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

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 $\chi(g)=\chi(e)=\dim \rho$.

• 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 ? – Dietrich Burde Apr 9 '15 at 19:20
• @DietrichBurde Indeed. Perhaps the theorem, of which this is a corollary, should be shared with us? – pjs36 Apr 9 '15 at 19:22
• The corollary was to the character table – Permian Apr 9 '15 at 19:35
• 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 $\hat{\rho}$. – Derek Holt Apr 9 '15 at 20:00

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.