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~
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~
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.
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.)
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 $2^n$.