= Solution
The set is the <ideal quotient> $K=(I:J)$. It contains zero. If $r,s\in K$, then $(r-s)J\subseteq I$, since $I$ is an <subgroup>. If $a\in R$ and $r\in K$, then $(ar)J=a(rJ)\subseteq I$ because $I$ is an <ideal>. Thus $K$ is an <ideal>.
Write $J=(a)$ and $K=(b)$. Because $I\subseteq J$, each $x\in I$ has the form $ar$. Then $rJ=r(aR)=xR\subseteq I$, so $r\in K$, and $x\in aK$. Conversely $aK\subseteq I$ by the definition of $K$. Therefore
$$
\boxed{I=aK=(ab),}
$$
so $I$ is a <principal ideal>. This argument uses only the commutative ring convention, not cancellation or an <integral domain> hypothesis.
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