[seqfan] Re: Symmetric groups

[W. Edwin Clark]

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
>
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