MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The theorem I am referring to in the title is this:

Theorem. If $p$ is a prime and $n$ is an integer with $n \geq p+3$, then the only primitive permutation groups on $n$ points containing a $p$-cycle are $A_{n}$ and $S_{n}$.

Without using the Classification of Finite Simple Groups, it seems hopeless to extend this to a statement detailing when it's possible to have a $p$-cycle in a primitive permutation group on $p+1$ points, since the Mathieu groups $M_{11}$, $M_{12}$ and $M_{24}$ arise this way when $p = 11$ or $23$.

The intermediate case is when $n=p+2$, and here there is a readily available family of examples: If $p = 2^{q}-1$ is a Mersenne prime, then the group $SL(2,2^{q})$ is primitive on the $2^{q}+1$ points of the projective line over $\mathbb{F_{2^{q}}}$. It contains $(2^{q}-1)$-cycles as the only elements fixing exactly 2 points. These groups can also be enlarged by adjoining field automorphisms, though the primality of $q$ means that the only larger group obtainable this way is the whole $\Sigma L(2,2^{q}) = SL(2,2^{q}):q$.

I have read that Burnside was able to prove that the only finite simple groups of even order in which every element has order 2 or odd order are $SL(2,2^{m})$ for some $m \geq 2$. My question is: Are methods not much stronger than the methods Burnside used to prove that result usable to prove no groups aside from those already mentioned act primitively on $p+2$ points and contain $p$-cycles?

share|cite|improve this question
I believe your group will be 2-transitive, since a point stabilizer will be transitive (even primitive), but of course, that just reduces to the case that needs CFSG. – Steve D Jan 9 '12 at 12:29
The existence of the $p$-cycle means that it will even be 3-transitive! But I suspect that you still need CFSG to finish this off. – Derek Holt Jan 9 '12 at 14:17
In fact, the stabilizer of two points is a permutation group of degree p, so by Burnside, it is either 2-transitive or else solvable of order pm, where m divides p-1. This means that the original group is either 4-transitive or has order of the form (p+2)(p+1)pm. I agree with Derek that to finish off the 4-transitive case probably requires CFSG, but I wouldn't be surprised if the not-4-transitive case can be done without the classification. Note that if m = p-1, the group cannot occur if p > 2 because in that case it would be sharply 4-transitive, and the only such are S_4, A_6 and M_11. – Marty Isaacs Feb 22 '12 at 21:52
Minor quibble: $S_{5} = \Sigma L(2,2^{2})$ is sharply 4-transitive, so "p > 2" should be strengthened to "p > 3". – DavidLHarden Feb 25 '12 at 18:29
In addition to the Mathieu groups mentioned by David, there are the examples $PSL(2,p)\le S_{p+1}$ in the case $n= p+1$; and solvable examples $\Gamma L(1,2^q)\le S_{2^q}$ where $p=2^q-1$ is a Mersenne prime. – Glasby Aug 22 '14 at 6:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.