# Prove that no finite abelian group is divisible.

A nontrivial abelian group $G$ is called divisible if for each $a \in G$ and each nonzero integer $k$ there exists an element $x \in G$ such that $x^k=a$. Prove that no finite abelian group is divisible.

I came across a prove that goes like this:

Let $G$ be a finite divisible abelian group. Then for each positive integer $k$ there is $x_k \in G$ such that $x_k^k = 1$. Note we may assume $x_k$ is minimal with respect to this property; i.e. the order of $x_k$ is $k$. Therefore, $G$ contains an element of every positive order and these must be distinct. Contradiction.

How can we assume $x_k$ has order $k$? If $G$ is finite of order $n$, then we must have $ord(x)|n$ for each $x \in G$. If $n<k$ then there is no $x\in G$ with order $k$. Right? Or did I go wrong somewhere?

Three steps (alternative to Pedro Tamaroff's nice answer) for a slightly more general result.

1. A quotient of a divisible group is divisible (the proof is very simple).

2. A finitely generated abelian group is a direct sum of cyclic groups (main theorem on finitely generated abelian groups).

3. No cyclic group is divisible (just a verification).

For any finite abelian group there is at least one integer $n$ such that $nG=1$. But for a divisible group, $kG=G$ for any $k$. Can you figure out how to find such $n$?

Here is my proof by contradiction.

Since $G$ is non trivial, exists an element $a\in G\setminus\{1\}$. For this element and $k=|G|$, exists and $x\in G$ such that $x^{|G|}=a$, but since $G$ is finite, $x^{|G|}=1$. This implies that a=1.

You are very close. This is a proof by contradiction. If an abelian group is divisible, then there must be, by definition of divisibility, some $x_k$ for each $k$.

This means, since $\operatorname{ord}(x_k)\mid |G|$, that $|G|$ must be divisible by more numbers than any integer could possibly be. Hence it cannot be finite.

You don't really need to assume that $\operatorname{ord}(x_k) = k$. For instance, there can only be one prime $k$ for which $\operatorname{ord}(x_k) \neq k$. You still get a contradiction.

• I get that divisibility implies existence of $x_k$, but it's possible that ord$(x_k)<k$, in which case $x_k^k$ still equals $e$ but now there is nothing that shows the existence of an element of order $k$ for each $k$ and hence no contradiction. – Sam Oct 3 '14 at 8:26
• As I said, if you focus on prime $k$, there can be only one $k$ where that is not the case. Remember that $1^i\neq 1$ unless $i$ is a multiple of some specific number, and that is only possible for at most one prime. – Arthur Oct 3 '14 at 8:31