# character tables and solubility

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 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

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 $\mathrm{GL}(2,\mathbb C)$.
• 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? – Peter A May 18 '15 at 10:03
• @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. – mesel May 18 '15 at 10:09