User Name Remember Me? Password

 Abstract Algebra Abstract Algebra Math Forum

 February 9th, 2017, 08:42 AM #1 Newbie   Joined: Feb 2017 From: Argentina Posts: 2 Thanks: 0 Primary groups and pure subgroups. Hi: Let G be a p-primary group that is not divisible. Assume there is $x \in G[p]$ that is divisible by $p^k$ but not by $p^{k+1}$ and let $x= p^k y$. I must prove that $\cap nG \subset n$. $p^{k+1}y= px= 0$. In case $(n,p)=1$, there is $b \in$ such that $y=nb$ and then $sy= snb= n(sb) \in n$. But when $(n,p) \ne 1$ I cannot find a way. Could you tell me how I should proceed? February 14th, 2017, 06:33 PM #2 Member   Joined: Jan 2016 From: Athens, OH Posts: 93 Thanks: 48 I'm not familiar with the notation G[p], but I assume you mean the following: Let p be a prime and G an abelian p group. Let k be a non-negative integer and suppose $x\in p^kG\setminus p^{k+1}G$ is of order p; say $p^ky=x$ but $p^{k+1}z=x$ has no solution z. Then is a pure subgroup; i.e. for any integer n, $\cap \,nG\subseteq n$. Proof. For any p group H and any integer n prime to p, nH=H. So it is sufficient to prove the statement for n of the form $p^m$ for any positive integer m. Now for any finite cyclic p group of order $p^{k+1}$, the subgroups of form a chain with respect to set inclusion. That is, every subgroup is of the form  and $\supset \supset \supset\cdots \supset\supset <0>$. 1. Suppose $m\leq k$. Then $p^{m-1}y\notin p^mG$: Otherwise there is $z\in G$ with $p^mz=p^{m-1}y$. Let $r=k-m+1$. Then $p^rz=p^{k+1}z=p^ky$, a contradiction. Since $H=\cap p^mG$ is a subgroup of , $p^{m-1}y\notin H$ and said subgroups form a chain, $H\subseteq$. 2. First $\cap p^{k+1}G=<0>$: if not, then again since the sugroups of form a chain, $\subset \cap p^{k+1}G\subset p^{k+1}G and x=p^ky\in p^{k+1}G$, a contradiction. Then for any $m\geq k+1$, $\cap \,p^mG\subseteq \cap \,p^{k+1}G=<0>=$. Tags groups, primary, pure, subgroups Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post beesee Abstract Algebra 6 January 31st, 2013 05:24 PM Artus Abstract Algebra 2 January 21st, 2013 07:27 AM mous99 Abstract Algebra 2 October 20th, 2012 01:38 PM gaussrelatz Abstract Algebra 9 July 9th, 2012 09:09 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top

Copyright © 2019 My Math Forum. All rights reserved.      