When I read in Group Theory of Scott. It has a question, I think it's hard. I have tried to solve it, but I can't.
Problem: "If G is a group whose order is odd and less than 1000, then G is solvable"
I would like to receive some feedback! Thanks so much!
I can do this with just the Sylow theorems with two exceptions, namely $|G| = 3 \cdot 5^2 \cdot 7 = 525$ and $|G| = 3 \cdot 5 \cdot 7^2 = 735$, which is immensely frustrating. Here's what I've got so far.
Case: one prime factor
Finite $p$-groups are nilpotent (because they have nontrivial centers), and so in particular solvable.
Case: two prime factors
Next let's consider groups of order $p^a q^b$. As Derek Holt says in the comments, these groups are all solvable by Burnside's theorem, but the usual proof requires some character theory. Fortunately, for groups of order under $1000$ we'll be able to get away with using just the Sylow theorems to show that such a group has a normal Sylow subgroup, since then it'll be an extension of a nilpotent group by a nilpotent group.
If $n_p$ is the number of Sylow $p$-subgroups, Sylow III tells us that $n_p \equiv 1 \bmod p$ and $n_p \mid q^b$, so we'll be happy if we can always prove that these conditions force $n_p = 1$. The second condition means $n_p = q^k$ for some $k \le b$, so we've proven the following
Lemma: If $G$ is a group of order $p^a q^b$ and the powers $q^k \bmod p, 1 \le k \le b$ are all not equivalent to $1 \bmod p$, then $G$ is solvable.
Suppose a group of odd order $p^a q^b \le 1000$ can't be proven solvable by this lemma (applied to either $p$ or $q$), and that $p < q$. Then some nontrivial power $q^m, k \le b$ of $q$ is congruent to $1 \bmod p$, and some other nontrivial power $p^n, n \le a$ of $p$ is congruent to $1 \bmod q$. Call such an order suspect.
If $p^a q^b$ is suspect we must have $a \ge 2$, since $q$ doesn't divide $p - 1$, being larger than it. In fact we must have $a \ge 3$, since if $p^2 \equiv 1 \bmod q$ then $q$ divides either $p + 1$ or $p - 1$, both of which are impossible because $q > p$ (here we use that $p$ is odd). Now we'll do some casework, which turns out to be a lot nicer than it could: it really helps that $1000$ is just not that big a number. We will repeatedly use the fact that if $p^a q^b \le 1000$ then $p^a \le \left\lfloor \frac{1000}{q^b} \right\rfloor$ and similarly for $q^b$.
Subcase: $b = 1$. Then $q \equiv 1 \bmod p$. If $p \ge 5$, then $q \equiv 1 \bmod 5$ and $q \le \frac{1000}{5^3} = 8$, so $q = 7$; contradiction. So $p = 3$. Then $q \equiv 1 \bmod 3$ and $q \le \lfloor \frac{1000}{3^3} \rfloor = 37$, so $q = 7, 13, 19, 31, 37$. We have $3^3 - 1 = 26$ which is divisible by $13$, so the order $3^3 \cdot 13$ is suspect. If $a \ge 4$ then $q \le \lfloor \frac{1000}{3^4} \rfloor = 12$, so $q = 7$. Since $3^2 - 1 = 8$ and $3^3 - 1 = 26$, the order of $3 \bmod 7$ is $6$, but $3^6 = 343$ is too large. So $3^3 \cdot 13 = 351$ is the only suspect order in this case.
Subcase: $b \ge 2$. Then $q \equiv \pm 1 \bmod p$ and $p \le \left\lfloor \sqrt[3]{ \frac{1000}{5^2} } \right\rfloor = 3$, so $p = 3$. Now $q \le \left\lfloor \sqrt[2]{ \frac{1000}{3^3} } \right\rfloor = 6$, so $q = 5$. So the only possible suspect order in this case is $3^3 \cdot 5^2 = 675$, because neither exponent can be increased. But the order of $3 \bmod 5$ is $4$, so this order isn't suspect.
So in fact $3^3 \cdot 13 = 351$ is the only suspect order! Now, if $G$ is a group with this order with no normal Sylow subgroups, then $n_3 = 13 \equiv 1 \bmod 3$ and $n_{13} = 27 \equiv 1 \bmod 13$. The latter means $G$ has $27$ Sylow $13$-subgroups, each a cyclic group of order $13$, and so $G$ must have at least $27 \cdot 12$ elements of order $13$, leaving only at most $27$ elements of different orders. But $G$ must also have $13$ Sylow $3$-subgroups, and two such subgroups must already occupy at least $2 \cdot 27 - 9 = 45$ elements of $G$, since their intersection has order at most $9$; contradiction.
We conclude that the Sylow theorems already suffice to prove that every group of odd order $p^a q^b \le 1000$ is solvable.
Case: three prime factors
This is the last case left, since $3 \cdot 5 \cdot 7 \cdot 11 = 1155 > 1000$. So, we now consider groups of odd order $p^a q^b r^c$ where $p < q < r$. C Monsour argues that $a \ge 3$ in this case, but the argument uses Burnside's normal $p$-complement theorem which I don't know the proof of, and I promised we can get away with just using the Sylow theorems. As before, it suffices to exhibit a normal Sylow $p$-subgroup, either so that we can reduce to the previous case, or because we can assume WLOG that $G$ is simple (if there is a counterexample then there is a simple counterexample).
First, another
Lemma: If $G$ is a group of order $pqr$, where $p < q < r$ are distinct primes (so $a = b = c = 1$), then $G$ is solvable.
Here we do not need the assumption that $|G|$ is odd.
Proof. If any of $n_p, n_q, n_r$ is equal to $1$ then $G$ has a normal Sylow subgroup (and groups of order a semiprime are solvable), so assume otherwise. Then $n_r = p, q, pq$ and $n_r \equiv 1 \bmod r$, so $n_r \ge r + 1 > p, q$, which means $n_r = pq$, so $G$ has $pq(r - 1)$ elements of order $r$, and hence $pq$ elements of other orders. Similarly $n_p \ge q$ and $n_q \ge r$, so $G$ has at least $(p - 1)q$ elements of order $p$ and at least $(q - 1)r$ elements of order $r$. But
$$(p - 1)q + (q - 1)r > pq$$
(which is equivalent, after some rearrangement, to the claim that $qr > q + r$, or $(q - 1)(r - 1) > 1$, which is true); contradiction. $\Box$
The rest of the argument will be a proof by contradiction starting from the assumption that $G$ is a simple group of odd order $p^a q^b r^c \le 1000$. Note that this means that none of $n_p, n_q, n_r$ can be equal to $1$, which we'll assume implicitly throughout.
By the above lemma, at least one of $a, b, c$ is at least $2$.
Also, if $p \ge 5$ then $|G| \ge 5^2 \cdot 7 \cdot 11 = 1925$, so $p = 3$.
Subcase: $a \ge 3$ (we will handle $a = 2$ separately). Then $q^b r^c \le \left\lfloor \frac{1000}{3^3} \right\rfloor = 37$, which gives $q^b r^c = 35$. So the only possible order here is $|G| = 3^3 \cdot 5 \cdot 7$.
In this case $n_3 = 7$. We can assume WLOG that $G$ is simple, so $G$ acts faithfully on its Sylow $3$-subgroups. But as C Monsour observed, this means $|G| \mid 7!$, but this is a contradiction since $3^3 \mid |G|$.
Subcase: $a = 2$. Then $q^b r^c \le \left\lfloor \frac{1000}{3^2} \right\rfloor = 111$. We have $5^2 \cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q \le \left\lfloor \frac{1000}{3^2 \cdot 11} \right\rfloor = 10$, so $q = 5, 7$.
If $q = 5$, then $r \le \left\lfloor \frac{1000}{3^2 \cdot 5} \right\rfloor = 22$, so $r = 7, 11, 13, 17, 19$. Also $n_r \mid 3^2 \cdot 5$ and $n_r \equiv 1 \bmod r$, which is only possible if $r = 11$ and $n_{11} = 3^2 \cdot 5 = 45$. So $|G| = 3^2 \cdot 5 \cdot 11 = 495$, and $|G|$ has $45 \cdot 10 = 450$ elements of order $11$ and $45$ elements of other orders. But $n_5 = 11$ and $n_3 = 55$, so $G$ has $11 \cdot 4 = 44$ elements of order $11$, leaving no room for elements of order $3$; contradiction.
Subcase: $b \ge 2$. Then $r^c \le \left\lfloor \frac{1000}{3 \cdot 5^2} \right\rfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q \ge 7$ then $r^c \le \left\lfloor \frac{1000}{3 \cdot 7^2} \right\rfloor = 6$; contradiction. So $q = 5$. So the possible orders here are $3 \cdot 5^2 \cdot (7, 11, 13)$.
If $r = 13$ then $n_{13} \mid 3 \cdot 5^2$ and $n_{13} \equiv 1 \bmod 13$ gives $n_{13} = 1$. If $r = 11$ then $n_{11} \mid 3 \cdot 5^2$ and $n_{11} \equiv 1 \bmod 11$ also gives $n_{11} = 1$. So $r = 7$ and $|G| = 3 \cdot 5^2 \cdot 7 = 525$.
In this case we have $n_7 = 15, n_5 = 21$, and $n_3 = 7, 25, 25 \cdot 7 = 175$. As above we can rule out $n_3 = 7$ because $|G| \nmid 7!$, so $n_3 = 25, 175$. At this point I'm stuck!
If $q = 7$ then $r \le \left\lfloor \frac{1000}{3^2 \cdot 7} \right\rfloor = 15$, so $r = 11, 13$. Also $n_r \mid 3^2 \cdot 7$ and $n_r \equiv 1 \bmod r$, which is impossible for either $r = 11$ or $r = 13$; contradiction.
Subcase: $c \ge 2$. Then $q^b \le \left\lfloor \frac{1000}{3 \cdot 7^2} \right\rfloor = 6$, so $q = 5, b = 1$. The only possible order here is $|G| = 3 \cdot 5 \cdot 7^2 = 735$; it is not possible to increase $a$, $r$, or $c$.
In this case we have $n_7 = 15, n_5 = 21$, and $n_3 = 7, 49$. As above we can rule out $n_3 = 7$ because $|G| \nmid 7!$, so $n_3 = 49$. At this point I'm stuck again!
A minimal non-solvable group of odd order must be non-abelian simple, so let us try to find such a group of order less than 1000. Burnside's $p^aq^b$ theorem, the order must be divisible by three primes. By Burnside's transfer theorem, no Sylow subgroup can be in the center of its normalizer. Since groups of order $p$ and $p^2$ are abelian and for odd primes do not have automorphisms of orders of larger primes, the order must be divisible by the cube of the smallest prime. $3^3\cdot 5\cdot 7$ is the only candidate less than 1000. Now consider the number of 3-Sylow subgroups and get a contradiction, since $3^3$ does not divide 7!.
This is a followup to my previous answer. I figured out how to get $|G| = 3 \cdot 5^2 \cdot 7$ and $|G| = 3 \cdot 5 \cdot 7^2$. The arguments are similar (and can probably be simplified) and the other answer is very long so I'll put them here.
First, some preliminary observations, inspired by C Monsour's answer. Sylow III gives that $n_p = \frac{|G|}{|N_G(P)|}$ where $N_G(P)$ is the normalizer of the Sylow $p$-subgroup $P$. In general, the normalizer of a subgroup $H$ acts by conjugation on $H$ with kernel the centralizer $Z_G(H)$; call the quotient $A_G(H)$, so that $|N_G(H)| = |A_G(H)| |Z_G(H)|$, and in particular
$$n_p = \frac{|G|}{|A_G(P)| |Z_G(P)|}.$$
$A_G(H)$ acts faithfully on $H$, and so its order must divide the order of $\text{Aut}(H)$. On the other hand, because $A_G(H)$ is a subquotient of $G$, its order must also divide the order of $G$, and in particular must be odd. In the arguments below we'll repeatedly use these and similar arguments to show that $|A_G(H)| = 1$ and hence that $N_G(H) = Z_G(H)$.
Case: $|G| = 3 \cdot 5^2 \cdot 7 = 525$. We have $n_3 = 25, 175$ and $n_5 = 21, n_7 = 15$ from the previous answer.
$n_7 = 15$ gives $|N_G(C_7)| = |A_G(C_7)| |Z_G(C_7)| = 35$. On the other hand, $|A_G(C_7)|$ divides $|\text{Aut}(C_7)| = 6$, so $|A_G(C_7)| = 1$ and hence $N_G(C_7) = Z_G(C_7)$ must be the cyclic group $C_{35}$.
Let's consider how many conjugates it has; we'll denote this number by $n_{35}$. We have $n_{35} = \frac{|G|}{|A_G(C_{35})| |Z_G(C_{35})|}$, where $Z_G(C_{35}) = C_{35}$ (any other elements in this centralizer must have already centralized $C_7$) and $|A_G(C_{35})|$ divides $|\text{Aut}(C_{35})| = \varphi(35) = 24$ and must be odd, so equals $1$ or $3$. In the latter case $N_G(C_{35})$ must have an element of order $3$, but such an element must have already normalized $C_7$. So $|A_G(C_{35})| = 1$ and $N_G(C_{35}) = C_{35}$, hence $n_{35} = 15$. Moreover, two conjugates of $C_{35}$ can't share any elements of order $35$, from which it follows that there are $15 \cdot 24 = 360$ elements of order $35$.
There are also $15 \cdot 6 = 90$ elements of order $7$ and at least $25 \cdot 2 = 50$ elements of order $3$; in total there are at least $360 + 90 + 50 = 490$ elements of orders $3, 7, 35$, leaving $35$ elements remaining. But a union of two distinct Sylow $5$-subgroups, which have elements of order $1, 5, 25$, has size at least $25 + 25 - 5 = 45$, since they intersect in a subgroup of size at most $5$; contradiction.
Case: $|G| = 3 \cdot 5 \cdot 7^2 = 735$. We have $n_3 = 49, n_5 = 21, n_7 = 15$ from the previous answer.
$|A_G(C_3)|$ divides $|\text{Aut}(C_3)| = 2$ and is odd, so $N_G(C_3) = Z_G(C_3)$. Since $n_3 = 49$, $|N_G(C_3)| = |Z_G(C_3)| = 15$ must be the cyclic group $C_{15}$.
As above we'll consider how any conjugates it must have. $|A_G(C_{15})|$ divides $|\text{Aut}(C_{15})| = 8$ and is odd, so $N_G(C_{15}) = Z_G(C_{15}) = C_{15}$, and so we find that $n_{15} = 49$, from which it follows that there are $49 \cdot 8 = 392$ elements of order $15$.
Similarly, $|A_G(C_5)|$ divides $|\text{Aut}(C_5)| = 4$ and is odd, so $N_G(C_5) = Z_G(C_5)$. Since $n_5 = 21$, $|N_G(C_5)| = |Z_G(C_5)| = 35$ is the cyclic group $C_{35}$. As before, $|A_G(C_{35})|$ divides $|\text{Aut}(C_{35})| = 24$ and is odd, so is equal to $1$ or $3$. As before, in the latter case $N_G(C_{35})$ must contain an element of order $3$, which contradicts $N_G(C_5) = C_{35}$, so $|A_G(C_{35})| = 1$ and $N_G(C_{35}) = C_{35}$, hence $n_{35} = 21$, from which it follows that there are $21 \cdot 24 = 504$ elements of order $35$.
So there are $392 + 504 = 896 > 735$ elements of orders either $15$ or $35$; contradiction.
Probably a lemma can be extracted from these arguments but it is too late in the day for me to see clearly what its hypotheses should be.