I'm preparing for an upcoming exam and tried doing some random rep.theory exercises which i found on the internet. The question of interest is stated here http://www.math.nus.edu.sg/~matgwt/teach/math202b-problem3.pdf Problem 10 (i) I discovered a gap in my knowledge. There are three Characters given which i know aren't any of the irreducible ones we are looking for ( I know this since the group we are working with is the PSl(2,7) and i know it's character table).

$(10)$ A group of order $168$ has $6$ conjugacy classes. $3$ representations of this group are known and their characters are given below, as well as the sizes of the $6$ conjugacy classes:

\begin{array}{c r r r r r r} &1&21&42&56&24&24\\ \alpha&14&2&0&-1&0&0\\ \beta&15&-1&-1&0&1&1\\ \gamma&16&0&0&-2&2&2\\ \end{array} (i) constuct the character table of the group. You may need to assume the fact that $\sqrt{7}$ is not contained in the field $\mathbb{Q}(\zeta_7),$ where $\zeta_7 = e^{2 \pi i/7}.$

So my question is how do i obtain irreducible characters from reducible ones(looked for a while on the internet and in my notes but i couldn't find anything usefull), and if u could demonstrate the way on doing this by showing me how to obtain one character i think i will be Abel to do the rest by myself than. Thanks for your help.

  • 1
    $\begingroup$ This problem apparently is very similar but not from the same source as MSE question 2768285 "How does knowing that $\sqrt{7}\notin\mathbb{Q}(e^{2\pi i/7})$ help construct this character table?" $\endgroup$ – Somos Jun 30 '18 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.