# Largest possible order of a nilpotent permutation group?

I'm trying to obtain a bound for the order of some finite groups, and part of it comes down to the order of a permutation group of degree $n$ that is nilpotent. I imagine these have to be much smaller than the full symmetric group, and a bound that is sub-exponential in $n$ would seem reasonable (given that permutation $p$-groups fall a long way short of having exponential order), but I haven't seen this written down anywhere.

I found one reference that looks promising:

P. Palfy, Estimations for the order of various permutation groups, Contributions to general algebra, 12 (Vienna, 1999), 37-49, Heyn, Klagenfurt, 2000.

However, I can't actually find the article anywhere online. Any suggestions?

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From springerlink.com/content/d6n24hgu6x180r64 , it follows that any nilpotent subgroup of $S_n$ has order less than $\sqrt{2n!}$. I can't access the article, but from the introduction, it seems this bound may be greatly improved within the paper; it talks about maximal nilpotent subgroups of $A_n$ corresponding to Sylow subgroups. –  Steve D Feb 18 '10 at 14:24

The paper of Vdovin mentioned by Steve shows that the nilpotent subgroups of the symmetric groups of maximal order are either the Sylow 2-subgroups P(n) of Sym(n), or P(n-3) x Alt(3) when n = 2(2k+1)+1.

Vdovin, E. P. "Large nilpotent subgroups of finite simple groups." Algebra Log. 39 (2000), no. 5, 526-546, 630; translation in Algebra and Logic 39 (2000), no. 5, 301-312. http://www.ams.org/mathscinet-getitem?mr=1805754 http://dx.doi.org/10.1007/BF02681614

It looks at the orbits of the center of the nilpotent group, and the action of the quotient on those orbits to give a reasonable bound. Then it shows that all nilpotent subgroups of maximal order are conjugate to the types mentioned.

The paper also handles alternating groups, groups of Lie type, and sporadic groups. For groups of Lie type, the maximal order nilpotent is usually a Sylow p-subgroup for p the characteristic. Sporadics are only handled briefly: the nilpotent subgroups of maximal order are always Sylows and they satisfy the same general bound as the other simple groups.

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Thanks, that sounds like exactly what I needed. –  Colin Reid Feb 18 '10 at 15:21
The 2-sylow of S_n has order 2^([n/2]+[n/4]+...) <= 2^n, where [x] is the integer part of x. So your conjectured exponential bound is correct. –  David Speyer Feb 18 '10 at 18:32