Finite Group with Nilpotent Subgroup of Prime Power Index is Solvable Let $G$ be a finite group, and assume that $H$ is a nilpotent subgroup whose index is a prime power. WLOG, we can say that the index of $H$ is the highest power of $p$ which divides the order of $G$. 
I want to show that $G$ is solvable, and I'm allowed to use Burnside's $p^aq^b$ theorem, as well as Hall's Theorem (which states that if every Sylow-$p$ subgroup has a complement, then $G$ is solvable).
So say $|G|=p^aq_1^{\alpha_1}\cdots q_n^{\alpha_n}$. Then since $H$ is nilpotent, it's the direct product of its Sylow subgroups, so $H=Q_1\times\cdots\times Q_n$. $H$ is obviously a complement for $G$'s Sylow-$p$ subgroup $P$, so I'll be done if I can show there's a complement to $Q_i$, the Sylow-$q_i$. Now, if we let $H_1=H/Q_i$, my first idea is to consider $PH_i$, which has the desired order, but of course there's no guarantee that this is a subgroup, since neither of the two subgroups need to be normal. 
I also tried assuming that $G$ is a minimal counterexample and thus simple and showing that some nontrivial proper subgroup is normal, but since $p$ does not have to be the largest prime, I can't argue that the action of $P$ on the $Q_i$'s by conjugation is trivial. 
Are either of these avenues fruitful, or should I try something else? 

The main argument in the character theoretic proof of Burnside's $p^aq^b$ theorem actually proves that a finite simple group cannot have a conjugacy class of prime power order bigger than $1$. In your problem, $H$ is nilpotent, so it has nontrivial centre. Then, for $1 \ne h \in Z(H)$, either $h \in Z(G)$, in which case $G$ is not simple, or $|{\rm Cl}_G(h)|$ is a nontrivial power of $p$, so again $G$ is not simple. Hence, by induction, $G$ is solvable.
