Is an ideal which is cyclic as a subgroup always principal?

Question

For $\mathbb Z$, all ideas are principal because they're cyclic as subgroups, hence each $x\in I= \left\langle n\right\rangle$ can be writte $x=mn$ for $m\in \mathbb N$. Luckily, $\mathbb N\subset \mathbb Z$ and so this is actually the multiplication of the ring $\mathbb Z$. Is there anything similar which works for more general (commutative) rings?

Answer 1

Identifying the image of $\Bbb Z$ in $R$ as $Z$, if $I$ is some ideal such that $aZ=I$, then trivially $I=aZ\subseteq aR\subseteq I$.

If you already get all elements of the ideal with addition, then multiplication won't give anything new.