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 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: 51 Thanks: 27 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 $m\geq k+1$, $\cap \,p^mG\subseteq \cap \,p^{k+1}G=<0>=$.

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