# How does knowing that $\sqrt{7}\notin\mathbb{Q}(e^{2\pi i/7})$ help construct this character table?

Here's a question that has haunted me since it appeared on a problem sheet in a Representation Theory course I attended as an undergraduate. I'll reproduce it exactly:

1. A group of order $168$ has $6$ conjugacy classes. Three representations of this group are known and have characters $\alpha$, $\beta$ and $\gamma$ summarised in the table below. Construct the character table of the group. You may assume if required that $\sqrt 7$ is not in the field generated by $\mathbb Q$ and a primitive $7^{\text{th}}$ root of unity.

\begin{array}{c r r r r r r} |[g]|&1&21&42&56&24&24\\ \hline \alpha&14&2&0&-1&0&0\\ \beta&15&-1&-1&0&1&1\\ \gamma&16&0&0&-2&2&2\\ \end{array}

I can do the question quite easily by using the various properties of character tables. But I don't need to use the fact that $\sqrt{7}\notin\mathbb{Q}(e^{2\pi i/7})$. So my question is:

What approach to the question did the writer have in mind when they allowed the student to assume that "$\sqrt 7$ is not in the field generated by $\mathbb Q$ and a primitive $7^{\text{th}}$ root of unity"?

• Something is not correct about the table. The sum of the squares of the character values of the identity class should be the order of the group, but the sum of $(14,15,16)$ squared is $677$ which is bigger than $120$ the size of this group. – Somos May 5 '18 at 21:25
• @Somos They aren't the characters of irreducible representations, they're just characters. So we're not completing the table based on three existing entries, we actually have to find all six irreducible characters. – Oscar Cunningham May 5 '18 at 21:28
• Ah, I had implicitly assumed they were irreducible. That makes the task harder and more challenging. – Somos May 5 '18 at 21:32
• In case it helps: the group is actually PSL(2,7) and its character table is on its Wikipedia page. – Oscar Cunningham May 5 '18 at 21:43
• It does. There is another group of order 168 with 8 conjugacy classes. – Somos May 5 '18 at 21:48

## 1 Answer

Okay, I finally worked it out. I'll give a complete solution to the exercise:

First of all we have the trivial character, which we'll denote $\chi_1$. We have $\left<\chi_1,\alpha\right>=\left<\chi_1,\beta\right>=\left<\chi_1,\gamma\right>=0$, so it isn't present in any of the characters we've been given.

Computing the inner product of each character with itself gives $\left<\alpha,\alpha\right>=2$, $\left<\beta,\beta\right>=2$ and $\left<\gamma,\gamma\right>=4$, so $\alpha$ and $\beta$ are each the sum of two distinct irreducible characters, and $\gamma$ is either the sum of four distinct irreducible characters or two copies of the same irreducible character. In fact it must be the latter, because $\left<\beta,\gamma\right>=2$, which would mean that both the irreducible characters in $\beta$ would have to appear in $\gamma$. But then the remaining two irreducible characters in $\gamma$ would only have dimension $1$ between them, contradiction. So $\chi_2=\gamma/2$ is an irreducible character of dimension $8$.

Then it turns out that $\left<\chi_2,\alpha\right>=1$ and $\left<\chi_2,\beta\right>=1$, so we can find two more irreducible characters: $\chi_3=\alpha-\chi_2$ and $\chi_4=\beta-\chi_2$, of dimensions $6$ and $7$. Using the fact that the squares of the dimensions of the irreducible representations must sum to 168, we can see that the two remaining irreducible characters must both have dimension $3$.

So our character table so far is given by:

\begin{array}{c r r r r r r} |[g]|&1&21&42&56&24&24\\ \hline \chi_1&1&1&1&1&1&1\\ \chi_2&8&0&0&-1&1&1\\ \chi_3&6&2&0&0&-1&-1\\ \chi_4&7&-1&-1&1&0&0\\ \chi_5&3&x_2&x_3&x_4&x_5&x_6\\ \chi_6&3&y_2&y_3&y_4&y_5&y_6\\ \end{array}

To work out the $x$s and $y$s we'll use column orthonormality. Using orthonormality of the second column with the first and with itself gives $x_2+y_2=-2$ and $|x_2|^2+|y_2|^2=2$, which imply $x_2=y_2=-1$. Likewise, $x_3+y_3=2$ and $|x_3|^2+|y_3|^2=2$ so $x_3=y_3=1$, and $x_4+y_4=0$ and $|x_4|^2+|y_4|^2=0$ so $x_3=y_3=0$.

In the fifth column we get $x_5+y_5=-1$ and $|x_5|^2+|y_5|^2=4$. These equations have multiple solutions. But since the complex conjugate of an irreducible character is an irreducible character, we know that either $x_5$ and $y_5$ are both real, or they are each other's complex conjugate. In the real case we have $x_5+y_5=-1$ and $x_5^2+y_5^2=4$, and hence $x_5=\frac{-1\pm\sqrt 7}2$ (without loss of generality we can take the "$+$" case, and then $y_5=\frac{-1-\sqrt 7}2$). In the unreal case we have $2\mathrm{Re}(x_5)=-1$ and $2|x_5|^2=4$, and hence $x_5=\frac{-1\pm\sqrt 7i}2$ (again we will take the "$+$" case, and hence $y_5=\frac{-1-\sqrt 7i}2$).

It's clear that $x_6$ and $y_6$ must take the same values as $x_5$ and $y_5$, but the other way around. So we have narrowed down the possibilities for the character table to two remaining choices, both of which fully satisfy all the orthogonality relations. The character table is either

\begin{array}{c r r r r r r} |[g]|&1&21&42&56&24&24\\ \hline \chi_1&1&1&1&1&1&1\\ \chi_2&8&0&0&-1&1&1\\ \chi_3&6&2&0&0&-1&-1\\ \chi_4&7&-1&-1&1&0&0\\ \chi_5&3&-1&1&0&\frac{-1+\sqrt 7}2&\frac{-1-\sqrt 7}2\\ \chi_6&3&-1&1&0&\frac{-1-\sqrt 7}2&\frac{-1+\sqrt 7}2\\ \end{array}

or

\begin{array}{c r r r r r r} |[g]|&1&21&42&56&24&24\\ \hline \chi_1&1&1&1&1&1&1\\ \chi_2&8&0&0&-1&1&1\\ \chi_3&6&2&0&0&-1&-1\\ \chi_4&7&-1&-1&1&0&0\\ \chi_5&3&-1&1&0&\frac{-1+\sqrt 7i}2&\frac{-1-\sqrt 7i}2\\ \chi_6&3&-1&1&0&\frac{-1-\sqrt 7i}2&\frac{-1+\sqrt 7i}2\\ \end{array}

and it remains to rule out one of these possibilities.

Here's the intended solution: By the Sylow Theorems we know that $C_7$ appears as a subgroup of our group either $1$ or $8$ times. None of these subgroups can intersect (except at the identity) since $C_7$ is generated by each of its nonidentity elements. So the number of elements of order $7$ is either $6$ or $48$. The elements in each conjugacy class must have the same order, so in fact it must be the case that the elements of order $7$ are precisely the elements of the two conjugacy classes of size $24$. The eigenvalues of a representation of an element of order $7$ must be $7$th roots of unity. So since $\boldsymbol{\sqrt{7}\notin\mathbb{Q}(e^{2\pi i/7})}$, the first character table above cannot be correct, and the answer must be the second one.

Or course I think it's actually easier to not use the extra assumption. For example you can calculate that (if we take $\chi_5$ and $\chi_6$ as in the first character table) $\left<\chi_5,\chi_5\otimes\chi_6\right>=1/2$, which is a contradiction since this value should be a natural number. This answers the question using only properties of characters, and without having to delve into the properties of the group.