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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.)

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2  
This clearly depends on how you define "easiest". For example, an early invocation of Burnside's Theorem would make your proof about 1/10 as long, and much easier for other people to read. –  S. Carnahan Feb 8 '11 at 5:50
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The classification of finite simple groups would be unthinkable without representation theory. Your constraint not to use character theory seems completely artificial and unreasonable to me. It's like if you demanded that any statement about real numbers should be proven without recourse to the complex ones. Anyway, in referring to double transitivity and such like, you are secretly using representation theory without admitting it. –  Alex B. Feb 8 '11 at 6:00
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Unlike Scott and Alex, I would agree that it is nice to have a proof that does not use character theory. In fact there is a very long and difficult proof of Burnside's $p^a q^b$ theorem that does not use character theory! But how do you know that a primitive subgroup of $S_9$ that contains a 3-cycle contains $A_9$? (It's true of course, but I am not sure where that is coming from in your proof.) –  Derek Holt Feb 8 '11 at 15:15

1 Answer 1

Here is an easier proof.

Starting as you did, we get $|G| = 2^a3^b$. Now let $N$ be minimal normal in $G$. Since the hypothesis on maximal subgroups is inherited by $G/N$, it follows (working by induction on $|G|$) that $G/N$ is solvable, so it suffices to show that $N$ is solvable. We can thus assume that $N$ is neither a 2-group or a 3-group. Let $P$ be a Sylow 3-subgroup of $G$, so $S = P \cap N$ is a Sylow $3$-subgroup of $N$, and thus $1 < S < N$, and hence $S$ is not normal in $G$. Let $M$ be a maximal subgroup of $G$ containing ${\bf N}_G(S)$. Since $P \subseteq M$, we see that $|G:M|$ is a power of $2$, so it is $2$ or $4$. By the Frattini argument, $NM = G$, so $|N:N \cap M|$ is $2$ or $4$. Then $N$ has a proper normal subgroup of index dividing $4!$. Then $N' < N$, so $N' = 1$ and $N$ is abelian. This completes the proof.

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