Are the primitive groups linearly primitive? A transitive permutation group $G \subset S_n$ is primitive if $G_1 \subset G$ is a maximal subgroup.
A finite group $G$ is linearly primitive if it has a faithful complex irreducible representation.  
Question: Are the primitive finite groups linearly primitive?
Remark: I've checked by a GAP computation that it's true for $n=[G:G_1] \le 200$ and $\vert G \vert \le 10^4$.

Yes, the following is a short self-contained proof: 
Let $G \subset S_n$ be a primitive permutation group; by definition $G_1$ is a core-free maximal subgroup of $G$.  Let $V$ be a non-trivial irreducible complex representation of $G$ such that the fixed point subspace  $V^{G_1} \neq 0$ (such a $V$ exists, see the comment).  By construction $G_1 \subset G_{(V^{G_1})} $ but by maximality $G_{(V^{G_1})} = G_1$ or $G$, if it is $G$ then $V^{G_1} = 0$ contradiction, so it is $G_1$.  Now $K=ker(\pi_V) \subset G_{(V^{G_1})} = G_1$, but $G_1$ is core-free in $G$ and $K$ is a normal subgroup of $G$. It follows that $K=\{ e \}$ and so $G$ is linearly primitive.  $\square$
