A nontrivial abelian group is called *divisible* if for each and each nonzero , there exists an element such that .

- Prove that is divisible.
- Prove that no finite abelian group is divisible.

- Let be a rational number and . Then , so that is a th multiple. Hence is divisible.
- Let be a finite divisible abelian group. In particular, for every positive natural number , there is an element such that . Note that we may assume is minimal with respect to this property- that is, that has order . Thus contains an element of every positive order, and these must be distinct- a contradiction since is finite.

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## Comments

The lemma seems false. Consider Z2. If x=0, then x to any power equals 0. But if x=1, then x to any odd power is 1.

I’m not sure what I was thinking when I wrote that- to fix the lemma so that it is true, it becomes trivial.

There’s probably a better way to prove this anyway. I’ll have a think about it.

Thanks!

Alright- I think it’s fixed now.

Let me know if you see a problem.

Thanks again!

Could you expand on the “Since A is finite, we must have x_k=x_l for some k<l ". I just don't see how that follows from A being finite.

Thanks!

We’re thinking of as a function , where is some “primitive” th root of . There is a primitive th root for all , but contains only finitely many elements. By the pigeonhole principle, some element has to be a primitive root for two different natural numbers- say and . This necessarily gives a contradiction by our definition of primitive root.

This proof is poorly written though, so I’ll try to clean it up.

Scratch that- there is an even simpler way. Letting , there exists an element of every finite order, which is a contradiction.

How about this for (2)? Let be a nontrivial abelian group of order . Then by Lagrange’s thm, every element of has an order dividing , so for all . Since is nontrivial, there exists a non-identity element . Then has no roots in .

I like it. Thanks!