# [seqfan] Re: Symmetric groups

jean-paul allouche jean-paul.allouche at imj-prg.fr
Mon Nov 28 07:28:50 CET 2016

The same argument can be used for larger n's.
There is only one group of order pq where p
and q are two primes such that p < q and
q \notequiv 1 \bmod p (this is true more
generally for the groups of order n such that
n and \phi(n) are coprime, where \phi is the
Euler totient).

best
jp allouche

PS By the way there is a similar
such that all groups of order n must
be abelian (see L. E. Dickson,
Trans. Amer. Math. Soc. 6 (1905), 198–204).
I did not check (yet) whether the corresponding
sequence of integers is in the OEIS: I bet it is!

Le 28/11/16 à 01:51, israel at math.ubc.ca a écrit :
> There is only one group of order 15, and it is cyclic.
> In order for a member of S_n to have order 15, n must be at least
> 8 (so you can have a disjoint 3-cycle and 5-cycle).
>
> Cheers,
> Robert
>
> On Nov 27 2016, W. Edwin Clark wrote:
>
>> No, S_5 does not have a subgroup of order 15 says GAP. Here's  GAP code
>> which gives the orders
>> of the subgroups of S_5:
>>
>> G:=SymmetricGroup(5);;
>> C:=ConjugacyClassesSubgroups(G);;
>>
>>                [ 1, 2, 3, 4, 5, 6, 8, 10, 12, 20, 24, 60, 120 ]
>>
>> 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.
>>>
>>>
>>>
>>> --
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
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>
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