Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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~

share|improve this question
5  
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
2  
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
add comment

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.

share|improve this answer
    
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
add comment

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.)

share|improve this answer
    
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
add comment

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 $2^n$.

share|improve this answer
add comment

Your Answer

 
discard

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.