Maximal subgroup and representations (principal part) Let $G$ be a finite group, $V$ an irreducible complex representation, $H$ a subgroup.
Let $V^H$ be the subspace of vectors of $V$ invariant under the action of $H$.    

Question: Is $dim(V^H) \le 1$ if $H$ is a maximal subgroup of $G$ ?    

Remark : If $H = \{ e \}$ then $G = \mathbb{Z}_p$ and $dim(V)=1$.
 A: $\dim(V^H) = [ \chi_H, 1_H ] = [ \chi, 1_H^G ]$ and it is a bit rare for a primitive permutation character to have a repeated factor, but not too uncommon.
There are several examples amongst simple groups: $G=L_2(11)$, $L_2(13)$, $L_2(17)$, $L_2(19)$, $L_3(3)$, $L_2(23)$, $L_2(25)$, $M_{11}$, $L_2(27)$, $L_2(29)$, $L_2(31)$, $Sz(8)$, $M_{12}$, $J_1$, $A_9$, $L_3(5)$, $J_2$, $L_2(109)$, $L_2(113)$, ..., $A_{10}$, $A_{12}$, $A_{13}$, ${}^2F_4(2)'$ for example.
For $G=L_2(11)$ we can take $H$ to be a Sylow 3-normalizer, and then one of the irreducible representations $W$ of dimension 10 (there are two, but only one works; the one with trace -1 on an element of order 6) works. This one might be a bit easier to check. In this example, one can even find $g$ so that $H \cap H^g = 1$ providing a counterexample to the “dual question.”
For $G=A_{13}$ we can take $H$ to be a Sylow 13-normalizer, and then almost every irreducible representation of $G$ works (all except the ones of degree 1, 12, 65, and 66).
