[seqfan] Re: Is A037289 multiplicative

[franktaw at netscape.net]

I had no sooner submitted this than I realized that there must indeed 
be only one Sylow p-subgroup for a commutative group.  By the second 
Sylow theorem, all Sylow p-subgroups are conjugate, and conjugation is 
the identity operation in a commutative group.

I'll update the sequence.

Franklin T. Adams-Watters

-----Original Message-----
From: franktaw <franktaw at netscape.net>
To: seqfan <seqfan at list.seqfan.eu>
Sent: Tue, Jul 10, 2012 7:58 am
Subject: [seqfan] Re: Is A037289 multiplicative


That isn't true in general; it is equivalent to the assertion that
there is only one Sylow p-subgroup.

Maybe it is true for commutative rings, however. Do you have a
reference?

Franklin T. Adams-Watters

-----Original Message-----
From: israel <israel at math.ubc.ca>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Tue, Jul 10, 2012 2:35 am
Subject: [seqfan] Re: Is A037289 multiplicative


Doesn't the Sylow p-subgroup consist of all elements x such that p^m x
= 0,
where p^m is the largest power of p in n? Those elements constitute an
ideal of R, since p^m (xy) = (p^m x) y = 0, and therefore in particular
a
subring.

Robert Israel
University of British Columbia


On Jul 9 2012, franktaw at netscape.net wrote:

>Looking at this sequence, there is a conjecture that the sequence is
>multiplicative. I can almost prove that it is.
>
>I need a lemma that, if R is a commutative ring of order n, and prime
p
>divides n, then there is a Sylow p-subgroup of the additive group of R
>that is in fact a subring if R. It is then easy to see that R must be
>the direct product of these subrings, and the result follows.
>
>(This is an equivalent assertion, by the way; if there is a
commutative
>ring with no such "Sylow p-subring", the sequence is not
multiplicative
>for that n.)
>
>Does anybody know of such a result? Or can you find a proof?
>
>Franklin T. Adams-Watters