There is a theorem which says:
Let $G$ be a finite group. Suppose that every maximal subgroup of $G$ has index equal to a prime or the square of a prime. Then $G$ is solvable.
Reading existing proofs and other results, I have cobbled together my own proof:
Proof. Suppose, to the contrary, that a counterexample exists. Let $G$ be a minimal counterexample. Since $G$ is nontrivial, let $p$ be the largest prime factor of $|G|$, and let $P$ be a Sylow $p$-subgroup of $G$.
If $P$ is normal in $G$, then the maximal subgroups of $G/P$ correspond to maximal subgroups of $G$ containing $P$. These have index equal to a prime or the square of a prime, so $G/P$ is a smaller group for which the condition holds and therefore $G/P$ is solvable. Then, since $P$ is also solvable, $G$ is solvable. Therefore we assume that $P$ is not normal in $G$.
Then the normalizer $N_{G}(P)$ is a proper subgroup of $G$. Then let $M$ be a maximal subgroup of $G$ containing $N_{G}(P)$. Then $[G:M] = \frac{[G:N_{G}(P)]}{[M:N_{G}(P)]} = \frac{[G:N_{G}(P)}{[M:N_{M}(P)]} \equiv \frac{1}{1} = 1 \mod{p}$.
Since $M$ is a proper subgroup of $G$, $[G:M] \geq p+1 > p$. Since $p$ is the largest prime factor of $|G|$, $[G:M]$ is not prime. Therefore there is a prime $q$ such that $q^{2} = [G:M]$. Since $q \neq p$ and $q | |G|$, $q < p$. Then $p \nmid q-1$ so $q^{2} \equiv 1 \mod{p}$ implies $q \equiv -1 \mod{p}$. Then, since $q < p$, we have $q = p-1$. Since $p$ and $q$ are both prime, we have $p=3$ and $q=2$.
Since $p=3$ and $p$ is the largest prime factor of $|G|$, $|G| = 2^{a}3^{b}$ for nonnegative integers $a,b$. This means all maximal subgroups of $G$ have index $2$, $3$, $4$, or $9$. The Frattini subgroup $\Phi(G)$ is nilpotent, so, since $G$ is a counterexample, $G/ \Phi(G)$ must be unsolvable. $\Phi(G)$ is the intersection of all maximal subgroups of $G$. Since the conjugate of a maximal subgroup is also a maximal subgroup, $\Phi(G)$ is the intersection of the cores of the maximal subgroups of $G$. If $N$ and $M$ are normal subgroups of $G$ with $G/N$ and $G/M$ solvable, then $G/(N \cap M)$ is also solvable. Therefore, since $G/ \Phi(G)$ is unsolvable, there is a maximal subgroup $K$ of $G$ such that $G/Core_{G}(K)$ is unsolvable. The quotient $G/Core_{G}(K)$ is the image, in the symmetric group $S_{[G:K]}$, of the action-on-cosets homomorphism based on the subgroup $K$. Therefore, if $G/Core_{G}(K)$ is unsolvable, the symmetric group $S_{[G:K]}$ must be unsolvable. This means that $[G:K] > 4$, so that $[G:K] = 9$.
Then $G/Core_{G}(K)$ is a transitive (in fact, primitive, since $K$ is maximal in $G$) unsolvable subgroup of the symmetric group $S_{9}$. Also, since $|G| = 2^{a}3^{b}$ for some nonnegative integers $a,b$, $|G/Core_{G}(K)| = 2^{\alpha}3^{\beta}$ for some nonnegative integers $\alpha , \beta$. From now on, we denote $G/Core_{G}(K)$ by $H$. The contradiction is obtained by showing no such subgroup of $S_{9}$ exists:
First of all, since $H$ is a subgroup of $S_{9}$, we have $\alpha \leq 7$ and $\beta \leq 4$ (from Lagrange's Theorem and the factorization of $9!$). If $\alpha = 7$, then $H$ is a primitive subgroup of $S_{9}$ containing a whole Sylow $2$-subgroup of $S_{9}$, and thus containing a transposition. Then $H = S_{9}$, contradicting $|H| = 2^{\alpha}3^{\beta}$. Therefore $\alpha \leq 6$. If $\beta = 4$, then $H$ is a primitive subgroup of $S_{9}$ containing a whole Sylow $3$-subgroup of $S_{9}$, and thus containing a $3$-cycle. Then $H$ contains $A_{9}$, for a similar contradiction. Therefore $\beta \leq 3$.
Since $H$ is transitive on $9$ points, $9 | |H|$ so $\beta \geq 2$. If $|H| | 864$ (equivalently, if $\alpha \leq 5$), then $H$ is solvable:
It suffices to prove this claim when $|H| = 864$. The number of Sylow $3$-subgroups of $H$ is $1$, $4$, or $16$. If this number is $1$ or $4$, then a Sylow $3$-subgroup normalizer is a solvable subgroup of $H$ having index at most $4$. Any group having a solvable subgroup of index at most $4$ is solvable, so $H$ is solvable. Therefore assume that the number of Sylow $3$-subgroups of $H$ is $16$.
$H$ acts transitively by conjugation on its Sylow $3$-subgroups. A Sylow $3$-subgroup $P$ fixes itself and acts without fixed points on the other Sylow 3-subgroups. Since $9 \nmid 16-1$, one of these suborbits must have exactly $3$ points in it. This gives us two Sylow $3$-subgroups $R, Q$ such that $[R: R \cap Q] = [Q: R \cap Q] = 3$. This means $R \cap Q$ is normalized by both $R$ and $Q$, so that $N_{H}(R \cap Q)$ has more than one Sylow $3$-subgroup of $H$. How many Sylow $3$-subgroups does it have? $4$ or $16$.
If $N_{H}(R \cap Q)$ has $16$ Sylow $3$-subgroups, then $|N_{H}(R \cap Q)|$ is a multiple of $16$ and $27$, so it is a multiple of $432$ and it is $432$ (in which case $N_{H}(R \cap Q)$ is normal in $H$ because its index is $2$) or $864$ (in which case $R \cap Q$ is normal in $H$, $R \cap Q$ is solvable because it is a $3$-group, and $H/(R \cap Q)$ is solvable because a Sylow $2$-subgroup of it is a solvable subgroup of index $3$, and a group with a solvable subgroup of index at most $4$ is solvable).
If $N_{H}(R \cap Q)$ has $4$ Sylow $3$-subgroups, then its order is a multiple of $27$ and $4$, so its order is a multiple of $108$. Then $[H : N_{H}(R \cap Q)] | 8$, and the solvability of $H$ follows (since a group with a solvable subgroup of index at most $4$ is solvable), except possibly when $N_{H}(R \cap Q)$ is a maximal subgroup of $H$.
If $N_{H}(R \cap Q)$ is maximal in $H$, let $H$ act on its cosets and let $L$ be the kernel of this homomorphism from $H$ to the symmetric group $S_{8}$.
Since $27 | |H|$ but $27$ does not divide $8!$, $3 | |L|$.
If $9 | |L|$, then $L$ is solvable because $L$ is a $3$-group and $H/L$ is solvable because a Sylow $2$-subgroup of $H/L$ is a solvable subgroup of index $1$ or $3$.
If $3$ exactly divides $|L|$, then $H/L$ is a primitive subgroup of $S_{8}$ whose order is a multiple of $9$. Then $H/L$ contains a whole Sylow $3$-subgroup of $S_{8}$ and thus $H/L$ contains a $3$-cycle. Then, since $H/L$ is primitive on $8$ points, $H/L$ contains $A_{8}$, contradicting $|H| = 864$.
Now it only remains to handle the cases when $\alpha = 6$, or, equivalently, when $|H| = 576$ or $1728$. If $|H| = 576$, then $H$ is solvable:
The number of Sylow $3$-subgroups of $H$ is $1$, $4$, $16$, or $64$. If the number is $1$ or $4$, then $H$ has a Sylow $3$-subgroup normalizer (itself obviously solvable) which has index $1$ or $4$ in $H$. Therefore $H$ is solvable. If $H$ has $64$ Sylow $3$-subgroups, each is self-normalizing. Since they are abelian (for they have order $9$), the Burnside $p$-complement Theorem applies to show that $H$ has a normal Sylow $2$-subgroup and so is solvable. Therefore assume that $H$ has exactly $16$ Sylow $3$-subgroups. As before, we may obtain two Sylow $3$-subgroups $Q, R$ such that $[Q: Q \cap R] = [R: Q \cap R] = 3$. Then let $x$ be chosen so that $ < x > = Q \cap R$. Then the centralizer $C_{H}(x)$ contains two Sylow $3$-subgroups of $H$, so it contains at least $4$ Sylow $3$-subgroups of $H$. In fact, the number of Sylow $3$-subgroups of $H$ it contains is $4$ or $16$. If it is $16$, then $C_{H}(x)$ is a subgroup of index $4$ in $H$. $C_{H}(x)$ is solvable because $C_{H}(x)/ < x >$ has a Sylow $2$-subgroup of index $3$, so $H$ is solvable. So assume that the number of Sylow $3$-subgroups of $H$ in $C_{H}(x)$ is $4$. In this case, $C_{H}(x)/ < x >$ is a group of order $12$ which has $4$ Sylow $3$-subgroups, so $C_{H}(x)/ < x > \cong A_{4}$. Then the subgroup $V$ of order $4$ in $A_{4}$ lifts to a subgroup $N$ of order $12$ in $C_{H}(x)$. Since $V$ is normal in $A_{4}$, $N$ is normal in $C_{H}(x)$. Moreover, since $N$ centralizes $x$, $N$ is abelian. Therefore $N$ has a characteristic subgroup $W$ of order $4$ which is normal in $C_{H}(x)$. But also $W$ is contained in a Sylow $2$-subgroup $P$ of $H$, and $N_{P}(W) > W$. Therefore $N_{H}(W)$ contains $C_{H}(x)$ which has order $36$, and $N_{P}(W)$, whose order is a multiple of $8$. Therefore $|N_{H}(W)|$ is a multiple of $72$, so it is $72$, $144$, $288$, or $576$. $N_{H}(W)/W$ has order dividing $144$, so it is solvable. Therefore $N_{H}(W)$ is solvable. If $|N_{H}(W)|$ is $144$, $288$, or $576$, then $H$ has a solvable subgroup of index at most $4$ and is therefore solvable. Therefore, assume $|N_{H}(W)| = 72$. Since the solvability of $H$ follows if $N_{H}(W)$ is not maximal in $H$, assume $N_{H}(W)$ is maximal in $H$. Then let $H$ act on the cosets of $N_{H}(W)$, and let $L$ be the kernel of the homomorphism obtained from $H$ to $S_{8}$. If $|L|$ is a nonmultiple of $3$, $H/L$ contains a whole Sylow $3$-subgroup of $S_{8}$ and so contains a $3$-cycle. Then $H/L$ is primitive on $8$ points and contains a $3$-cycle, so $H/L$ contains $A_{8}$, contradicting $|H| = 576$. So assume that $|L|$ is a multiple of $3$. Then $H/L$ is a transitive subgroup of $S_{8}$, so $8 | |H/L|$. This means that $|L| | 72$, so $|L| | 864$ and $L$ is solvable. Then, since $3 | |L|$, $H/L$ has a Sylow $2$-subgroup of index at most $3$ and so is solvable. Therefore $H$ is solvable, as claimed.
We now come to the final case, in which it will be shown that $S_{9}$ has no primitive subgroup $H$ with $|H| = 1728$: If $H$ has a subgroup of index $2$, then it is solvable and therefore $H$ is solvable. Therefore we assume that $H$ has no subgroup of index $2$, so that $H < S_{9}$ implies $H < A_{9}$. Since $H$ is a transitive group on $9$ points, a point stabilizer in $H$ has order $192$. Since all of the subgroups of order $192$ in $A_{8}$ are conjugate, in $S_{8}$, to the stabilizer, in $A_{8}$, of the partition $12|34|56|78$ of the $8$ indices, we conclude that $H$ is doubly transitive on $9$ points and that any $2$-point stabilizer has a third fixed point. How many sets of $3$ points arise as the fixed point sets of $2$-point stabilizers in $H$? This number is $\frac{\binom{9}{2}}{\binom{3}{2}} = 12$. Since $H$ is doubly transitive, all the $2$-point stabilizers in $H$ are conjugate in $H$. Therefore $H$ also acts transitively on the $12$ sets of $3$ that arise as fixed point sets of $2$-point stabilizers. This means that the stabilizer $M$, in $H$, of one of these sets of $3$ has order $144$. How does $M$ act on the other $6$ points? It can't act faithfully, since $S_{6}$ has no subgroup of order $144$ (for $S_{n}$ has no subgroup of index strictly between $2$ and $n$, except when $n = 4$). Then a nontrivial element of $M$ fixing the $6$ points can only be a $3$-cycle or a transposition, so that $H$ is a primitive permutation group containing a transposition or a 3-cycle and thus $H$ contains $A_{9}$, contradicting $|H| = 1728$.
My question is: what is the easiest way to prove this solvability theorem? (The Burnside $p^{a}q^{b}$ Theorem is too magical via character theory, and too hard without it, for my taste.)
