2
$\begingroup$

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

$\endgroup$

1 Answer 1

Reset to default
9
$\begingroup$

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 $\mathrm{GL}(2,\mathbb C)$.

$\endgroup$
6
  • $\begingroup$ +1, excellent answer. I have taken the liberty to do a little LaTeX editing. $\endgroup$
    – Andreas Caranti
    May 17, 2015 at 16:56
  • $\begingroup$ @AndreasCaranti: Thank you very much :) $\endgroup$
    – mesel
    May 17, 2015 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, 2015 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, 2015 at 10:09
  • $\begingroup$ ah yes and this kernel is trivial and therefore isomorphic to the required subgroup? @mesel $\endgroup$
    – Peter A
    May 18, 2015 at 10:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .