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I need help with an exercise from Kaplansky's Infinite Abelian Groups (Section 9, Exercise 27). He states the problem as follows:

Let $G$ be a reduced primary group which is not of bounded order. Prove that $G$ has cyclic direct summands of arbitrarily high order.

This also is an exercise in Fuchs' Infinite Abelian Groups (Section 27, Exercise 1).

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See also the similar question:… – Jack Schmidt Jun 29 '11 at 0:15
I've merged your accounts (by the way, flagging a post for moderator attention is indeed the right way of going about it). – Zev Chonoles Jun 29 '11 at 7:31
@Zev Chonoles: Thank you. – Anononym Jun 29 '11 at 8:31
up vote 1 down vote accepted

Let $A$ be such a group, and let $B_i = \{a\in A\ |\ |a|=p\text{ and } h(a)=i\}$. Note that at least one of the $B_i$ is non-empty by Lemma 8 in section 9. Also, if there existed an $N$ such that for all $m>N$, $B_m$ was empty, then for all $a\in A$, we would have $p^{N+1}a=0$, so $A$ would have bounded order. Thus infinitely many of the $B_i$ are non-empty; now simply mimic the proof of Theorem 9, using Lemma 7 and Theorem 7.

EDIT - Sorry, I left out a couple details, which I don't mind filling in. First, $h(a)$ means the height of $a$. Second, if such an $N$ existed as above, then every element of order $p$ in $p^{N+1}A$ would have infinite height in $A$. It is easy to see this implies it has infinite height in $p^{N+1}A$. Thus $p^{N+1}A$ is divisible; since $A$ is reduced, it is $0$.

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Awesome, thanks for the quick answer! – Anononym Jun 29 '11 at 0:32

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