# Dummit and Foote $2.4.20$, divisible group

Q. Prove that if $$A$$ and $$B$$ are nontrivial abelian groups, then $$A \times B$$ is divisible if and only if both $$A$$ and $$B$$ are divisible groups.

My approach

Let $$A, B$$ be two divisible groups, and consider any element $$(a, b) \in A \times B$$. Since $$A, B$$ are divisible, for each $$k \in \mathbb{Z} \backslash\{0\}$$, there exists $$x \in A, y \in B$$ such that $$x^{k}=a$$ and $$y^{k}=b$$. Therefore, for each $$k \in \mathbb{Z} \backslash\{0\}$$, there exists $$(x, y) \in A \times B$$ such that $$(x, y)^{k}=\left(x^{k}, y^{k}\right)=(a, b)$$. So, $$A \times B$$ is divisible.

Is it okay? Also give me hints or solution. Thank you...

Definition of divisible group(from wiki) :-in the field of group theory, a divisible group is an abelian group in which every element can, in some sense, be divided by positive integers, or more accurately, every element is an nth multiple for each positive integer n.

• This part of the problem seems to me to be solved correctly. Now it remains to prove that if $A\times B$ is divisible, then $A$ and $B$ are divisible. Right? May 5, 2022 at 2:51