Is a p-Sylow subgroup of the underlying additive group of a finite commutative ring an ideal?

Question

Let $R$ be a finite commutative ring. Let $H$ be a $p$-Sylow subgroup of the additive group $(R,+)$. Is $H$ an ideal in $R$?

For ex: R= Z/6Z is a finite commutative ring. H={0,3} is a 2-sylow subgroup which in itself is an ideal of the ring. Is this true for any finite commutative ring R?

Answer 1

As $(R,+)$ is an abelian group , so any subgroup is normal , so for $p$ prime , it has a unique Sylow $p$ subgroup , so $H$ is the unique $p$-Sylow subgroup . Now to show $H$ is an ideal , let $a\in H , r\in R$ , then o$(a)=p^k$ for some $k\ge 0$ . Then $p^k(ra)=r(p^ka)=0$ , so o$(ra)|p^k$ , so o$(ra)=p^m$ for some $m \ge 0$ , so by Second Sylow theorem , $ra $ is contained in a $p$-Sylow subgroup , hence the unique $p$-Sylow subgroup $H$ , thus $ra \in H$

Answer 2

Let $H$ be a $p$-Sylow subgroup of $(R,+)$ ; let$|R|=p^mn$ , where $p$ does not divide $n$ ; then

$|H|=p^m$. As $(R,+)$ is abelian , so $H$ is normal in $(R,+)$ and $|R/H|=n$ . Now let $r\in R , a\in H$ ,

then $ra+H \in (R/H,+)$ , so $nra+H=n(ra+H)=H$ , so $nra \in H$ . Now as

g.c.d.$(p^m,n)=1 $ , so $\exists s,t \in \mathbb Z$ such that $sp^m+tn=1$ , so

$ra=t(nra)+s(p^mra)=t(nra)+sr(p^ma)=t(nra)\in H$ , where $p^ma=0$ because $a\in H $ ,

and $ |H|=p^m$ . Thus $H$ is an ideal