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I am trying to solve problem 6.2.16 from Dummit and Foote, namely

Prove there are no non-abelian simple groups of odd order $< 10000$.

I did something similar for order $<100$, where I showed the only non-abelian simple group of order $< 100$ is $A_5$, but this was done by looking at many different forms of order in terms of prime arrangements and picking a few special cases aside to rule everything out but order $60$. The order is too big to do this here. The only thought I have is that $10000=100^2$ and so taking $G$ to be a minimal counterexample with $|G|>100$ we can write $|G|=ab$ with $a<100, b\geq 100$. Apart from that not much that I feel could be constructive. Any help would be appreciated for getting to an elegant proof that doesn't grind through every prime arrangement.

Shaun
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Little Narwhal
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    A deep result (which you probably are not allowed to use) is that groups with odd order are always solvable, hence can only be simple if the order is an odd prime which means however that the group is abelian. – Peter Sep 04 '21 at 11:10
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    @Peter yeah I'm not proving the Feit-Thompson theorem haha. – Little Narwhal Sep 04 '21 at 11:12
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    I think that it's bound to reduce to examining cases. I don't know if it helps but I think that you can show that the largest prime divisor of a counterexample is at most 57. ;-) – ancient mathematician Sep 04 '21 at 11:39
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    @ancientmathematician Ah! The Grothendieck prime reappears! – Kapil Sep 04 '21 at 11:53
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    Are you allowed to use Burnside's theorem that groups of order $p^aq^b$ are solvable for primes $p,q$? That can be used to siginificantly reduce the number of cases to be considered. – Derek Holt Sep 04 '21 at 15:19
  • @DerekHolt I think not as though the result is mentioned in the book earlier, the proof is left to a later part of the book: usually when this is the case the reader is told they are allowed to use such a theorem (in a similar vein I could just use the feit-thompson theorem) – Little Narwhal Sep 04 '21 at 15:21
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    It seems very reasonable to allow using Burnside's theorem here, as well as Burnside's transfer theorem. Then arguing as in https://math.stackexchange.com/a/484777/23805 you know either $|G|$ is divisible by $3^3$ or $|G| = 5^3 \cdot 7 \cdot 11$. – Sean Eberhard Sep 04 '21 at 15:33
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    I think Burnside's $p^aq^b$ theorem, Burnside's transfer theorem, and the Sylow theorems reduce the possibilities to $3^3 \cdot 7 \cdot 13$, $3^3 \cdot 13 \cdot 19$, and $3^4 \cdot 7 \cdot 13$. – Sean Eberhard Sep 04 '21 at 15:58
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    @LittleNarwhal You cannot reasonably say that using Burnside's $p^aq^b$ theorem is in the same vein to using the Feit-Thompson theorem. Burnside's result is a classical application of character theory, which is regularly taught in advanced undergraduate courses, whereas the proof of the Feit-Thompson theorem is over 200 pages, and even specialists in the area would need a long time to read and understand the proof completely. – Derek Holt Sep 04 '21 at 16:02
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    @SeanEberhard Those three orders can be eliminated easily with Sylow's Theorem. For example, for $n=3^4 \cdot 7 \cdot 13$, we ger $n_{13}=27$ and $n_7 = 351$. The first implies there is an element of order $91$ and the second that there is no such element. – Derek Holt Sep 04 '21 at 16:09
  • @DerekHolt fair enough but character theory is addressed about 600 pages later in the book so I do still find the idea unreasonable. – Little Narwhal Sep 04 '21 at 19:39
  • There are a number of specific cases such as orders $pq$ and $p^2q$ that can be handled just by Sylow's theorems. Possibly some of them have featured in earlier results or exercises from Dummit and Foote, so you could use those. – Derek Holt Sep 04 '21 at 22:10

1 Answers1

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If $n$ is the order of a nonabelian finite simple group then $n$ has at least 3 prime factors (Burnside's $p^a q^b$ theorem) and Sylow's theorem implies that the number $n_p$ of Sylow $p$-subgroups is a nontrivial divisor of $n / p^{v_p(n)}$ congruent to $1$ mod $p$.

Moreover, Burnside's transfer theorem implies

  • if $v_p(n) = 1$ then $n / n_p$ shares a factor with $p-1$,
  • if $v_p(n) = 2$ then $n / n_p$ shares a factor with $(p-1)(p+1)$.

Proof: Let $P \leq G$ be a Sylow $p$-subgroup. Assuming $v_p(n) \leq 2$, $P$ is abelian. By Burnside's transfer theorem (see the answer at https://math.stackexchange.com/a/484777/23805 for discussion), $N_G(P)$ acts nontrivially on $P$. Since $P$ is abelian, $P \leq C_G(P)$, so $N_G(P)/C_G(P)$ is isomorphic to a nontrivial subgroup of $\operatorname{Aut}(P)$ of $p'$-order. Hence $|N_G(P)| = n / n_p$ and $|\operatorname{Aut}(P)|$ share a prime factor other than $p$. Now consider the possibilities for $P$ and $\operatorname{Aut}(P)$. $\square$

Trawling through all odd $n \leq 10000$ (I recommend using technology) shows that we have reduced to three possibilities:

  • $3^3 \cdot 5^2 \cdot 7$.
  • $3^{4} \cdot 7 \cdot 13$.
  • $3^{3} \cdot 5^{2} \cdot 11$.

Probably @DerekHolt can rule these out using Sylow's theorem in clever ways.

Sean Eberhard
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  • Thanks, I won't mark this as answered quite yet in case someone finds a proof not using Burnside, but ill settle for it otherwise. Ill figure out the three special cases on my own. – Little Narwhal Sep 04 '21 at 19:29
  • @LittleNarwhal If you aren't happy with Burnside's $p^aq^b$ but are happy with the transfer theorem then the same argument can still be used and you just have four additional possibilities: $3^4 \cdot 13, 3^5 \cdot 13, 3^5 \cdot 5^2, 3^6 \cdot 13$. – Sean Eberhard Sep 04 '21 at 22:49
  • the problem is Burnside's transfer theorem also won't really work out since only a weaker version of it is proved for solvable groups in a previous exercise, and I know the proof of it uses transfer theory which is not at all studied in the book. – Little Narwhal Sep 06 '21 at 09:36