## A finite direct product is divisible if and only if the direct factors are divisible

Let $A$ and $B$ be nontrivial abelian groups. Prove that $A \times B$ is divisible if and only if $A$ and $B$ are divisible.

$(\Rightarrow)$ Suppose $A \times B$ is divisible, and let $a \in A$. Since $A \times B$ is divisible, for every $k \in \mathbb{Z}^+$ there exists an element $(x,y) \in A \times B$ such that $(x,y)^k = (x^k,y^k) = (a,1)$. Thus for every $k$ there exists $x \in A$ such that $x^k = a$; hence $A$ is divisible. By a similar argument, $B$ is divisible.

$(\Leftarrow)$ Suppose $A$ and $B$ are divisible, and let $(a,b) \in A \times B$. Let $k \in \mathbb{Z}^+$. Since $A$ and $B$ are divisible, there exist $x \in A$ and $y \in B$ such that $x^k = a$ and $y^k = b$; then $(x,y)^k = (x^k,y^k) = (a,b)$. Thus $A \times B$ is divisible.