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?
Best Answer
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.