[seqfan] Re: Symmetric groups
nikos.ap at gmail.com
Mon Nov 28 15:55:33 CET 2016
Except that for n = 11, 33 divides n!, and the first time S_n could have a
subgroup of order 33, is for n = 14. So the sequence up to 13 should be
15, 15, 15, 28, 35, 35, 31, 31, 31.
On Sun, Nov 27, 2016 at 8:43 PM Nikos Apostolakis <nikos.ap at gmail.com>
> 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
> On Sun, Nov 27, 2016 at 8:07 PM W. Edwin Clark <wclark at mail.usf.edu>
> 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:
> On Sun, Nov 27, 2016 at 7:47 PM, Nikos Apostolakis <nikos.ap at gmail.com>
> > 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
> > > smallest symmetric group with a subgroup of order n. If not, where is
> > > first exception? (8 in S_4?) Is the sequence so described in the OEIS?
> > > not, it should be added.
> > >
> > The dihedral group is a subgroup of S_4 (<(1,2,3,4), (2,4)> ) and has
> > 8.
> > As noted, GAP can be used to calculate all conjugacy classes of
> > For n = 5, there are no subgroups of order 15, 30, 40. For n =6 there
> > no subgroups of orders 15, 30, 40, 45, 80, 90, 144, 180, 240.
> > Nikos
> > --
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