## 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 $k$th 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.

• 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.

Thanks!

• 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.

Thanks!

• 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” $k$th root of $a$. There is a primitive $k$th 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!