# [seqfan] Re: Symmetric group S_n as product of at most A225788(n) cyclic subgroups.

L. Edson Jeffery lejeffery2 at gmail.com
Sun May 26 00:56:49 CEST 2013

```On Sat, May 25, 2013 at 3:47 PM, W. Edwin Clark <wclark at mail.usf.edu> wrote:

>Concerning the sequence a(n) = smallest k such that the symmetric group S_n
>is a product of  k cyclic subgroups:

>I calculate that for n from 1 to 7 we have a = 1,1,2,3,3,4,4. I used GAP
>and a more or less brute force approach. Perhaps someone can do better?
>The computation depends on the number of distinct cyclic subgroups of S_n
>--given by http://oeis.org/A051625  Note that for n = 8 there are 14170
>cyclic subgroups. Using the fact that we can fix one of the cyclic
>subgroups to be generated by an element in one of the p(8) = 15 conjugacy
>classes of S_8, this still means that to prove that a(8) is not 4 (see
>lower bound below) requires checking possibly 15*14170^3  = 42677680695000
>products  of  4 cyclic subgroups of S_8.  Or am I missing something?

>Using the idea in the paper by Miklos Abert  mentioned in A225788 it is
>easy to see that if m is the maximal order of an element of of S_n (
http://oeis.org/A000793 ) then m^a(n) >= n! and hence a(n) >= log[m](n!).

>The sequence log[m](n!), m = A000783(n):
>      1,1,2,3,3,4,4,4,5,5,6,5,6,6,6,7,7,7,7,8,8,9,8,9,9,9,9,9,10,
>at the beginning matches the sequence a, and apparently is not in the OEIS
>--Edwin

Edwin, I think you should submit your sequence, even though only the first
seven terms are verified. (More terms can be appended as they are worked
out.)

I don't know if the following will help or not. We have that not all
abelian groups are cyclic, but every cyclic group is abelian. Now,
according to another proof by Abért, if

f(n) = min(k | S_n is a product of k abelian subgroups),

then

f(2) = 1 and f(n+1) <= f(n) + 1.

So, does it follow that if

a(n) = min(j | S_n is a product of j cyclic subgroups),

then a(n) <= f(n)?

Ed Jeffery
```