly real, so its values on $x_3$ and $x_4=x_3^{-1}$ are the same.

$$\begin{matrix} &x_1&x_2&x_3&x_3^{-1}\\ \chi_1&1&1&1&1 \\\chi_2&1&1&\zeta&\zeta^2 \\\chi_3&1&1&\zeta^2&\zeta \\\chi_4&3&-1&c&c \end{matrix}$$

Column orthogonality with (say) $x_1$ gives the result. If you don't want to work with any complex numbers, note that orthogonality of columns 1 and 3 gives $\alpha+3c=0$ for some $\alpha$, and columns 2 and 3 gives $\alpha-c=0$. Thus $c=0$.

$$\begin{matrix} &x_1&x_2&x_3&x_3^{-1}\\ \chi_1&1&1&1&1 \\\chi_2&1&1&\zeta&\zeta^2 \\\chi_3&1&1&\zeta^2&\zeta \\\chi_4&3&-1&0&0 \end{matrix}$$