Odd order groups with order less than 1000 are solvable











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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!










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    What did you try?
    – lhf
    Nov 13 at 17:24










  • This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
    – lhf
    Nov 13 at 17:25






  • 3




    You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
    – Derek Holt
    Nov 13 at 17:52






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    Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
    – rschwieb
    Nov 13 at 17:59

















up vote
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down vote

favorite
1












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!










share|cite|improve this question




















  • 2




    What did you try?
    – lhf
    Nov 13 at 17:24










  • This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
    – lhf
    Nov 13 at 17:25






  • 3




    You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
    – Derek Holt
    Nov 13 at 17:52






  • 1




    Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
    – rschwieb
    Nov 13 at 17:59















up vote
4
down vote

favorite
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up vote
4
down vote

favorite
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1





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!










share|cite|improve this question















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!







abstract-algebra group-theory finite-groups solvable-groups






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edited Nov 15 at 16:08









Martin Sleziak

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asked Nov 13 at 17:17









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  • 2




    What did you try?
    – lhf
    Nov 13 at 17:24










  • This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
    – lhf
    Nov 13 at 17:25






  • 3




    You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
    – Derek Holt
    Nov 13 at 17:52






  • 1




    Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
    – rschwieb
    Nov 13 at 17:59
















  • 2




    What did you try?
    – lhf
    Nov 13 at 17:24










  • This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
    – lhf
    Nov 13 at 17:25






  • 3




    You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
    – Derek Holt
    Nov 13 at 17:52






  • 1




    Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
    – rschwieb
    Nov 13 at 17:59










2




2




What did you try?
– lhf
Nov 13 at 17:24




What did you try?
– lhf
Nov 13 at 17:24












This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
– lhf
Nov 13 at 17:25




This follows from the Feit–Thompson theorem, but I doubt that the book means to use that...
– lhf
Nov 13 at 17:25




3




3




You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
– Derek Holt
Nov 13 at 17:52




You have to do every order individually! Are you allowed to use the Burnside Theorem that groups of order $p^aq^b$ (with $p,q$ prime) are solvable?
– Derek Holt
Nov 13 at 17:52




1




1




Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
– rschwieb
Nov 13 at 17:59






Please include your tries at a solution that you alluded to. After all, why let that work go to waste? If in fact you didn't do anything, see this advice on how to add other context to make your question more than just a problem dump. You're welcome for the feedback.
– rschwieb
Nov 13 at 17:59












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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 $p^a le leftlfloor frac{1000}{q^b} rightrfloor$ 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 $p le leftlfloor sqrt[3]{ frac{1000}{5^2} } rightrfloor = 3$, so $p = 3$. Now $q le leftlfloor sqrt[2]{ frac{1000}{3^3} } rightrfloor = 6$, 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 $q^b r^c le leftlfloor frac{1000}{3^3} rightrfloor = 37$, 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 $q^b r^c le leftlfloor frac{1000}{3^2} rightrfloor = 111$. We have $5^2 cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q le leftlfloor frac{1000}{3^2 cdot 11} rightrfloor = 10$, so $q = 5, 7$.



If $q = 5$, then $r le leftlfloor frac{1000}{3^2 cdot 5} rightrfloor = 22$, 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 $r^c le leftlfloor frac{1000}{3 cdot 5^2} rightrfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q ge 7$ then $r^c le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$; 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 $r le leftlfloor frac{1000}{3^2 cdot 7} rightrfloor = 15$, 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 $q^b le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$, 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!






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    Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
    – Qiaochu Yuan
    Nov 14 at 5:40










  • If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
    – j.p.
    Nov 14 at 7:00






  • 1




    Okay, I have the last two cases but my arguments are incredibly annoying.
    – Qiaochu Yuan
    Nov 14 at 7:15


















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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. $3^3cdot 5cdot 7$ 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!.






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  • I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
    – Qiaochu Yuan
    Nov 14 at 4:41






  • 1




    @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
    – Tobias Kildetoft
    Nov 14 at 6:06










  • Aha, of course. Thanks!
    – Qiaochu Yuan
    Nov 14 at 6:09


















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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.






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    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 $p^a le leftlfloor frac{1000}{q^b} rightrfloor$ 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 $p le leftlfloor sqrt[3]{ frac{1000}{5^2} } rightrfloor = 3$, so $p = 3$. Now $q le leftlfloor sqrt[2]{ frac{1000}{3^3} } rightrfloor = 6$, 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 $q^b r^c le leftlfloor frac{1000}{3^3} rightrfloor = 37$, 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 $q^b r^c le leftlfloor frac{1000}{3^2} rightrfloor = 111$. We have $5^2 cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q le leftlfloor frac{1000}{3^2 cdot 11} rightrfloor = 10$, so $q = 5, 7$.



    If $q = 5$, then $r le leftlfloor frac{1000}{3^2 cdot 5} rightrfloor = 22$, 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 $r^c le leftlfloor frac{1000}{3 cdot 5^2} rightrfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q ge 7$ then $r^c le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$; 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 $r le leftlfloor frac{1000}{3^2 cdot 7} rightrfloor = 15$, 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 $q^b le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$, 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!






    share|cite|improve this answer



















    • 1




      Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
      – Qiaochu Yuan
      Nov 14 at 5:40










    • If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
      – j.p.
      Nov 14 at 7:00






    • 1




      Okay, I have the last two cases but my arguments are incredibly annoying.
      – Qiaochu Yuan
      Nov 14 at 7:15















    up vote
    6
    down vote













    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 $p^a le leftlfloor frac{1000}{q^b} rightrfloor$ 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 $p le leftlfloor sqrt[3]{ frac{1000}{5^2} } rightrfloor = 3$, so $p = 3$. Now $q le leftlfloor sqrt[2]{ frac{1000}{3^3} } rightrfloor = 6$, 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 $q^b r^c le leftlfloor frac{1000}{3^3} rightrfloor = 37$, 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 $q^b r^c le leftlfloor frac{1000}{3^2} rightrfloor = 111$. We have $5^2 cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q le leftlfloor frac{1000}{3^2 cdot 11} rightrfloor = 10$, so $q = 5, 7$.



    If $q = 5$, then $r le leftlfloor frac{1000}{3^2 cdot 5} rightrfloor = 22$, 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 $r^c le leftlfloor frac{1000}{3 cdot 5^2} rightrfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q ge 7$ then $r^c le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$; 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 $r le leftlfloor frac{1000}{3^2 cdot 7} rightrfloor = 15$, 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 $q^b le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$, 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!






    share|cite|improve this answer



















    • 1




      Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
      – Qiaochu Yuan
      Nov 14 at 5:40










    • If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
      – j.p.
      Nov 14 at 7:00






    • 1




      Okay, I have the last two cases but my arguments are incredibly annoying.
      – Qiaochu Yuan
      Nov 14 at 7:15













    up vote
    6
    down vote










    up vote
    6
    down vote









    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 $p^a le leftlfloor frac{1000}{q^b} rightrfloor$ 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 $p le leftlfloor sqrt[3]{ frac{1000}{5^2} } rightrfloor = 3$, so $p = 3$. Now $q le leftlfloor sqrt[2]{ frac{1000}{3^3} } rightrfloor = 6$, 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 $q^b r^c le leftlfloor frac{1000}{3^3} rightrfloor = 37$, 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 $q^b r^c le leftlfloor frac{1000}{3^2} rightrfloor = 111$. We have $5^2 cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q le leftlfloor frac{1000}{3^2 cdot 11} rightrfloor = 10$, so $q = 5, 7$.



    If $q = 5$, then $r le leftlfloor frac{1000}{3^2 cdot 5} rightrfloor = 22$, 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 $r^c le leftlfloor frac{1000}{3 cdot 5^2} rightrfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q ge 7$ then $r^c le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$; 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 $r le leftlfloor frac{1000}{3^2 cdot 7} rightrfloor = 15$, 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 $q^b le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$, 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!






    share|cite|improve this answer














    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 $p^a le leftlfloor frac{1000}{q^b} rightrfloor$ 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 $p le leftlfloor sqrt[3]{ frac{1000}{5^2} } rightrfloor = 3$, so $p = 3$. Now $q le leftlfloor sqrt[2]{ frac{1000}{3^3} } rightrfloor = 6$, 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 $q^b r^c le leftlfloor frac{1000}{3^3} rightrfloor = 37$, 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 $q^b r^c le leftlfloor frac{1000}{3^2} rightrfloor = 111$. We have $5^2 cdot 7 = 175 > 111$, so $b = c = 1$. We also have $q le leftlfloor frac{1000}{3^2 cdot 11} rightrfloor = 10$, so $q = 5, 7$.



    If $q = 5$, then $r le leftlfloor frac{1000}{3^2 cdot 5} rightrfloor = 22$, 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 $r^c le leftlfloor frac{1000}{3 cdot 5^2} rightrfloor = 13$, so $r = 7, 11, 13$ and $c = 1$. If $q ge 7$ then $r^c le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$; 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 $r le leftlfloor frac{1000}{3^2 cdot 7} rightrfloor = 15$, 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 $q^b le leftlfloor frac{1000}{3 cdot 7^2} rightrfloor = 6$, 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!







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 14 at 5:27

























    answered Nov 14 at 4:55









    Qiaochu Yuan

    274k32578914




    274k32578914








    • 1




      Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
      – Qiaochu Yuan
      Nov 14 at 5:40










    • If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
      – j.p.
      Nov 14 at 7:00






    • 1




      Okay, I have the last two cases but my arguments are incredibly annoying.
      – Qiaochu Yuan
      Nov 14 at 7:15














    • 1




      Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
      – Qiaochu Yuan
      Nov 14 at 5:40










    • If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
      – j.p.
      Nov 14 at 7:00






    • 1




      Okay, I have the last two cases but my arguments are incredibly annoying.
      – Qiaochu Yuan
      Nov 14 at 7:15








    1




    1




    Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
    – Qiaochu Yuan
    Nov 14 at 5:40




    Apparently there are many facts that will rule out the cases I'm stuck on (groupprops.subwiki.org/wiki/…) but I don't know how to prove any of them.
    – Qiaochu Yuan
    Nov 14 at 5:40












    If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
    – j.p.
    Nov 14 at 7:00




    If two $p$-Sylow subgroups intersect non-trivially, it often helps to look at the normalizer of this non-trivial intersection by applying Sylow's theorems to it.
    – j.p.
    Nov 14 at 7:00




    1




    1




    Okay, I have the last two cases but my arguments are incredibly annoying.
    – Qiaochu Yuan
    Nov 14 at 7:15




    Okay, I have the last two cases but my arguments are incredibly annoying.
    – Qiaochu Yuan
    Nov 14 at 7:15










    up vote
    3
    down vote













    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. $3^3cdot 5cdot 7$ 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!.






    share|cite|improve this answer























    • I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
      – Qiaochu Yuan
      Nov 14 at 4:41






    • 1




      @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
      – Tobias Kildetoft
      Nov 14 at 6:06










    • Aha, of course. Thanks!
      – Qiaochu Yuan
      Nov 14 at 6:09















    up vote
    3
    down vote













    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. $3^3cdot 5cdot 7$ 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!.






    share|cite|improve this answer























    • I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
      – Qiaochu Yuan
      Nov 14 at 4:41






    • 1




      @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
      – Tobias Kildetoft
      Nov 14 at 6:06










    • Aha, of course. Thanks!
      – Qiaochu Yuan
      Nov 14 at 6:09













    up vote
    3
    down vote










    up vote
    3
    down vote









    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. $3^3cdot 5cdot 7$ 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!.






    share|cite|improve this answer














    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. $3^3cdot 5cdot 7$ 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!.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 14 at 18:44

























    answered Nov 13 at 23:30









    C Monsour

    5,562224




    5,562224












    • I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
      – Qiaochu Yuan
      Nov 14 at 4:41






    • 1




      @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
      – Tobias Kildetoft
      Nov 14 at 6:06










    • Aha, of course. Thanks!
      – Qiaochu Yuan
      Nov 14 at 6:09


















    • I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
      – Qiaochu Yuan
      Nov 14 at 4:41






    • 1




      @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
      – Tobias Kildetoft
      Nov 14 at 6:06










    • Aha, of course. Thanks!
      – Qiaochu Yuan
      Nov 14 at 6:09
















    I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
    – Qiaochu Yuan
    Nov 14 at 4:41




    I think I've almost understood this argument, but I'm not quite sure how you get that the exponent of the smallest prime is at least $3$. I can show that if it's at most $2$ then the index of the centralizer of a Sylow $p$-subgroup $P$ (where $p$ is the smallest prime dividing $|G|$) in its normalizer must be either $1$ or $p$. The index being $1$ is ruled out by Burnside's transfer theorem; how do you rule out the index being $p$?
    – Qiaochu Yuan
    Nov 14 at 4:41




    1




    1




    @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
    – Tobias Kildetoft
    Nov 14 at 6:06




    @Qiaochu The centralizer contains the Sylow subgroup, so the centralizer already has order divisible by the largest possible power of the prime.
    – Tobias Kildetoft
    Nov 14 at 6:06












    Aha, of course. Thanks!
    – Qiaochu Yuan
    Nov 14 at 6:09




    Aha, of course. Thanks!
    – Qiaochu Yuan
    Nov 14 at 6:09










    up vote
    0
    down vote













    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.






    share|cite|improve this answer

























      up vote
      0
      down vote













      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.






      share|cite|improve this answer























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        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.






        share|cite|improve this answer












        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.







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        answered Nov 14 at 8:08









        Qiaochu Yuan

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