# [seqfan] Re: Symmetric groups

Nikos Apostolakis nikos.ap at gmail.com
Mon Nov 28 02:43:42 CET 2016

```Following Robert's observation, a subgroup of order 35 is cyclic and the
first S_n that has a subgroup of order 35 is S_13.  So for n from 5 to 12
the sequence should be 15, 15, 15, 28, 35, 35, 35, 35

Nikos

On Sun, Nov 27, 2016 at 8:07 PM W. Edwin Clark <wclark at mail.usf.edu> wrote:

> For n from 5 to 9,  GAP gives the sequence
>                                           15, 15, 15, 28, 35
> where a(n) is the smallest m dividing n! for which S_n has no subgroup of
> order m. Perhaps someone can extend it.
>
> A brief web search finds only this result answering the original question:
>                                show-that-there-is-no-
> subgroup-of-s-n-of-order-n-1-n
> <
> http://math.stackexchange.com/questions/1687845/show-that-there-is-no-subgroup-of-s-n-of-order-n-1-n
> >
>
> On Sun, Nov 27, 2016 at 7:47 PM, Nikos Apostolakis <nikos.ap at gmail.com>
> wrote:
>
> > On Sun, Nov 27, 2016 at 4:14 PM Frank Adams-Watters <
> franktaw at netscape.net
> > >
> > wrote:
> >
> > > If n divides m!, does the symmetric group S_m always have a subgroup of
> > > order n?
> > >
> > > If so, a comment should be added to A002034 that a(n) is the genus of
> the
> > > smallest symmetric group with a subgroup of order n. If not, where is
> the
> > > first exception? (8 in S_4?) Is the sequence so described in the OEIS?
> If
> > > not, it should be added.
> > >
> >
> > The dihedral group is a subgroup of S_4 (<(1,2,3,4), (2,4)> ) and has
> order
> > 8.
> >
> > As noted, GAP can be used to calculate all conjugacy classes of
> subgroups.
> > For n = 5, there are no subgroups of order 15, 30, 40.  For n =6 there
> are
> > no subgroups of orders 15, 30, 40, 45, 80, 90, 144, 180, 240.
> >
> >  Nikos
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>

```