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?