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Let $G$ be a degree $n$ group, i.e., a subgroup of the symmetric group $S_n$. Let $p(G)$ be the number of $n$-cycles in $G$ divided by the size of $G$.

Examples:

  1. If $G$ is a cyclic transitive group, then $p=\varphi(n)/n$.
  2. If $G=S_n$, then $p=1/n$.
  3. (If $G$ is not transitive, then $p=0$)

The question is whether $p(G)\leq \varphi(n)/n$ for every degree $n$ group?

Note:

  1. One can see that $p(G)=k/n$, where $k$ is the number of conjugacy classes of $n$-cycles, so the answer is YES if $n$ is prime.
  2. Numerical testing shows the answer is YES for $n\leq 30$ and for primitive groups for $n\leq 1000$.
  3. There are non-cyclic groups achieving the bound $\varphi(n)/n$, e.g., the wreath product of cyclic groups.

Edit: Recently Joachim König solved this using the classification both in the induction basis as Michael Giudici mentioned and also in the induction step. I guess we should wait for the paper which is now in refereeing process.

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Idle question: if G is a finite group endowed with an n-dimensional complex representation rho, do we expect the same bound on the number of elements h of G such that the restriction of G to the cyclic subgroup <h> is the regular representation of <h>? Sometimes results about characters are easier to prove than results about permutations. –  JSE Jan 11 '10 at 16:40
    
If you already have some code checking the conjecture for given $n$'s, can you check the stronger version that sets $p\left(G\right)$ to be the number of elements in $G$ of order $n$ ? –  darij grinberg Jan 11 '10 at 23:48
    
Another remark: the "cycle index" of a permutation group is a generating function keeping track of the cycle types of all elements of G; so this might be a good keyword for identifying papers or people possessing some insight into this problem. –  JSE Jan 12 '10 at 1:49
    
@darij, why this is stronger? if you know how many elements of order $n$ there are, does it gives you information about $n$-cycles, or just a bound? –  Lior Bary-Soroker Jan 12 '10 at 3:36
    
Because $n$-cycles have order $n$... but it's only stronger if it's true ;) –  darij grinberg Jan 12 '10 at 13:06
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2 Answers

It is true for all primitive groups: The primitive groups of degree n containing an n-cycle were independently classified in

Li, Cai Heng The finite primitive permutation groups containing an abelian regular subgroup. Proc. London Math. Soc. (3) 87 (2003), no. 3, 725--747. )

and

Jones, Gareth A. Cyclic regular subgroups of primitive permutation groups. J. Group Theory 5 (2002), no. 4, 403--407.

They are the groups G such that

-$C_p\leqslant G\leqslant AGL(1,p)$ for p a prime

-$A_n$ for n odd, or $S_n$

-$PGL(d,q)\leqslant G \leqslant P\Gamma L(d,q)$: here there is a unique class of cyclic subgroups generated by an n-cycle except for $G=P\Gamma L(2,8)$ in which case there are two.

-$(G,n)=(PSL(2,11),11), (M_{11},11), (M_{23},23)$

All these groups satisfy the bound.

Gordon Royle has pointed out to me that the bound does not hold for elements of order n. The smallest examples which do not meet the bound are of degree 12 and are the groups numbered 263 and 298 in the catalogue of groups of degree 12 in Magma.

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Do you use the classification of finite simple groups? In any rate, I tried quite a lot to reduce the problem to primitive groups but failed... –  Lior Bary-Soroker Jan 13 '10 at 14:34
    
According to the MathSciNet review, Heng uses the classification. For some reason, MIT won't let me get at the Journal of Group Theory right now, so I don't know about Jones. –  David Speyer Jan 13 '10 at 17:48
    
A classical result of Burnside states that a primitive group of degree n which contains an n-cycle is either contained in AGL(1,p) or is 2-transitive. The classification of 2-transitive groups (which relies on the classification of finite simple groups) then gives the result. –  Michael Giudici Jan 14 '10 at 1:00
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This answer exists to record a false approach I had.

Let $G$ be a subgroup of $S_n$. Call a subgroup of $G$ "maximal cyclic" if it is generated by an $n$-cycle. False Statement: Any two maximally cyclic subgroups are conjugate.

This is true in a number of cases, and implies the claimed result. However, as Dmitri led me to realize, it is false for the alternating group $A_9$.

I used to have a longer answer explaining this. I deleted it and replaced it with this answer because it got an upvote which, as I understand it, removed this excellent question from the unanswered list. So PLEASE DON'T VOTE THIS UP! Let's see if someone can come up with a real answer!

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As a piece of insurance (and, okay, because it's somewhat amusing to do so), I am downvoting your answer. I will remove it upon request, or when it becomes clear that it is not necessary. –  Pete L. Clark Dec 28 '09 at 14:04
    
Thanks for the insurance! –  David Speyer Dec 28 '09 at 14:32
    
Actually, could you try removing the downvote? This question isn't on the unanswered page now, and it was before your downvote, so I suspect that your vote is at fault. –  David Speyer Dec 28 '09 at 20:03
    
Okay, done. (But I had to edit your answer in order to do it. Something is a little screwy here...) –  Pete L. Clark Dec 28 '09 at 21:46
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