0
$\begingroup$

Character table of $GL_3(\mathbb{F}_2)$.

\begin{array}{|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & 7A & 7B \\ \hline \chi_1 & 1 & 1 & 1 & 1 & 1 & 1\\ \hline \chi_2 & 6 & 2 & 0 & 0 & -1 & -1 \\ \hline \chi_3 & 8 & 0 & -1 & 0 & 1 & 1 \\ \hline \chi_4 & 7 & -1 & 1 & -1 & 0 & 0 \\ \hline \chi_5 & 3 & -1 & 0 & 1 & \frac{1+\sqrt{-7}}{2} & \frac{1-\sqrt{7}}{2} \\ \hline \chi_6 & 3 & -1 & 0 & 1 & \frac{1-\sqrt{7}}{2} & \frac{1+\sqrt{-7}}{2} \\ \hline \end{array}

Corollary. $GL_3(\mathbb{F}_2)$ is simple.

Proof. If not, we would have a normal subgroup $N \triangleleft G, N \neq \{e\},N \neq G$. Then take any irreducible representation $\rho \neq \mathbb{I}$ of $G/N$ and lift it to $G \implies$ irreducible representation of $G$ with $\chi(g)=\chi(e)=\dim \rho$ for all $g \in N$. But we have $\chi(g) \neq \chi(e)$ for all $\chi \neq \mathbb{I}, g \neq e$.

I cannot see how it is justified that after lifting it will be an irreducible representation of $G$ with $\chi(g)=\chi(e)=\dim \rho$.

$\endgroup$
  • 1
    $\begingroup$ Would be nice, if three lines are enough to show that $GL(3,2)$ is a simple group. What results in representation theory did you prove so far before the Corollary ? $\endgroup$ – Dietrich Burde Apr 9 '15 at 19:20
  • $\begingroup$ @DietrichBurde Indeed. Perhaps the theorem, of which this is a corollary, should be shared with us? $\endgroup$ – pjs36 Apr 9 '15 at 19:22
  • $\begingroup$ The corollary was to the character table $\endgroup$ – Permian Apr 9 '15 at 19:35
  • 2
    $\begingroup$ In general, if $\rho$ is a representation of $G/N$ then you can define a representation $\hat{\rho}$ of $G$ by $\hat{\rho}(g) = \rho(gN)$. Clearly $\rho$ and $\hat{\rho}$ have the same degrees and if $\rho$ is irreducible the so is $\hat{\rho}$. $\endgroup$ – Derek Holt Apr 9 '15 at 20:00
1
$\begingroup$

Note that if $\rho\colon G/N \to \mathrm{GL}_n(\mathbb C)$ is your nontrivial representation then it's lift is given by the composition $\widetilde\rho\colon G \to G/N \to \mathrm{GL}_n(\mathbb C)$. If $g \in N$ then $gN = 1N$ in $G/N$ so $\widetilde\rho(g) = \rho(gN) = \rho(1N) = I_n$ where $I_n \in \mathrm{GL}_n(\mathbb C)$ is the identity matrix. Thus the character of $\widetilde\rho$ takes the value $\mathrm{trace}(I_n) = n = \dim\rho$ on $g$.

Next note that $\rho$ irreducible means there are no proper nontrivial invariant subspaces of $\mathbb C^n$, so if $V \subseteq \mathbb C^n$ is any proper nontrivial subspace then it's not invariant, so there exists a $v \in V$ and a $gN \in G/N$ such that $\rho(gN)v \notin V$. But then $\widetilde\rho(g)v = \rho(gN)v \notin V$ so $\widetilde\rho$ is also irreducible.

$\endgroup$

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.