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 $\vert G \vert \le 10^4$.
 A: Yes. The O'Nan-Scott Theorem classifies finite primitive groups into a number of different types, such as affine type, almost simple type, product type, etc. The groups in all but one of these types have a unique minimal normal subgroup $N$, which implies that they are linearly primitive, since the kernel of any unfaithful representation must contain $N$, and so any faithful representation must have an irreducible constituent whose kernel does not contain $N$ and is consequently faithful.
The groups $G$ in the remaining type have two isomorphic minimal normal subgroups $N_1$ and $N_2$, both of which are direct products of one or more isomorphic nonabelian simple groups. The smallest such example is $A_5 \times A_5$ acting on the cosets of a diagonal subgroup.
But direct products of nonabelian simple groups are linearly primitive, because the tensor product of nontrivial irreducible representations of their direct factors is a faithful irreducible representation. So $N_1 \times N_2$ has a faithful irreducible representation $\rho$ and any irreducible constituent of the induced representation $\rho^G$ is faithful for $G$.
A: 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.  $\square$
