This question if from "Contemporary Abstract Algebra" -Galian (9th ed.) Chapter 18 #2. It is an assigned homework problem. I'm hitting a wall trying to crack this problem. I'm sure it is relatively simple but I'm missing some key component. This is where my thinking and research has taken me. Please help get me over the hurdle.
Let $D$ be an integral domain and $a$ and $b$ be non-zero elements of $D$ such that $a$ and $b$ are associates. Then $a=ub$ and $b=va$ for some units $u,v\in D$. Clearly, $u$ and $v$ are inverses, but I can't make the mental connection that $ub\in \langle b\rangle$ or $va\in \langle a\rangle$.
Associates defined as: Elements $a$ and $b$ of integral domain $D$ are associates if $a=ub$ where $u$ is a unit of $D$.
Train of thought:
$a=vua \implies a\sim ua$. (this seems trivial since $a=ua$ given $u$ must have an inverse.)
There is no indication that $D$ must be finite, so we're not guaranteed that any elements in $\langle a \rangle$ contain any units or that the subgroup of $D$ generated by $a$ (or $b$) is itself an integral domain. If $\langle a\rangle$ is an integral domain then it becomes simple to show that $b=ua\in \langle a\rangle$.
Please help me along.
If $a$ and $b$ are associates:
Let $D$ be an integral domain with non-zero elements $a,b\in D$ such that $a$ and $b$ are associates.
For some units $u,v\in D$, $a=ub$ and $b=va$ (by definition of associates).
$a=ub\implies a\in\langle b\rangle$ and $\langle a\rangle\subseteq\langle b\rangle$.
$b=va\implies b\in\langle a\rangle$ and $\langle b\rangle\subseteq\langle a\rangle$.
Taken together: $\langle b\rangle\subseteq\langle a\rangle$ and $\langle a\rangle\subseteq\langle b\rangle\implies \langle a\rangle=\langle b\rangle$
If $a=pb$ and $b=qa$ for some $a,b,p,q\in D$ (where $D$ is an integral domain):
$a=pb\implies a=pqa\implies pq=1$ which implies that $p$ and $q$ are inverses and by definition are units in $D$. Thus $a\sim b$ and as shown above $\langle a\rangle=\langle b\rangle$.