Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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).

share|improve this question
    
See also the similar question: math.stackexchange.com/questions/47115/… –  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
add comment

1 Answer

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 $0$.

share|improve this answer
    
Awesome, thanks for the quick answer! –  Anononym Jun 29 '11 at 0:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.