1
$\begingroup$

I am currently going through a past exam paper for a group theory module and am unable to answer the following section of a question. The copy of my lecture notes doesn't seem to have a section on soluble groups but I would still like to know that if i am given the character table of this group enter image description here

The how would i determine whether the group is:

  • soluble?
  • Simple?
  • Isomorphic to a subgroup of $GL(n, \mathbb{C})$?

Any help would be much appreciated and thanks in advance

|cite|improve this question|||||
$\endgroup$
7
$\begingroup$

Note that number of the linear character is $1$ hence $\lvert G/G'\rvert =1$. Thus, $G$ is not solvable.

$g_2\in \ker \chi_4$ hence $G$ is not simple.

$\ker\chi_2=1$ hence $G$ is isomorphic to a subgroup of $\mathrm{GL}(2,\mathbb C)$.

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ +1, excellent answer. I have taken the liberty to do a little LaTeX editing. $\endgroup$ – Andreas Caranti May 17 '15 at 16:56
  • $\begingroup$ @AndreasCaranti: Thank you very much :) $\endgroup$ – mesel May 17 '15 at 17:23
  • $\begingroup$ thank you @mesel for your answer, how did you conclude that the kernel of $x_2 = 1$ ? the kernel is all the elements $g\in G$ such that $\rho(g) =1$ but where is this info on the table? $\endgroup$ – Peter A May 18 '15 at 10:03
  • $\begingroup$ @PeterA: $Ker(\chi_2)=\{g\in G| \chi_2(g)=\chi_2(1) \})$. I think you are confused about the defination of kernel of character. $\endgroup$ – mesel May 18 '15 at 10:09
  • $\begingroup$ ah yes and this kernel is trivial and therefore isomorphic to the required subgroup? @mesel $\endgroup$ – Peter A May 18 '15 at 10:12

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.