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P. P. Palfy proved that a primitive solvable subgroup of $S_n$ has order bounded by $24^{-1/3} n^{3.24399\dots}$ (in: Pálfy, P. P. A polynomial bound for the orders of primitive solvable groups. J. Algebra 77 (1982), no. 1, 127–137. )

This is sharp (and notice that for $n$ prime, the affine group of the line (all transformations of the form $x \rightarrow a x + b$ is solvable, and is of order $O(n^2)$

I seem to be unable to find any sort of analogue for nilpotent groups (there are results on large nilpotent groups, but these are presumably [by Palfy's theorem!) NOT primitive, nor even transitive).

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Nilpotent groups are not even semiprimitive, see, Lemma $2.3$. –  Dietrich Burde Jul 23 '13 at 14:41
@Dietrich: thanks! But notice that I hedged by saying "some analogue"... Maybe just transitive? –  Igor Rivin Jul 23 '13 at 14:45
you mean that the ORDER is bounded (above), not the degree, right? –  Nick Gill Jul 23 '13 at 15:12
If you take $n=2^k$, then a Sylow $2$-subgroup of $S_n$ will be transitive... and it is rather large. It has order $$2^{1+2+4+8+\cdots 2^{k-1}}=2^{2^k-1}\sim \exp(n).$$ So I feel like you'll need more than just transitivity to say anything very much... –  Nick Gill Jul 23 '13 at 15:19
@NickGill Yes, fixed (re degree). Otherwise, yes, you are right, but what is the right condition to get a sensible result? –  Igor Rivin Jul 23 '13 at 15:25

2 Answers 2

This answer fleshes out observations made in comments above. The result below is an analogue of Palfy's theorem for nilpotent primitive groups, as requested.

Prop. Let $N<S_n$ be a nilpotent primitive group. Then $n$ is prime and $N$ is cyclic of order $n$.

Proof. Since $N$ is nilpotent, a minimal normal subgroup $E$ of $N$ is elementary-abelian of order $p^a$. Since $N$ is primitive it must be transitive and, since $E$ is abelian, it must act regularly - so $n=p^a$. Indeed $E$ must be the unique minimal normal subgroup of $N$ because if there another, $E'$ say, then $E'$ would centralize $E$ and $EE'$ would be a transitive abelian subgroup of $S_n$ of order greater than $n$, which is impossible.

Now, since $E$ is unique, we conclude that $C_N(E)=E$. In particular if $g$ is an element of order coprime to $p$, then $g\not\in C_N(E)$. But this contradicts the fact that $N$ is nilpotent. Thus $N$ is a $p$-group. Then $N$ acts on $E$ via linear transformations and so $N/E$ is a $p$-subgroup of $GL_a(p)$ and, in particular, fixes a 1-dimensional subspace of $E$. This subspace is normal in $N$ and hence, since $N$ is primitive, is transitive. i.e. $E$ is itself 1-dimensional, i.e. $E=C_p$ with $n=p$. But, since $N$ is a $p$-group inside $S_p$ we conclude that $N=E=C_p$ as required. QED

In fact, rather than primitivity, all I've used here is that $N$ has no intransitive normal subgroups. This property is called quasiprimitivity - it is a little weaker than primitivity.

There is an interesting sort of strong converse to this result which also sheds light on the original question.

Prop. Let $n$ be a prime and $N<S_n$ be a nilpotent transitive group. Then $N$ is cyclic of order $n$.

Proof. Since $N$ is transitive it contains a cyclic subgroup $C$ of order $n$. But $C_{S_n}(C)=C$ and so $N$ must be an $n$-group. But then $N=C$, as required. QED

This result, along with the example given above in the comments - a Sylow $2$-subgroup when $n=2^k$ - demonstrate that bounding the order of a nilpotent transitive group is strongly dependent on the prime factorization of $n$. It's not clear to me whether there is a natural stronger condition than transitivity that will hold for any $n$...

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I suggest you read the following paper by Pálfy. I don't have it handy here and it doesn't seem to be downloadable, but I recall he surveys similar results on the orders of nilpotent permutation groups.

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

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