Let $G$ be solvable and assume that every $\chi\in\text{Irr}(G)$ is quasi-primitive. Show that $G$ is abelian. 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.
 A: 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.
A: Here is another way to prove it.
A QPC-group is a finite group all whose irreducible complex characters are quasi-primitive. Let us start with a lemma.
Lemma 1 Let $G$ be a QPC-group and let $A \unlhd G$ be abelian. Then $A \subseteq Z(G)$.
Proof Let $\chi \in Irr(G)$, we must have $\chi_A =\chi(1)\lambda$ for some
linear $\lambda \in Irr(A)$ since $\chi$ is quasi-primitive. Hence $A \subseteq Z(\chi)$. We conclude that $A \subseteq \{Z(\chi):\chi \in Irr(G) \}$. Since $Z(G) = \bigcap\{Z(\chi): \chi \in Irr(G)\}$ (CTFG, Corollary(2.28)), $A \subseteq Z(G)$ follows.
Lemma 2 Let $G$ be a $p$-group, then it contains a normal subgroup $A$ with $A=C_G(A)$ (note that such a subgroup must be abelian).
Proof Let $A$ be maximal among the abelian normal subgroups of $G$. Such an $A$ exists since $Z(G)$ is non-trivial. Write $C=C_G(A)$ and suppose $A \lneq C$. It is easy to show $C$ is normal in $G$. Now by standard $p$-group theory (see Lemma 1.23 in I.M. Isaacs, Finite Group Theory), we can find $B \unlhd G$, with $A \leq B \leq C$, and index$[B:A]=p$. Since $B \subseteq C$, we have $A \subseteq Z(B)$. Hence $B$ must be abelian (apply that if $|B/Z(B)| \text{ divides }p$, then $B$ is abelian). This contradicts the maximality of $A$ and the proof is finished.
Corollary If $G$ is a $p$-group and also a $QPC$-group, then it is abelian.
Proof Lemma 2 provides the existence of an $A \lhd G$, with $A=C_G(A)$. But Lemma 1 yields $A \subseteq Z(G)$, and this implies $A=C_G(A)=G$, hence $G$ is abelian.
Theorem A solvable $QPC$ group is abelian.
Proof Let $A=G^{(k-1)}$ be the prelast term of the derived series of $G$ that ends at $G^{(k)}=1$ and assume $k \gt 1$. Then $A$ is abelian, non-trivial and by Lemma 1, $A \subseteq Z(G)$. Now we use induction on $|G|$. It is easy to see that the quotient group $G/Z(G)$ is also a $QPC$-group, hence by induction $G/Z(G)$ is abelian, so $G' \subseteq Z(G)$ and it follows that $G$ is nilpotent (of class two). So $G$ is the direct product of its Sylow subgroups and each of the factors is a $QPC$-group. The corollary above gives $G$ is abelian.
