No finite abelian group is divisible

A nontrivial abelian group A is called divisible if for each a \in A and each nonzero k \in \mathbb{Z}^+, there exists an element x \in A such that x^k = a.

  1. Prove that (\mathbb{Q}, +) is divisible.
  2. Prove that no finite abelian group is divisible.

  1. Let a = \frac{m}{n} be a rational number and k \in \mathbb{Z}^+. Then k \cdot \frac{n}{km} = a, so that a is a kth multiple. Hence \mathbb{Q} is divisible.
  2. Let A be a finite divisible abelian group. In particular, for every positive natural number k, there is an element x_k \in A such that x_k^k = 1. Note that we may assume x_k is minimal with respect to this property- that is, that x_k has order k. Thus A contains an element of every positive order, and these must be distinct- a contradiction since A is finite.
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  • Math Student  On December 4, 2010 at 11:16 am

    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.

    • nbloomf  On December 4, 2010 at 11:44 am

      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.


      • nbloomf  On December 4, 2010 at 11:53 am

        Alright- I think it’s fixed now.

        Let me know if you see a problem.

        Thanks again!

  • fauxgt4  On April 3, 2011 at 8:05 pm

    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.


    • nbloomf  On April 4, 2011 at 6:57 am

      We’re thinking of x as a function \mathbb{N}^+ \rightarrow A, where x_k is some “primitive” kth root of a. There is a primitive kth root for all k, but A contains only finitely many elements. By the pigeonhole principle, some element has to be a primitive root for two different natural numbers- say k and \ell. 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.

      • nbloomf  On April 4, 2011 at 7:05 am

        Scratch that- there is an even simpler way. Letting a = 1, there exists an element of every finite order, which is a contradiction.

  • Gobi Ree  On November 16, 2011 at 8:03 pm

    How about this for (2)? Let A be a nontrivial abelian group of order n. Then by Lagrange’s thm, every element of A has an order dividing n, so x^n=1 for all x \in A. Since A is nontrivial, there exists a non-identity element c \neq 1. Then x^n=c has no roots in A.

    • nbloomf  On November 17, 2011 at 10:59 am

      I like it. Thanks!

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