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What are the solvable subgroup of $S_n$?

I know that when $n \geq 5$, both $S_n$ and $A_n$ are not solvable. But, how "large" can a solvable permutation group be when $n$ is given.

Many thanks~

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The second question seems much more tractable than the first. Of course any finite solvable group is a subgroup of $S_n$ for some $n$... –  Qiaochu Yuan May 1 '11 at 17:42
Thanks for taking time to comment. I am sorry that I didn't make it clear. It is true that every finite group is a subroup of $S_n$ for some $n$. What I want to know is when $n$ is given, what the structures of solvable subgroups of $S_n$ are, and how "large" can they be. –  ShinyaSakai May 1 '11 at 17:54

3 Answers 3

up vote 13 down vote accepted

I'll assume you mean: how large can the order of a solvable subgroup of the symmetric group on n points?

The largest nilpotent subgroup of the symmetric group of degree 2n is the Sylow 2-subgroup, an iterated wreath product of a very simple nilpotent group, the symmetric group on 2 points. The largest nilpotent subgroup of a general symmetric group is just a direct product of these and sometimes the alternating group on three points.

Similarly, the largest solvable subgroup of the symmetric group of degree 4n is the iterated wreath product of a very compact solvable permutation group, the symmetric group on 4 points. Using similar ideas, the associated bound for the order can be shown to hold for all symmetric groups:

In Dixon (1967), it is shown that if G ≤ Sym(n) is a solvable permutation group of degree n, then |G| ≤ k(n−1), where k≈2.88 is the cube root of 24.

Often one is interested in transitive or even primitive groups. Then much smaller bounds are available, but the actual maximum order of a solvable transitive or primitive group of degree n depends as much on the arithmetic properties of n as on its size. For instance, if n is prime and G is transitive, then |G| ≤ n(n−1) is a much smaller upper bound (attained by AGL(1,n)). If G is solvable and primitive of degree n, then |G| ≤ n4 (which even holds without the solvable hypothesis, by Prager–Saxl (1980)). Pálfy (1982) gives even better bounds.

Dixon, John D. "The Fitting subgroup of a linear solvable group." J. Austral. Math. Soc. 7 (1967) 417–424. MR230814 DOI:10.1017/S1446788700004353

Pálfy, P. P. "A polynomial bound for the orders of primitive solvable groups." J. Algebra 77 (1982), no. 1, 127–137. MR665168 DOI:10.1016/0021-8693(82)90281-2.

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Note that Dixon's bound is attained (as Dixon himself remarked), whenever $n = 4^d$, using iterated wreath products of $S_4$ with itself. –  Geoff Robinson Jul 10 '11 at 9:51

If $n = p$ is prime, then any transitive solvable subgroup of $S_p$ is contained in the group of affine transformations $x \mapsto ax + b$ (here $a \in \mathbb F_p^{\times}$ and $b \in \mathbb F_p$), thought of as a group of transformations from $\mathbb F_p$ (a set of order $p$) to itself. (This result goes back to Galois himself, and was one of the motivations for his invention of finite fields.)

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Could you kindly tell me where can i get account of the result of galois regarding prime degree polynomials...@MattE –  sayak Apr 28 at 15:29
Dear Sayak, Have you looked in E. Artin's book on Galois theory? I'm told that it is in there, although I haven't studied that book myself. I learnt about it from Hans Wussing's book on the history of group theory. Regards, –  Matt E Apr 28 at 23:03

Here's a lower bound. For a prime $p$ and an integer $n$ let $\nu_p(n)$ denote the greatest power of $p$ dividing $n$. Recall that $$\nu_p(n!) = \sum_{k \ge 1} \left\lfloor \frac{n}{p^k} \right\rfloor \approx \frac{n}{p-1}.$$

It follows that the Sylow $p$-subgroup of $S_n$ (which, being a $p$-group, is automatically solvable) has order about $p^{ \frac{n}{p-1} }$. For fixed $n$ this is maximized when $p = 2$, giving a solvable subgroup of $S_n$ of order about $2^n$.

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