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stf92 February 9th, 2017 09:42 AM

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 $<y> \cap nG \subset n<y>$. $p^{k+1}y= px= 0$. In case $(n,p)=1$, there is $b \in <y>$ such that $y=nb$ and then $sy= snb= n(sb) \in n<y>$. But when $(n,p) \ne 1$ I cannot find a way. Could you tell me how I should proceed?

johng40 February 14th, 2017 07:33 PM

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 <y> is a pure subgroup; i.e. for any integer n, $<y>\cap \,nG\subseteq n<y>$.
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 <y> of order $p^{k+1}$, the subgroups of <y> form a chain with respect to set inclusion. That is, every subgroup is of the form $<p^sy>$ and $<y>\supset <py>\supset <p^2y>\supset\cdots \supset<p^ky>\supset <0>$.
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 $H=<y>\cap p^mG$ is a subgroup of <y>, $p^{m-1}y\notin H$ and said subgroups form a chain, $H\subseteq<p^my>$.
2. First $<y>\cap p^{k+1}G=<0>$: if not, then again since the sugroups of <y> form a chain, $<p^ky>\subset <y>\cap p^{k+1}G\subset p^{k+1}G and x=p^ky\in p^{k+1}G$, a contradiction. Then for any $m\geq k+1$, $<y>\cap \,p^mG\subseteq <y>\cap \,p^{k+1}G=<0>=<p^my>$.


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