Question: Let $G$ be solvable and assume that every $\chi\in\text{Irr}(G)$ is quasi-primitive. Show that $G$ is abelian.
Thoughts: First, we say that irreducible characters whose restriction to every normal subgroup is homogenous is called a quasi-primitive character. With that being said, characters which are multiples of an irreducible character are called homogeneous characters. So, if we want to show that $G$ is abelian, one way to do this is to show that every irreducible character of $G$ is linear.
So, let $N\trianglelefteq G$. Since $\chi$ is quasiprimitive, we know that $\chi_N$ is homogenous. That is, $e\chi_N=\theta$ for some $\theta\in\text{Irr}(N)$ and some nonzero number $e$. Now, since $G$ is solvable, and $N$ is a normal subgroup of $G$, then both $N$ and $G/N$ are solvable. Now, we know that the inner product, $[\chi_N,\theta]\neq 0$ (by Clifford this is equal to $e$), and so by problem 6.7 in Isaacs C.T.F.G, we have that $\chi(1)/\theta(1)$ must divide $|G:N|=p^a$ for some prime $p$ and natural number $a$. I figure this part may come in handy, because it gives us the degree of $\chi$, and if we can show that $\chi(1)=1$ then we'd be done. Now, Theorem 6.9 in the same book tells us that if $\chi(1)$ is a power of the prime $p$ for every irreducible character of $G$, then $G$ has a normal abelian $p$-complement. So, I wonder if (and maybe this isn't a good way to go about it) we can say that $|G:N|=p$, and then somehow use some divisibility stuff to get that $\chi(1)=1$.
Again, this may not be the correct way to go about the problem, but I am not seeing anything else that is useful. Any help is appreciated 🙂
A quick note: ideally, there would be a way to salvage my idea above. Since this question is in chapter 6 of Isaacs, though interesting, I wouldn't want the solution to be dependent on something from after chapter 6.
Best Answer
Suppose that every irreducible character of $G$ is quasi-primitive, and let $N$ be any non-trivial abelian normal subgroup of $G$. (One exists since $G$ is soluble.) It is also true that any character of $G/N$ is quasiprimitive, so by induction $G/N$ is abelian. We claim that $N\leq Z(G)$, using Brauer's permutation lemma (6.32). Since $\chi_{N}$ is homogeneous, the induced action on $\mathrm{Irr}(N)$ is trivial. Hence $G$ acts trivially on $N$ as well, i.e., $N\leq Z(G)$.
But now let $x\in G\setminus Z(G)$; then $\langle Z(G),x\rangle$ is an abelian normal subgroup of $G$, hence is central, a contradiction to the existence of $x$. Hence $G=Z(G)$, as needed.