Non-solvable group without subgroups except p-groups and Hall subgroups Does it exist a non-solvable finite group which all proper, nontrivial subgroups are p-groups and Hall subgroups?
 A: I do not know whether it can be proven in a more elementary way but we can show that such groups must be solvable by Thomson-Glaberman theorems.
Let $G$ be a minimal counterexample. Then we can assume that $G$ is simple. Let $p$ be odd prime dividing the order of $G$ and $J(P)$ denotes the Thompson subgroup of $G$ where $P\in Syl_p(G)$.
Let $T=N_G(Z(J(P)))$. If $T$ is a $p$-group then $G$ is $p$-nilpotent by Glaberman Theorem. Thus, $T$ is a Hall subgroup. Let $q\neq p$ be a prime dividing the order of $T$ and $Q$ be a subgroup of order $q$ in $T$. Then the group $Z(J(P)Q$ is also a Hall subgroup of $G$. Thus, $Q\in Syl_q(G).$
Now, we see that $G$ has cyclic Sylow subgroups of prime orders for some primes. Let $r$ be the smallest one, that is, $R\in Syl_r(G)$ is cyclic and if $s$ is a prime smaller than $r$ then Sylow $S$ subgroup is not cyclic.
If $N_G(R)=C_G(R)$ then $G$ is $r$-nilpotent by Burnside theorem. Thus, assume that $N_G(R)\neq C_G(R)$. By assumtion, both $N_G(R)$ and $C_G(R)$ are Hall subgroups of $G$. In particular, $C_G(R)$ is a normal Hall subgroup of $N_G(R)$. By Schur-Zassenhaus theorem, there is complement $Y$ for $C_G(R)$ in $N_G(R)$. Since $Y\cap R=1$, $Y$ act faithfully on $R$. Since $Aut(R)\cong Z_{r-1}$, $Y\leq \mathbb Z_{r-1}$. Morever, since $Y$ is an Hall subgroup of a Hall subgroup, $Y$ is a Hall subgroup of $G$. Since every Sylow subgroup of $Y$ is cyclic and $r$ was choosen the smallest one, this leads a contradiction.
Edit: I noticed that Frobenius normal $p$-complement theorem (Which is much weaker than Thompson-Glaberman theorems ) is enough to conclude that $G$ has some cyclic Sylow subgroup of prime order and rest of them proceed in a similar way. Thus, this can be a nice exercise after getting familiar with theorems of Transfer and  Fusion.
