# 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?

• Well there is a theorem of Kegel that a finite group that is a product of two nilpotent subgroups is solvable, so it follows from that. But you are talking about what you are allowed to use, so is this an exercise? If so, it might be helpful to give more context, such as what were the earlier exercises. – Derek Holt Jun 22 '14 at 19:10
• This is from Dummit and Foote, and it's a Representation Theory section that used Character Theory to prove Burnside's Theorem and then used standard Group Theory plus Burnside to prove Hall's Theorem. The other exercises were to prove that a finite group all of whose maximal subgroups have prime or prime squared index is solvable, to prove the problem in the OP for "abelian" instead of "nilpotent," and to prove Burnside's Theorem when the Sylow $p$ and $q$ subgroups are abelian. – Nishant Jun 22 '14 at 19:17
• Does the theorem of Kegel have an easier proof when one of the groups is a $p$ group? – Nishant Jun 22 '14 at 19:37

## 1 Answer

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.

• Ah, I missed that lemma...just to be sure, does the hypothesis of there being a nilpotent subgroup of prime power index still hold when we quotient out by a normal subgroup? – Nishant Jun 22 '14 at 22:33
• Yes, $HN/N$ is a quotient of a nilpotent group, so it's still nilpotent, and $|G/N:HN/N| = |G:HN|$ is still a power of $p$. (It could be $p^0=1$, but then $G/N$ would be nilpotent.) – Derek Holt Jun 22 '14 at 22:35
• And if $G/N$ is nilpotent, then since $N$ is also, both are solvable, and thus $G$ is solvable. Thanks! – Nishant Jun 22 '14 at 22:41