4
$\begingroup$

I would like to know if there is some results concerning about the following question:

When could a $p$-Sylow subgroup of a finite ring $R$ be a subring?

In other words, is it possible to induce the multiplication of the ring on the $p$-Sylow? If yes, are there conditions to guarantee this?

  • 0
    It is certainly (and trivially) true for finite fields, so yes, it is possible (and being a field is an example of a condition that would guarantee this).2012-08-21
  • 0
    @tomasz. So the interesting cases are for non fields. Any idea?2012-08-21
  • 0
    Actually, do you have any examples when it is *not* true?2012-08-21
  • 0
    No, I started to think about this just today. I am looking for. Let's continue.2012-08-21

1 Answers 1

3

For $r \in R$ fixed, the maps $s \mapsto rs$ and $s \mapsto sr$ are endomorphisms of the abelian group structure on $R$. Now use the fact that for an abelian group $G$ with Sylow subgroup $P$, any endomorphism $\phi$ of $G$ stabilizes $P$: $\phi(P) \subseteq P$ (this follows from standard Sylow theory). Therefore the Sylow subgroups are actually (two sided) ideals in $R$.

  • 1
    Nice answer! This shows, @Sigur, if you want your subrings to share identity with the superring, then the $p$-Sylows will usually not be subrings. If you don't care if the subring shares identity (or don't even care if it has identity) then more is possible. For example, you could take two fields, one of order 27 and one of order 8, and their product would have "Sylow p subrings". Now to think up an example where they don't have identity...2012-08-21
  • 0
    @rschwieb, let me see: considering $\mathbb{Z}_{p^r}\times \mathbb{Z}_{q^s}$ the ring with canonical operations, we have the identity $(\bar 1,\bar 1)$. But we have $\mathbb{Z}_{p^n}\times \{0\}$ as a $p$-Sylow subring with identity $(\bar 1,\bar 0)$, which is different of the identity of the ring. Do you agree?2012-08-21
  • 0
    @Sigur Yes, sure.2012-08-21